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Samacheer Kalvi 12th Physics Solutions Chapter 10 Communication Systems

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Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 10 Communication Systems

Samacheer Kalvi 12th Physics Communication Systems Textual Evaluation Solved

Samacheer Kalvi 12th Physics Communication Systems Multiple Choice Questions

Question 1.
The output transducer of the communication system converts the radio signal into …………. .
(a) Sound
(b) Mechanical energy
(c) Kinetic energy
(d) None of the above
Answer:
(a) Sound.

Question 2.
The signal is affected by noise in a communication system …………. .
(a) At the transmitter
(b) At the modulator
(c) In the channel
(d) At the receiver
Answer:
(c) In the channel.

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Question 3.
The variation of frequency of carrier wave with respect to the amplitude of the modulating signal is called …………. .
(a) Amplitude modulation
(b) Frequency modulation
(c) Phase modulation
(d) Pulse width modulation
Answer:
(b) Frequency modulation.

Question 4.
The internationally accepted frequency deviation for the purpose of FM broadcasts …………. .
(a) 75 kHz
(h) 68 kHz
(c) 80 kHz
(d) 70 kHz
Answer:
(a) 75 kHz.

Question 5.
The frequency range of 3 MHz to 30 MHz is used for …………. .
(a) Ground wave propagation
(b) Space wave propagation
(c) Sky wave propagation
(d) Satellite communication
Answer:
(c) Sky wave propagation.

Samacheer Kalvi 12th Physics Communication Systems Short Answer Questions

Question 1.
Give the factors that are responsible for transmission impairments.
Answer:

  •  Attenuation
  • Distortion (Harmonic)
  • Noise

Question 2.
Distinguish between wireline and wireless communication? Specify the range of electromagnetic waves in which it is used.
Answer:

Wireline communication Wireless communication
It is a point-to-point
communication.
It is a broadcast mode communication.
It uses mediums like wires, cable and optical fibres. It uses free space as a communication medium.
These systems cannot be used for long distance transmission as they are connected. These systems can be used for long distance transmission.
Ex: telephone, intercom and
cable TV.
Ex: mobile, radio or TV broadcasting and satellite communication.

Question 3.
Explain centre frequency or resting frequency in frequency modulation.
Answer:
When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency and is called as centre frequency or resting frequency.

Question 4.
What does RADAR stand for?
Answer:
Radar basically stands for Radio Detection and Ranging System.

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Question 5.
What do you mean by Internet of Things?
Answer:
Internet of Things (IoT), it is made possible to control various devices from a single device. Example: home automation using a mobile phone.

Samacheer Kalvi 12th Physics Communication Systems Long Answer Questions

Question 1.
What is modulation? Explain the types of modulation with necessary diagrams.
Answer:
Modulation:
For long distance transmission, the low frequency baseband signal (input signal) is superimposed onto a high frequency radio signal by a process called modulation. There are 3 types of modulation based on which parameter is modified. They are (i) Amplitude modulation, (ii) Frequency modulation, and (iii) Phase modulation.

(i) Amplitude Modulation (AM):
If the amplitude of the carrier signal is modified according to the instantaneous amplitude of the baseband signal, then it is called amplitude modulation. Here the frequency and the phase of the carrier signal remain constant. Amplitude modulation is used in radio and TV broadcasting.

The signal shown in figure (a) is the message signal or baseband signal that carries information, figure (b) shows the high frequency carrier signal and figure (c) gives the amplitude modulated signal. We can see clearly that the carrier wave is modified in proportion to the amplitude of the baseband signal
Samacheer Kalvi 12th Physics Solutions Chapter 10 Communication Systems-1

(ii) Frequency Modulation (FM):
The frequency of the carrier signal is modified according to the instantaneous amplitude of the baseband signal in frequency modulation. Here the amplitude and the phase of the carrier signal remain constant. Increase in the amplitude of the baseband signal increases the frequency of the carrier signal and vice versa. This leads to compressions and rarefactions in the frequency spectrum of the modulated wave. Louder signal leads to compressions and relatively weaker signals to rarefactions.
Samacheer Kalvi 12th Physics Solutions Chapter 10 Communication Systems-2
When the amplitude of the baseband signal is zero in Figure (a), the frequency of the modulated signal is the same as the carrier signal. The frequency of the modulated wave increases when the amplitude of the baseband signal increases in the positive direction (A,C). The increase in amplitude in the negative half cycle (B,D) reduces the frequency of the modulated wave (Figure (c)).

(iii) Phase Modulation (PM)
The instantaneous amplitude of the baseband signal modifies the phase of the carrier signal keeping the amplitude and frequency constant is called phase modulation. This modulation is used to generate frequency modulated signals. It is similar to frequency modulation except that the phase of the carrier is varied instead of varying frequency.
Samacheer Kalvi 12th Physics Solutions Chapter 10 Communication Systems-3
The carrier phase changes according to increase or decrease in the amplitude of the baseband signal. When the modulating signal goes positive, the amount of phase lead increases with the amplitude of the modulating signal. Due to this, the carrier signal is compressed or its frequency is increased.

On the other hand, the negative half cycle of the baseband signal produces a phase lag in the carrier signal. This appears to have stretched the frequency of the carrier wave. Hence similar to frequency modulated wave, phase modulated wave also comprises of compressions and rarefactions. When the signal voltage is zero (A, C and E) the carrier frequency is unchanged.

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Question 2.
Elaborate on the basic elements of communication system with the necessary block diagram.
Elements of an electronic communication system:
1. Information (Baseband or input signal):
Information can be in the form of a sound signal like speech, music, pictures, or computer data Information can be in the form of a sound signal like speech, music, pictures, or computer data which is given as input to the input transducer.
Samacheer Kalvi 12th Physics Solutions Chapter 10 Communication Systems-4

2. Input transducer:
A transducer is a device that converts variations in a physical quantity (pressure, temperature, sound) into an equivalent electrical signal or vice versa. In communication system, the transducer converts the information which is in the form of sound, music, pictures or computer data into corresponding electrical signals. The electrical equivalent of the original information is called the baseband signal. The best example for the transducer is the microphone that converts sound energy into electrical energy.

3. Transmitter:
It feeds the electrical signal from the transducer to the communication channel. It consists of circuits such as amplifier, oscillator, modulator, and power amplifier. The transmitter is located at the broadcasting station.

4. Amplifier:
The transducer output is very weak and is amplified by the amplifier.

5. Oscillator:
It generates high-frequency carrier wave (a sinusoidal wave) for long distance transmission into space. As the energy of a wave is proportional to its frequency, the carrier wave has very high energy.

6. Modulator:
It superimposes the baseband signal onto the carrier signal and generates the modulated signal.

7. Power amplifier:
It increases the power level of the electrical signal in order to cover a large distance.

8. Transmitting antenna:
It radiates the radio signal into space in all directions. It travels in the form of electromagnetic waves with the velocity of light (3 x 108 ms-1).

9. Communication channel:
Communication channel is used to carry the electrical signal from transmitter to receiver with less noise or distortion. The communication medium is basically of two types: wireline communication and wireless communication.

10. Noise:
It is the undesirable electrical signal that interfaces with the transmitted signal. Noise attenuates or reduces the quality of the transmitted signal. It may be man-made (automobiles, welding machines, electric motors etc .) or natural (lightening, radiation from sun and stars and environmental effects). Noise cannot be completely eliminated. However, it can be reduced using various techniques.

11. Receiver:
The signals that are transmitted through the communication medium are received with the help of a receiving antenna and are fed into the receiver. The receiver consists of electronic circuits like demodulator, amplifier, detector etc. The demodulator extracts the baseband signal from the carrier signal. Then the baseband signal is detected and amplified using amplifiers. Finally, it is fed to the output transducer.

12. Repeaters:
Repeaters are used to increase the range or distance through which the signals are sent. It is a combination of transmitter and receiver. The signals are received, amplified, and retransmitted with a carrier signal of different frequency to the destination. The best example is the communication satellite in space.

13. Output transducer:
It converts the electrical signal back to its original form such as sound, music, pictures or data. Examples of output transducers are loudspeakers, picture tubes, computer monitor, etc.

14. Attenuation:
The loss of strength of a signal while propagating through a medium is known as attenuation. Range It is the maximum distance between the source and the destination up to which the signal is received with sufficient strength.

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Question 3.
Explain the three modes of propagation of electromagnetic waves through space. Propagation of electromagnetic waves:
Answer:
The electromagnetic wave transmitted by the transmitter travels in three different modes to reach the receiver according to its frequency range:

  1. Ground wave propagation (or) surface wave propagation (nearly 2 kHz to 2 MHz)
  2. Sky wave propagation (or) ionospheric propagation (nearly 3 MHz to 30 MHz)
  3. Space wave propagation (nearly 30 MHz to 400 GHz)

1. Ground wave propagation:
If the electromagnetic waves transmitted by the transmitter glide over the surface of the earth to reach the receiver, then the propagation is called ground wave propagation. The corresponding waves are called ground waves or surface waves.

Both transmitting and receiving antennas must be close to the earth. The size of the antenna plays a major role in deciding the efficiency of the radiation of signals. During transmission, the electrical signals are attenuated over a distance. Some reasons for attenuation are as follows:
Samacheer Kalvi 12th Physics Solutions Chapter 10 Communication Systems-5

1. Increasing distance:
The attenuation of the signal depends on (i) power of the transmitter (ii) frequency of the transmitter, and (iii) condition of the earth surface.

2. Absorption of energy by the Earth:
When the transmitted signal in the form of EM wave is in contact with the Earth, it induces charges in the Earth and constitutes a current. Due to this, the earth behaves like a leaky capacitor which leads to the attenuation of the wave.

3. Tilting of the wave:
As the wave progresses, the wavefront starts gradually tilting according to the Skywave curvature of the Earth. This increase in the tilt decreases the electric field strength of the wave. Finally, at some distance, the surface wave dies out due to energy loss.

The frequency of the ground waves is mostly less than 2 MHz as high frequency waves undergo more absorption of energy at the earth’s atmosphere. The medium wave signals received during the day time use surface wave propagation. It is mainly used in local broadcasting, radio navigation, for ship-to-ship, ship-to-shore communication and mobile communication.

2. Sky Wave Propagation:
The mode of propagation in which the electromagnetic waves radiated from an antenna, directed upwards at large angles gets reflected by the ionosphere back to earth is called sky wave propagation or ionospheric propagation. The corresponding waves are called sky waves.
Samacheer Kalvi 12th Physics Solutions Chapter 10 Communication Systems-6
The frequency range of EM waves in this mode of propagation is 3 to 30 MHz. EM waves of frequency more than 30 MHz can easily penetrate through the ionosphere and does not undergo reflection. It is used for short wave broadcast services. Medium and high frequencies are for long-distance radio communication.

Extremely long distance communication is also possible as the radio waves can undergo multiple reflections between the earth and the ionosphere. A single reflection helps the radio waves to travel a distance of approximately 4000 km.

Ionosphere acts as a reflecting surface. It is at a distance of approximately 50 km and spreads up to 400 km above the Earth surface. Due to the absorption of ultraviolet rays, cosmic ray, and other high energy radiations like a, p rays from sun, the air molecules in the ionosphere get ionized.

This produces charged ions and these ions provide a reflecting medium for the reflection of radio waves or communication waves back to earth within the permitted frequency range. The phenomenon of bending the radio waves back to earth is nothing but the total internal reflection.

3. Space wave propagation:
The process of sending and receiving information signal through space is called space wave communication. The electromagnetic waves of very high frequencies above 30 MHz are called as space waves. These waves travel in a straight line from the transmitter to the receiver. Hence, it is used for a line of sight communication (LOS).
Samacheer Kalvi 12th Physics Solutions Chapter 10 Communication Systems-7
For high frequencies, the transmission towers must be high enough so that the transmitted and received signals (direct waves) will not encounter the curvature of the earth and hence travel with less attenuation and loss of signal strength. Certain waves reach the receiver after getting reflected from the ground.

Question 4.
What do you know about GPS? Write a few applications of GPS.
Answer:
GPS stands for Global Positioning System. It is a global navigation satellite system that offers geolocation and time information to a GPS receiver anywhere on or near the Earth. GPS system works with the assistance of a satellite network. Each of these satellites broadcasts a precise signal like an ordinary radio signal.

These signals that convey the location data are received by a low-cost aerial which is then translated by the GPS software. The software is able to recognize the satellite, its location, and the time taken by the signals to travel from each satellite. The software then processes the data it accepts from each satellite to estimate the location of the receiver.

Applications:
Global positioning system is highly useful many fields such as fleet vehicle management (for tracking cars, trucks and buses), wildlife management (for counting of wild animals) and engineering (for making tunnels, bridges etc).

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Question 5.
Give the applications of ICT in mining and agriculture sectors.
Answer:
(i) Agriculture:
The implementation of information and communication technology (ICT) in agriculture sector enhances the productivity, improves the living standards of farmers and overcomes the challenges and risk factors.
(a) ICT is widely used in increasing food productivity and farm management.
(b) It helps to optimize the use of water, seeds and fertilizers etc.
(c) Sophisticated technologies that include robots, temperature and moisture sensors, aerial images, and GPS technology can be used.
(d) Geographic information systems are extensively used in farming to decide the suitable place for the species to be planted.

(ii) Mining:
(a) ICT in mining improves operational efficiency, remote monitoring and disaster locating system.
(b) Information and communication technology provides audio-visual warning to the trapped underground miners.
(c) It helps to connect remote sites.

Question 6.
Modulation helps to reduce the antenna size in wireless communication – Explain. Antenna size:
Answer:
Antenna is used at both transmitter and receiver end. Antenna height is an important parameter to be discussed. The height of
the antenna must be a multiple of \(\frac { λ }{ 4 }\),
h = \(\frac { λ }{ 4 }\) ….. (1)
where λ is wavelength ( λ = \(\frac { c}{ v }\)), c is the velocity of light and v is the frequency of the signal to be transmitted.

An example:
Let us consider two baseband signals. One signal is modulated and the other is not modulated. The frequency of the original baseband signal is taken as v = 10 kHz while the modulated signal is v = 1 MHz. The height of the antenna required to transmit the original baseband signal of frequency v = 10 kHz is
h1 = \(\frac { λ }{ 4 }\) = \(\frac { c }{ 4v }\) = \(\frac { 3\times { 10 }^{ 8 } }{ 4\times 10\times { 10 }^{ 3 } } \) = 7.5 km …. (2)
The height of the antenna required to transmit the modulated signal of frequency v = 10 kHz is
h1 = \(\frac { λ }{ 4 }\) = \(\frac { c }{ 4v }\) = \(\frac { 3\times { 10 }^{ 8 } }{ 4\times 10\times { 10 }^{ 6 } } \) = 75 m …. (3)
Comparing equations (2) and (3), we can infer that it is practically feasible to construct an antenna of height 75 m while the one with 7.5 km is not possible. It clearly manifests that modulated signals reduce the antenna height and are required for long distance transmission.

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Question 7.
Fiber optic communication is gaining popularity among the various transmission media -justify.
Answer:
The method of transmitting information from one place to another in terms of light pulses through an optical fiber is called fiber optic communication. It is in the process of replacing wire transmission in communication systems. Light has very high frequency (400 THz – 790 THz) than microwave radio systems.  The fibers are made up of silica glass or silicon dioxide which is highly abundant on Earth.

Now it has been replaced with materials such as chalcogenide glasses, fluoroaluminate crystalline materials because they provide larger infrared wavelength and better transmission capability. As fibers are not electrically conductive, it is preferred in places where multiple channels are to be laid and isolation is required from electrical and electromagnetic interference.

Applications:
Optical fiber system has a number of applications namely, international communication, inter-city communication, data links, plant and traffic control and defense applications.

Merits:

  1. Fiber cables are very thin and weight lesser than copper cables.
  2. This system has much larger bandwidth. This means that its information canying capacity is larger.
  3. Fiber optic system is immune to electrical interferences.
  4. Fiber optic cables are cheaper than copper cables.

Demerits:

  1. Fiber optic cables are more fragile when compared to copper wires.
  2. It is an expensive technology.

Samacheer Kalvi 12th Physics Communication Systems Additional Questions

Samacheer Kalvi 12th Physics Communication Systems Multiple Choice Questions

Question 1.
Which of the following frequencies will be suitable for beyond the horizon communication using sky waves?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Answer:
(b) 10 MHz
Hint:
Frequency of 10 KHz will require very large radiating antenna while frequencies 1GHz and 1000 GHz will penetrate the ionosphere and cannot be reflected by it.

Question 2.
Frequency in the UHF range normally propagate by means of:
(a) Ground waves
(b) sky waves
(c) surface waves
(d) space waves
Answer:
(d) space waves.
Hint:
Frequency in the UHF (0.3 to 3 GHz) range normally propagate by means of space waves. Their sky wave reflection from ionosphere is not possible.

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Question 3.
Antenna is
(a) inductive
(b) capacitive
(c) resistive above its resonant frequency
(d) resistive at resonant frequency
Answer:
(d) resistive at resonant frequency.
Hint:
An antenna is tuned circuit consisting of an inductance and a capacitance. At resonant frequency, it is resistive.

Question 4.
In frequency modulated wave
(a) frequency varies with time
(b) amplitude varies with time
(c) both frequency and amplitude vary with time
(d) both frequency and amplitude are constant
Answer:
(a) frequency varies with time.
Hint:
In frequency modulated wave, frequency of the carrier wave varies in accordance with the modulating signal.

Question 5.
Laser light is considered to be coherent because it consist of
(a) many wavelengths
(b) uncoordinated wavelengths
(c) coordinated waves of exactly the same wavelength
(d) divergent beam
Answer:
(c) coordinated waves of exactly the same wavelength
Hint:
Laser light consists of waves of save wavelength exactly in same phase. So it is highly coherent.

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Question 6.
The waves used by artificial satellites for communication purposes are
(a) microwaves
(b) AM radiowaves
(c) FM radiowaves
(d) X-rays
Answer:
(a) microwaves
Hint:
Microwaves are used in artificial satellites for communication purposes.

Question 7.
An oscillator is producing FM waves of frequency 2 kHz with a variation of 10 kHz. What is the modulation index?
(a) 0.67
(b) 5.00
(c) 0.20
(d) 1.5
Answer:
(b) 5.00
Hint:
mf = \(\frac { ∆f }{ f }\) = \(\frac { 10kHz }{ 2kHz }\) = 5

Question 8.
A laser beam is used for locating distant object because it
(a) has small angular spread
(b) is not absorbed
(c) is coherent
(d) is monochromatic
Answer:
(a) has small angular spread
Hint:
Laser beam has very small angular spread.

Question 9.
In short wave communication, waves of which of the following frequencies will be reflected back by the ionospheric layer having electron density 1011 m-3?
(a) 2 MHz
(b) 10 MHz
(c) 12 MHz
(d) 18 MHz
Answer:
(a) 2 MHz
Hint:
Critical frequency, fc = 9 (Nmax)1/2
= 9(1011)1/2 = 2.8 x 106 Hz = 2.8 MHz
∴ The wave of frequency 2 MHz will be reflected by the ionosphere.

Question 10.
The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to
(a) h1/2
(b) h
(c) 3/2
(d) h2
Answer:
(a) h1/2
Hint:
d = \(\sqrt { 2Rh } \) ; d ∝ h1/2

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Question 11.
If the highest modulating frequency of the wave is 5 kHz, the number of stations that can be accommodated in a 150 kHz band width is
(a) 15
(b) 10
(c) 5
(d) none of these
Answer:
(a) 15
Samacheer Kalvi 12th Physics Solutions Chapter 10 Communication Systems-8= \(\frac { 150khz }{ 2×5kHz }\) = 15

Question 12.
In communication with help of antenna if height is doubled, then the range covered with which was initially r would become
(a) \(\sqrt { 2r } \)
(b) 3r
(c) 4r
(d) 5r
Answer:
(a) \(\sqrt { 2r } \)
Hint:
Initial range, r – \(\sqrt { 2Rh } \)
When height of antenna is doubled, r’ = \(\sqrt { 2R\times 2h } \) – \(\sqrt { 2r } \)

Question 13.
A laser beam is used for carrying out surgery, because it
(a) is highly monochromatic
(b) is highly coherent
(e) is highly directional
(d) can be sharply focused
Answer:
(d) can be sharply focused
Hint:
A laser beam is highly monochromatic, directional and coherent and hence it can be sharply focused for carrying out surgery.

Question 14.
Ozone layer is present in
(a) troposphere
(b) stratosphere
(c) ionosphere
(d) mesosphere
Answer:
(b) stratosphere

Question 15.
Ozone layer blocks the radiation of wavelength
(a) less than 3 x 10-7 m
(b) equal to 3 x 10-7 m
(c) more than 3 x 10-7 m
(d) none of these
Answer:
(a) less than 3 x 10-7 m
Hint:
Ozone layer blocks ultraviolet radiation from the sun for this radiation λ < 3 x 10-7 m.

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Question 16.
What is the cause of Green house effect?
(a) Infrared rays
(b) ultraviolet rays
(c) X-rays
(d) radiowaves
Answer:
(a) Infrared rays

Question 17.
Ozone layer in atmosphere is useful, because it
(a) stops ultraviolet radiation
(b) stops green house effect
(c) stops increase in temperature of atmosphere
(d) absorbs pollutent gases
Answer:
(a) stops ultraviolet radiation

Question 18.
Biological importance of ozone layer is
(a) ozone layer controls O2 / H2 ratio in atmosphere
(b) it stops ultraviolet rays
(c ) ozone layer reduces green house
(d) ozone layer reflects radio waves
Answer:
(b) it stops ultraviolet rays
Hint:
Ozone layer stops ultraviolet radiation.

Question 19.
The principle used in the transmission of signals through an optical fibre is
(a) total internal reflection
(b) refraction
(c) dispersion
(d) interference
Answer:
(a) total internal reflection
Hint:
Signals propagate through an optical fibre by suffering repeated total internal reflections.

Question 20.
LANDSAT series of satellites move in near polar orbits at an altitude of
(a) 3600 km
(b) 3000 km
(c) 918 km
(d) 512 km
Answer:
(c) 918 km
Hint:
LANDSAT satellites are polar satellites which move in polar orbit at a height of 918 km above the surface of the earth.

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Question 21.
Which of the following is not a transducer?
(a) loudspeaker
(b) amplifier
(c) microphone
(d) all of these
Answer:
(b) amplifier
Hint:
Any device which converts energy from one form to another is called a transducer. Loudspeaker and microphone are transducers but not an amplifier.

Question 22.
If a radio receiver amplifies all the signal frequencies equally well, it is said to have high
(a) fidelity
(b) distortion
(c) sensitivity
(d) selectivity
Answer:
(a) fidelity
Hint:
If a radio receiver amplifies all the signal frequencies equally well, it is said to have high fidelity.

Question 23.
The sky wave propagation is suitable for radio waves of frequency
(a) from 2 MHz to 50
(b) upto 2MHz
(c) from 2 MHz to 30 MHz
(d) from 2MHz to 20 MHz
Answer:
(c) from 2 MHz to 30 MHz

Question 24.
Refractive index of ionosphere is
(a) zero
(b) more than one
(c) less than one
(d) one
Answer:
(c) less than one
Hint:
Ionosphere is the upper most layer of earth’s atmosphere having veiy low density. Its refractive index is less than one.

Question 25.
When radiowaves pass through ionosphere phase difference between space current and capacitive displacement current is
(a) 0 rad
(b) 3π/2 rad
(c) π/2 rad
(d) π rad
Answer:
(a) 0 rad
Hint:
The phase difference between space current and capacitive displacement current is zero.

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Question 26.
Advantages of optical fibres are
(a) high bandwidth and EM interference
(b) low bandwidth and EM interference
(c) high bandwidth, low transmission capacity and no EM interference
(d) high bandwidth, high data transmission capacity and no EM interference
Answer:
(d) high bandwidth, high data transmission capacity and no EM interference
Hint:
Optical fibres have high bandwidth, high data transmission capacity and are free from electromagnetic interference.

Question 27.
A TV tower has a height of 100 m. What is the maximum distance upto which the TV transmission can be received? R = 8 x 106 m
(a) 34.77 km
(b) 32.70 km
(c) 40 km
(d) 40.70 km
Answer:
(c) 40 km
Hint:
d = \(\sqrt { 2hr } \) = \(\sqrt { 2\times 100\times 8\times { 10 }^{ 6 } } \) = 40,000 m = 40 km

Question 28.
Modem is a device which performs
(a) modulation
(b) demodulation
(c) rectification
(d) modulation and demodulation
Answer:
(d) modulation and demodulation

Question 29.
Which of the following device is full duplex?
(a) Mobile phone
(b) walky-talky
(c) loud-speaker
(d) radio
Answer:
(a) Mobile phone
Hint:
A mobile phone is a full duplex device by which two persons can talk and hear each other at the same time.

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Question 30.
For a radio signal to travel 150 km from the transmitter to a receiving antenna, it takes
(a) 5 x 10-4 second
(b) 4.5 x 10-3 second
(c) 5 x 10-8 second
(d) 4.5 x 10-6 second
Answer:
(a) 5 x 10-4 second
Hint:
t = \(\frac { s }{ v }\) = \(\frac { 150\times { 10 }^{ 3 }m }{ 3\times { 10 }^{ 8 }{ ms }^{ -1 } } \) = 5 x 10-4 s.

Samacheer Kalvi 12th Physics Communication Systems Short Answer Questions

Question 1.
What is a communication system?
Answer:
The set up used to transmit information from one point to another is called a communication system.

Question 2.
Write down the advantages and limitations of amplitude modulation (AM)?
Answer:
Advantages of AM:

  1. Easy transmission and reception
  2. Lesser bandwidth requirements
  3. Low cost

Limitations of AM:

  1. Noise level is high
  2. Low efficiency
  3. Small operating range

Question 3.
Write down the advantages and limitations of frequency modulation (FM)?
Answer:
Advantages of FM:

  1. Large decrease in noise. This leads to an increase in signal-noise ratio.
  2. The operating range is quite large.
  3. The transmission efficiency is very high as all the transmitted power is useful.
  4. FM bandwidth covers the entire frequency range which humans can hear. Due to this, FM radio has better quality compared to AM radio.

Limitations of FM:

  1. FM requires a much wider channel.
  2. FM transmitters and receivers are more complex and costly.
  3. In FM reception, less area is covered compared to AM.

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Question 4.
Write down the advantages of phase modulation (PM)?
Answer:
Advantages of PM:

  1. FM signal produced from PM signal is very stable.
  2. The centre frequency called resting frequency is extremely stable.

Question 5.
Define bandwidth?
Answer:
The frequency range over which the baseband signals or the information signals such as voice, music, picture, etc. is transmitted is known as bandwidth.

Question 6.
Define bandwidth of transmission system?
Answer:
The range of frequencies required to transmit a piece of specified information in a particular channel is called channel bandwidth or the bandwidth of the transmission system.

Question 7.
Define skip distance.
Answer:
The shortest distance between the transmitter and the point of reception of the sky wave along the surface is called as the skip distance.

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Question 8.
What is skip zone or skip area.
Answer:
There is a zone in between where there is no reception of electromagnetic waves neither ground nor sky, called as skip zone or skip area.

Question 9.
What is mean by fibre optic communication?
Answer:
The method of transmitting information from one place to another in terms of light pulses through an optical fiber is called fiber optic communication.

Question 10.
Write down the application of ICT in Fisheries?
Answer:
Fisheries:
(a) Satellite vessel monitoring system helps to identify fishing zones.
(b) Use of barcodes helps to identify time and date of catch, species name, quality of fish.

Samacheer Kalvi 12th Physics Communication Systems Long Answer Questions

Question 1.
Explain the concept of satellite communication? Write its applications.
Answer:
The satellite communication is a mode of communication of signal between transmitter and receiver via satellite. The message signal from the Earth station is transmitted to the satellite on board via an uplink (frequency band 6 GHz), amplified by a transponder and then retransmitted to another earth station via a downlink (frequency band 4 GHz).
Samacheer Kalvi 12th Physics Solutions Chapter 10 Communication Systems-9
The high-frequency radio wave signals travel in a straight line (line of sight) may come across tall buildings or mountains or even encounter the curvature of the earth. A communication satellite relays and amplifies such radio signals via transponder to reach distant and far off places using uplinks and downlinks. It is also called as a radio repeater in sky. The applications are found to be in all fields and are discussed below.

Applications:
Satellites are classified into different types based on their applications. Some satellites are discussed below.

  1. Weather Satellites:
    They are used to monitor the weather and climate of Earth. By measuring cloud mass, these satellites enable us to predict rain and dangerous storms like hurricanes, cyclones etc.
  2. Communication satellites:
    They are used to transmit television, radio, internet signals etc. Multiple satellitesare used for long distances.
  3. Navigation satellites:
    These are employed to determine the geographic location of ships, aircrafts or any other object.

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Question 2.
Explain the mobile communication? Write its applications.
Answer:
Mobile communication is used to communicate with others in different locations without the use of any physical connection like wires or cables. It allows the transmission over a wide range of area without the use of the physical link. It enables the people to communicate with each other regardless of a particular location like office, house, etc. It also provides communication access to remote areas.

It provides the facility of roaming:
that is. the user may move from one place to another without the need of compromising on the communication. The maintenance and cost of installation of this communication network are also cheap.

Applications:

  1. It is used for personal communication and cellular phones offer voice and data connectivity with high speed.
  2. Transmission of news across the globe is done within a few seconds.
  3. Using Internet of Things (IoT), it is made possible to control various devices from a single device.
    Example: home automation using a mobile phone.
  4. It enables smart classrooms, online availability of notes, monitoring student activities etc. in the field of education.

Question 3.
What do you know about INTERNET? Write its few applications?
Answer:
Internet is a fast growing technology in the field of communication system with multifaceted tools. It provides new ways and means to interact and connect with people. Internet is the largest computer network recognized globally that connects millions of people through computers. It finds extensive applications in all walks of life.

Applications:

  1. Search engine:
    The search engine is basically a web-based service tool used to search for information on World Wide Web.
  2. Communication:
    It helps millions of people to connect with the use of social networking: emails, instant messaging services and social networking tools.
  3. E-Commerce:
    Buying and selling of goods and services, transfer of funds are done over an electronic network.

Samacheer Kalvi 12th Physics Communication Systems Additional problems

Question 1.
A radio can tune to any station in 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Solution:
Frequency of 7.5 MHz belongs to SW band. The corresponding wavelength is
λ = \(\frac { c }{ υ }\) = \(\frac { 3\times { 10 }^{ 8 } }{ 7.5\times { 10 }^{ 6 } } \) = 40m
Frequency of 12 MHz belongs to HF band, the corresponding wavelength is
λ = \(\frac { c }{ υ }\) = \(\frac { 3\times { 10 }^{ 8 } }{ 12\times { 10 }^{ 6 } } \) = 25m
Corresponding to the given frequency band, the wavelength band is 25 m – 40 m band.

Question 2.
A TV transmitting antenna is 125 m tall. How much service area can this transmitting antenna cover, if the receiving antenna is at the ground level? Radius of earth = 6400 km.
Solution:
hT = 125 m and R = 6400 x 103 m
d = \(\sqrt { 2{ h }_{ T }R } \) = \(\sqrt { 2{ h }_{ T }R } \) = 40 x 103m = 40 km
Area covered A = πd2 = 3.14 x (40)2 = 5024 km2

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Question 3.
A transmitting antenna at the top of a tower has a height 32 m and that of the receiving antenna is 100 m. What is the maximum distance between them for satisfactory communication in LOS mode? Given radius of earth 6.4 x 106 m.
Solution:
hT = 32 m, hR = 100 m
R = 6.4 x 106 m
dm = \(\sqrt { 2R{ h }_{ T } } \) = \(\sqrt { 2R{ h }_{ R } } \)
= \(\sqrt { 2\times 604\times { 10 }^{ 6 }\times 32 } \) + \(\sqrt { 2\times 604\times { 10 }^{ 6 }\times 100 } \)
= (64 x 102√10) + 98 x 103√10) = 144 x 102√10 machines
dm = 45. 5 km

Samacheer Kalvi 12th Physics Solutions Chapter 11 Recent Developments in Physics

Students can Download Physics Chapter 11 Recent Developments in Physics Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 11 Recent Developments in Physics

Samacheer Kalvi 12th Physics Recent Developments in Physics Textual Evaluation Solved

Samacheer Kalvi 12th Physics Recent Developments in Physics Multiple Choice Questions

Question 1.
The particle size of ZnO material is 30 nm. Based on the dimension it is classified as –
(a) Bulk material
(b) Nanomaterial
(c) Soft material
(d) Magnetic material.
Answer:
(b) Nanomaterial

Question 2.
Which one of the following is the natural nanomaterial?
(a) Peacock feather
(b) Peacock beak
(c) Grain of sand
(d) Skin of the Whale.
Answer:
(a) Peacock feather

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Question 3.
The blue print for making ultra durable synthetic material is mimicked from-
(a) Lotus leaf
(b) Morpho butterfly
(c) Parrot fish
(d) Peacock feather.
Answer:
(c) Parrot fish

Question 4.
The method of making nanomaterial by assembling the atoms is called-
(a) Top down approach
(h) Bottom up approach
(c) Cross down approach
(d) Diagonal approach.
Answer:
(b) Bottom up approach

Question 5.
“Sky wax” is an application of nano product in the field of-
(a) Medicine
(b) Textile
(c) Sports
(d) Automotive industry.
Answer:
(c) Sports

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Question 6.
The materials used in Robotics are-
(a) Aluminium and silver
(b) Silver and gold
(c) Copper and gold
(d) Steel and aluminium.
Answer:
(d) Steel and aluminum.

Question 7.
The alloys used for muscle wires in Robots are-
(a) Shape memory alloys
(b) Gold copper alloys
(c) Gold silver alloys
(d) Two dimensional alloys.
Answer:
(a) Shape memory alloys

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Question 8.
The technology used for stopping the brain from processing pain is-
(a) Precision medicine
(b) Wireless brain sensor
(c) Virtual reality
(d) Radiology.
Answer:
(c) Virtual reality

Question 9.
The particle which gives mass to protons and neutrons are-
(a) Higgs particle
(b) Einstein particle
(c) Nanoparticle
(d) Bulk particle.
Answer:
(a) Higgs particle

Question 10.
The gravitational waves were theoretically proposed by-
(a) Conrad Rontgen
(b) Marie Curie
(c) Albert Einstein
(d) Edward Purcell.
Answer:
(c) Albert Einstein

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Samacheer Kalvi 12th Physics Recent Developments in Physics Short Answer Questions

Question 1.
Distinguish between Nanoscience and Nanotechnology?
Answer:
1. Nanoscience:

  • Nanoscience is the science of objects with typical sizes of 1 – 100 nm. Nano means one – billionth of a metre that is 10-9 m.
  • If matter is divided into such small objects the mechanical, electrical, optical, magnetic and other properties change.

2. Nanotechnology:

  • Nanotechnology is a technology involving the design, production, characterization, and applications of nano structured materials.

Question 2.
What is the difference between Nano materials and Bulk materials?
Answer:

  1. The solids are made up of particles. Each of the particle has a definite number of atoms, which might differ from material to material. If the particle of a solid is of size less than 100 nm, it is said to be a ‘nano solid’.
  2. When the particle size exceeds 100 nm, it is a ‘bulk solid’. It is to be noted that nano and bulk solids may be of the same chemical composition.
  3. For example, ZnO can be both in bulk and nano form.
  4. Though chemical composition is the same, nano form of the material shows strikingly different properties when compared to its bulk counterpart.

Question 3.
Give any two examples for “Nano” in nature.
Answer:
1.  Single strand DNA:
A single strand of DNA, the building block of all living things, is about three nanometers wide.

2. Morpho Butterfly:
The scales on the wings of a morpho butterfly contain nanostructures that change the way light waves interact with each other, giving the wings brilliant metallic blue and green hues. Mimic in laboratories – Manipulation of colours by adjusting the size of nano particles with which the materials are made.

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Question 4.
Mention any two advantages and disadvantages of Robotics.
Answer:

  1. Advantages of Robotics:
    • The robots are much cheaper than humans.
    • Robots never get tired like humans. It can work for 24 x 7. Hence absenteeism in work place can be reduced.
    • Robots are more precise and error free in performing the task.
  2. Disadvantages of Robotics:
    • Robots have no sense of emotions or conscience.
    • They lack empathy and hence create an emotionless workplace.
    • If ultimately robots would do all the work, and the humans will just sit and monitor them, health hazards will increase rapidly.

Question 5.
Why steel is preferred in making Robots?
Answer:
Steel is several time stronger. In any case, because of the inherent strength of metal, robot bodies are made using sheet, bar, rod, channel, and other shapes.

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Question 6.
What are black holes?
Answer:
Black holes are end stage of stars which are highly dense massive object. Its mass ranges from 20 times mass of the sun to 1 million times mass of the sun. It has very strong gravitational force such that no particle or even light can escape from it. The existence of black holes is studied when the stars orbiting the black hole behave differently from the other starts. Every galaxy has black hole at its center. Sagittarius A* is the black hole at the center of the Milky Way galaxy.

Question 7.
What are sub atomic particles?
Answer:

  1. The three main subatomic particles that form an atom are protons, neutrons and electrons.
  2. Subatomic particles are particles that are smaller than the atom, proton and neutron are made up of quarks which is interact through gluons.
  3. Subatomic particle having two types of particles, they are elementary particle and composite particle.

Samacheer Kalvi 12th Physics Recent Developments in Physics Long Answer Questions

Question 1.
Discuss the applications of Nanomaterials in various fields?
Answer:
(i) Automotive industry:

  • Lightweight construction
  • Painting (fillers, base coat, clear coat)
  • Catalysts
  • Tires (fillers)
  • Sensors
  • Coatings for window screen and car bodies

(ii) Chemical industry:

  • Fillers for paint systems
  • Coating systems based on nanocomposites
  • Impregnation of papers
  • Switchable adhesives
  • Magnetic fluids

(iii) Engineering

  • Wear protection for tools and machines (anti blocking coatings, scratch resistant coatings on plastic parts, etc.)
  • Lubricant – free bearings

(iv) Electronic industry

  • Data memory
  • Displays
  • Laser diodes
  • Glass fibres
  • Optical switches
  • Filters (IR-blocking)
  • Conductive, antistatic coatings

(v) Construction:

  • Construction materials
  • Thermal insulation
  • Flame retardants
  • Surface – functionalised building materials for wood, floors, stone, facades, tiles, roof tiles, etc.
  • Facade coatings
  • Groove mortar

(vi) Medicine:

  • Drug delivery systems
  • Contrast medium
  • Prostheses and implants
  • Agents in cancer therapy
  • Active agents
  • Medical rapid tests
  • Antimicrobial agents and coatings

(vii) Textile / fabrics / non – wovens:

  • Surface – processed textiles
  • Smart clothes

(viii) Energy:

  • Fuel cells
  • Solar cells
  • Batteries
  • Capacitors

(ix) Cosmetics:

  • Sun protection
  • Lipsticks
  • Skin creams
  • Tooth paste

(x) Food and drinks:

  • Package materials
  • Additives
  • Storage life sensors
  • Clarification of fruit juices

(xi) Household:

  • Ceramic coatings for irons
  • Odors catalyst
  • Cleaner for glass, ceramic, floor, windows

(xii) Sports / outdoor:

  • Ski wax
  • Antifogging of glasses / goggles
  • Antifouling coatings for ships / boats
  • Reinforced tennis rackets and balls.

Question 2.
What are the possible harmful effects of usage of Nanoparticles? Why?
Answer:
Possible harmful effects of usage of Nanoparticles:

1. The research on the harmful impact of application of nanotechnology is also equally important and fast developing. The major concern here is that the nanoparticles have the dimensions same as that of the biological molecules such as proteins. They may easily get absorbed onto the surface of living organisms and they might enter the tissues and fluids of the body.

2. The adsorbing nature depends on the surface of the nanoparticle. Indeed, it is possible to deliver a drug directly to a specific cell in the body by designing the surface of a nanoparticle so that it adsorbs specifically onto the surface of the target cell.

3. The interaction with living systems is also affected by the dimensions of the nanoparticles. For instance, nanoparticles of a few nanometers size may reach well inside biomolecules, which is not possible for larger nanoparticles.

4. Nanoparticles can also cross cell membranes. It is also possible for the inhaled nanoparticles to reach the blood, to reach other sites such as the liver, heart or blood cells.

5. Researchers are trying to understand the response of living organisms to the presence of nanoparticles of varying size, shape, chemical composition and surface characteristics.

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Question 3.
Discuss the functions of key components in Robots?
Answer:
The robotic system mainly consists of sensors, power supplies, control systems, manipulators and necessary software. Most robots are composed of 3 main parts:

    1. The Controller: Also known as the “brain” which is run by a computer program. It gives commands for the moving parts to perform the job.
    2. Mechanical parts: Motors, pistons, grippers, wheels, and gears that make the robot move, grab, turn, and lift.
    3. Sensors: To tell the robot about its surroundings. It helps to determine the sizes and shapes of the objects around, distance between the objects, and directions as well.
      Samacheer Kalvi 12th Physics Solutions Chapter 11 Recent Developments in Physics Q3

Question 4.
Elaborate any two types of Robots with relevant examples?
Answer:
(i) Human Robot: Certain robots are made to resemble humans in appearance and replicate the human activities like walking, lifting, and sensing, etc.

  1. Power conversion unit:
    Robots are powered by batteries, solar power, and hydraulics.
  2. Actuators:
    Converts energy into movement. The majority of the actuators produce rotational or linear motion.
  3. Electric motors:
    They are used to actuate the parts of the robots like wheels, arms, fingers,
    legs, sensors, camera, weapon systems etc. Different types of electric motors are used. The most often used ones are AC motor, Brushed DC motor, Brushless DC motor, Geared DC motor, etc.
  4. Pneumatic Air Muscles:
    They are devices that can contract and expand when air is pumped inside. It can replicate the function of a human muscle. They contract almost 40% when the air is sucked inside them.
  5. Muscle wires:
    They are thin strands of wire made of shape memory alloys. They can contract by 5% when electric current is passed through them.
  6. Piezo Motors and Ultrasonic Motors:
    Basically, we use it for industrial robots.
  7. Sensors:
    Generally used in task environments as it provides information of real – time knowledge.
  8. Robot locomotion:
    Provides the types of movements to a robot. The different types are:

    • Legged
    • Wheeled
    • Combination of Legged and Wheeled Locomotion
    • Tracked slip / skid.

(ii) Industrial Robots:
Six main types of industrial robots:

  1. Cartesian
  2. SCARA (Selective Compliance Assembly Robot Arm)
  3. Cylindrical
  4. Delta
  5.  Polar
  6. Vertically articulated

Six – axis robots are ideal for:

  1. Arc Welding
  2. Spot Welding
  3. Material Handling
  4. Machine Tending
  5. Other Applications

Question 5.
Comment on the recent advancement in medical diagnosis and therapy.
Answer:
The recent advancement in medical diagnosis and therapy:

  1. Virtual reality
  2. Precision medicine
  3. Health wearables
  4. Artificial organs
  5.  3 – D printing
  6. Wireless brain sensors
  7. Robotic surgery
  8. Smart inhalers

1. Virtual reality:
Medical virtual reality is effectively used to stop the brain from processing pain and cure soreness in the hospitalized patients. Virtual reality has enhanced surgeries by the use of 3D models by surgeons to plan operations. It helps in the treatment of Autism, Memory loss, and Mental illness.

2. Precision medicine:
Precision medicine is an emerging approach for disease treatment and prevention that takes into account individual variability in genes, environment, and lifestyle for each person. In this medical model it is possible to customise healthcare, with medical decisions, treatments, practices, or products which are tailored to the individual patient.

3. Health wearables:
A health wearable is a device used for tracking a wearer’s vital signs or health and fitness related data, location, etc. Medical wearables with artificial intelligence and big data provide an added value to healthcare with a focus on diagnosis, treatment, patient monitoring and prevention.

Note: Big Data:Extremely large data sets that may be analysed computationally to reveal patterns, trends, and associations, especially relating to human behaviour and interactions.

4. Artificial organs:
An artificial organ is an engineered device or tissue that is implanted or integrated into a human. It is possible to interface it with living tissue or to replace a natural organ. It duplicates or augments a specific function or functions of human organs so that the patient may return to a normal life as soon as possible.

5. 3D printing:
Advanced 3D printer systems and materials assist physicians in a range of operations in the medical field from audiology, dentistry, orthopedics and other applications.

6. Wireless brain sensors:
Wireless brain sensors monitor intracranial pressure and temperature and then are
absorbed by the body. Hence there is no need for surgery to remove these devices.

7. Robotic surgery:
Robotic surgery is a type of surgical procedure that is done using robotic systems. Robotically – assisted surgery helps to overcome the limitations of pre – existing minimally invasive surgical procedures and to enhance the capabilities of surgeons performing open
surgery.

8. Smart inhalers:
Inhalers are the main treatment option for asthma. Smart inhalers are designed with health systems and patients in mind so that they can offer maximum benefit. Smart inhalers use bluetooth technology to detect inhaler use, remind patients when to take their medication and gather data to help guide care.

Samacheer Kalvi 12th Physics Recent Developments in Physics Additional Questions

Samacheer Kalvi 12th Physics Recent Developments in Physics Multiple Choice Questions

Question 1.
An automatic apparatus or device that performs functions ordinarily ascribed to human or
operate with what appears to be almost human intelligence is called ……………. .
(a) Robot
(b) Human
(c) Animals
(d) Reptiles.
Answer:
(a) Robot.

Question 2.
The laws of Robotics are ……………. .
(a) a robot may not injure a human being
(b) a robot must obey the order given by human
(c) a robot must protect its own existence
(d) both b and c.
Answer:
(d) both b and c.
Hint:
A robot may not injure a human being or through in action, allow human being to be harmed.

Question 3.
The basic components of robot are ……………. .
(a) mechanical linkage
(b) sensors and controllers
(c) user interface and power conversion unit
(d) All the above.
Answer:
(d) All the above.

Question 4.
What is the name for information sent from robot sensors to robot controllers ……………. .
(a) temperature
(b) pressure
(c) feedback
(d) signal.
Answer:
(c) feedback

Question 5.
Which of the following uses radio frequency to produce nano – particles ……………. .
(a) Plasma arching
(b) Chemical vapour deposition
(c) Sol-gel technique
(d) Electro deposition.
Answer:
(a) Plasma arching

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Question 6.
Which of the following the atoms do not move from each other ……………. .
(a) Shape memory alloys
(b) Nano materials
(c) Dielectrics
(d) Static materials.
Answer:
(b) Nano materials

Question 7.
The diameter of the nano wire is about ……………. .
(a) 10-6 m
(b) 10-3 m
(c) 10-8 m
(d) 10-9 m.
Answer:
(d) 10-9 m.

Question 8.
A suspended nano wire is a wire that is produced in ……………. .
(a) Air medium
(b) Vaccum
(c) Low vaccum chamber
(d) High vaccum chamber.
Answer:
(d) High vaccum chamber.

Question 9.
For nano metres whose diameters less than …………….are used as welding purposes.
(a) 10 nm
(b) 20 nm
(c) 30 nm
(d) 40 nm.
Answer:
(a) 10 nm

Question 10.
Nano wires are used in ……………. .
(a) 10 nm
(b) Resistors
(c) Capacitors
(d) Transducers.
Answer:
(a) 10 nm

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Question 11.
Generally what is the material of needle electrodes ……………. .
(a) Stainless steel
(b) Copper
(c) Lead
(d) Iron.
Answer:
(a) Stainless steel

Question 12.
……………. introduced is used to hold patients head and guide the placements of electrodes.
(a) Monotaxic
(b) Stereotonic
(c) Stereotaxic
(d) Monotonic.
Answer:
(c) Stereotaxic

Question 13.
Who is the father of the modem robotics industry formed the world’s first robotic company in 1956 ……………. .
(a) Joliot
(b) Cormark
(c) Engelberger
(d) Edward purcell.
Answer:
(c) Engelberger

Question 14.
In 1954, ……………. invented the first digitally operated programmable robot called unimate.
(a) Edward purcell
(b) George Devol
(c) Engel berger
(d) Joliot.
Answer:
(b) George Devol

Question 15.
The phenomenon of artificial radioactivity was invented by ……………. .
(a) Joliot and Irene curie
(b) Felix Bloch and Edward purcell
(c ) Connack and Hounsfield
(d) Wilhelm conrad – Rontgen.
Answer:
(a) Joliot and Irene curie

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Samacheer Kalvi 12th Physics Recent Developments in Physics Short Answer Question

Question 1.
What is physics?
Answer:
Physics is the basic building block for Science, Engineering, Technology and Medicine.

Question 2.
Write down the applications of Nano technology?
Answer:

  • Energy storage
  • Metallurgy and materials
  • Optical engineering and communication
  • Agriculture and food
  • Biotechnology
  • Defense and security Electronics
  • Biomedical and drug delivery
  • Cosmetics and paints
  • Textile.

Question 3.
What is robotics?
Answer:
Robotics is an integrated study of mechanical engineering, electronic engineering, computer engineering, and science.

Question 4.
What is meant by ‘Robot’? Write its uses?
Answer:
Robot is a mechanical device designed with electronic circuitry and programmed to perform a specific task. These automated machines are highly significant in this robotic era where they can take up the role of humans in certain dangerous environments that are hazardous to people like defusing bombs, finding survivors in unstable ruins, and exploring mines and shipwrecks.

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Question 5.
What is the meaning of particle physics?
Answer:
Particle physics deals with the theory of fundamental particles of nature and it is one of the active research areas in physics. Initially it was thought that atom is the fundamental entity of matter.

Question 6.
Define cosmology?
Answer:
Cosmology is the branch that involves the origin and evolution of the universe. It deals with formation of stars, galaxy etc.

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Students can Download Physics Chapter 1 Nature of Physical World and Measurement Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement TextBook Questions Solved

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Multiple Choice Questions

Question 1.
One of the combinations from the fundamental physical constants is \(\frac{h c}{\mathrm{G}}\). The unit of this expression is
(a) Kg2
(b) m3
(c) S-1
(d) m
Answer:
(a) Kg2
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 2.
If the error in the measurement of radius is 2%, then the error in the determination of volume of the sphere will be …….
(a) 8%
(b) 2%
(c) 4%
(d) 6%
Answer:
(d) 6%

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 3.
If the length and time period of an oscillating pendulum have errors of 1 % and 3% respectively
then the error in measurement of acceleration due to gravity is …… [Related to AMPMT 2008]
(a) 4%
(b) 5%
(c) 6%
(d) 7%
Answer:
(d) 7%

Question 4.
The length of a body is measured as 3.51 m, if the accuracy is 0.01mm, then the percentage error in the measurement is ……
(a) 351%
(b) 1%
(c) 0.28%
(d) 0.035%
Answer:
(c) 0.28%

Question 5.
Which of the following has the highest number of significant figures?
(a) 0.007 m2
(b) 2.64 × 1024 kg
(c) 0.0006032 m2
(d) 6.3200 J
Answer:
(d) 6.3200 J

Question 6.
If π = 3.14, then the value of π2 is …..
(a) 9.8596
(b) 9.860
(c) 9.86
(d) 9.9
Answer:
(c) 9.86

Question 7.
Which of the following pairs of physical quantities have same dimension?
(a) force and power
(b) torque and energy
(c) torque and power
(d) force and torque
Answer:
(b) torque and energy

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 8.
The dimensional formula of Planck’s constant h is [AMU, Main, JEE, NEET]
(a) [ML2T-1]
(b) [ML2T-3]
(c) [MLTT-1]
(d) [MLTT3-3]
Answer:
(a) [ML2T-1]

Question 9.
The velocity of a particle v at an instant t is given by v = at + bt2. The dimensions of b is ……
(a) [L]
(b) [LT-1]
(c) [LT-2]
(d) [LT-3]
Answer:
(d) [LT-3]

Question 10.
The dimensional formula for gravitational constant G is [Related to AIPMT 2004]
(a) [ML-3T-2]
(b) [M-1L3T-2]
(c) [M-1L-3T-2]
(d) [ML-3T2]
Answer:
(b) [M-1L3T-2]

Question 11.
The density of a material in CGS system of units is 4 g cm-3. In a system of units in which unit of length is 10 cm and unit of mass is 100 g, then the value of density of material will be ……
(a) 0.04
(b) 0.4
(c) 40
(d) 400
Answer:
(c) 40

Question 12.
If the force is proportional to square of velocity, then the dimension of proportionality constant
is [JEE-2000] ……
(a) [MLT0]
(b) [MLT-1]
(c) [ML-2T]
(d) [ML-1T0]
Answer:
(d) [ML-1T0]

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 13.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 1213 [MainAIPMT2011]
(a) length
(b) time
(c) velocity
(d) force
Answer:
(c) velocity

Question 14.
Planck’s constant (h), speed of light in vaccum (c) and Newton’s gravitational constant (G) are taken as three fundamental constants. Which of the following combinations of these has the dimension of length? [NEET 2016 (Phase II)]
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 1214
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 7

Question 15.
A length-scale (l) depends on the permittivity (ε) of a dielectric material, Boltzmann constant (kB), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression for l is dimensionally correct?. [JEE (advanced) 2016]
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 8
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 9

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Short Answer Questions

Question 1.
Briefly explain the types of physical quantities.
Answer:
Physical quantities are classified into two types. There are fundamental and derived quantities. Fundamental or base quantities are quantities which cannot be expressed in terms of any other physical quantities. These are length, mass, time, electric current, temperature, luminous intensity and amount of substance.
Quantities that can be expressed in terms of fundamental quantities are called derived quantities. For example, area, volume, velocity, acceleration, force.

Question 2.
How will you measure the diameter of the Moon using parallax method?
Answer:
Let θ is the angular diameter of moon
d – is the distance of moon from earth, from figure, θ = \(\frac{\mathrm{D}}{d}\)
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 10
Diameter of moon D = d.θ
by knowing θ, d, diameter of moon can be calculated

Question 3.
Write the rules for determining significant figures.
Answer:
Rules for counting significant figures:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 12
Note: 1 Multiplying or dividing factors, which are neither rounded numbers nor numbers representing measured values, are exact and they have infinite numbers of significant figures as per the situation.
For example, circumference of circle S = 2πr, Here the factor 2 is exact number. It can be written as 2.0, 2.00 or 2.000 as required.
Note: 2 The power of 10 is irrelevant to the determination of significant figures.
For example x = 5.70 m = 5.70 × 102 cm = 5.70 × 103 mm = 5.70 × 10-3 km.
In each case the number of significant figures is three.

Question 4.
What are the limitations of dimensional analysis?
Answer:
Limitations of Dimensional analysis
1. This method gives no information about the dimensionless constants in the formula like 1, 2, ……… π, e, etc.
This method cannot decide whether the given quantity is a vector or a scalar.
This method is not suitable to derive relations involving trigonometric, exponential and logarithmic functions.
It cannot be applied to an equation involving more than three physical quantities.
It can only check on whether a physical relation is dimensionally correct but not the correctness of the relation. For example, using dimensional analysis, Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 13 is dimensionally correct whereas the correct relation is Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 132

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 5.
Define precision and accuracy. Explain with one example.
Answer:
The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity. Precision of a measurement is a closeness of two or more measured values to each other.
The true value of a certain length is near 5.678 cm. In one experiment, using a measuring instrument of resolution 0.1 cm, the measured value is found to be 5.5 cm. In another experiment using a measuring instrument of greater resolution, say 0.01 cm, the length is found to be 5.38 cm. We find that the first measurement is more accurate as it is closer to the true value, but it has lesser precision. On the contrary, the second measurement is less accurate, but it is more precise.

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Long Answer Questions

Question 1.
(i) Explain the use of screw gauge and vernier caliper in measuring smaller distances.
Answer:
(i) Measurement of small distances: screw gauge and vernier caliper Screw gauge:
The screw gauge is an instrument used for measuring accurately the dimensions of objects up to a maximum of about 50 mm. The principle of the instrument is the magnification of linear motion using the circular motion of a screw. The least count of the screw gauge is 0.01 mm. Vernier caliper: A vernier caliper is a versatile instrument for measuring the dimensions of an object namely diameter of a hole, or a depth of a hole. The least count of the vernier caliper is 0.1 mm.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 20
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 21
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 22

(ii) Write a note on triangulation method and radar method to measure larger distances. Triangulation method for the height of an accessible object;
Let AB = h be the height of the tree or tower to be measured. Let C be the point of observation at distance x from B. Place a range finder at C and measure the angle of elevation, ACB =
θ as shown in figure.
From right angled triangle ABC,
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 23
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 24
(or) height h = x tan θ
Knowing the distance x, the height h can be determined.
RADAR method
The word RADAR stands for radio detection and ranging. A radar can be used to measure accurately the distance of a nearby planet such as Mars. In this method, radio waves are sent from transmitters which, after reflection from the planet, are detected by the receiver. By measuring, the time interval (t) between the instants the radio waves are sent and received, the distance of the planet can be determined as where v is the speed of the radio wave. As the time taken (t) is for the distance covered during the forward and backward path of the radio waves, it is divided by 2 to get the actual distance of the object. This method can also be used to determine the height, at which an aeroplane flies from the ground.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 25

Question 2.
Explain in detail the various types of errors.
Answer:
The uncertainty in a measurement is called an error. Random error, systematic error and gross error are the three possible errors.
(i) Systematic errors: Systematic errors are reproducible inaccuracies that are consistently *, in the same direction. These occur often due to a problem that persists throughout the experiment. Systematic errors can be classified as follows
(1) Instrumental errors: When an instrument is not calibrated properly at the time of manufacture, instrumental errors may arise. If a measurement is made with a meter scale whose end is worn out, the result obtained will have errors. These errors can be corrected by choosing the instrument carefully.
(2) Imperfections in experimental technique or procedure: These errors arise due to the limitations in the experimental arrangement. As an example, while performing experiments with a calorimeter, if there is no proper insulation, there will be radiation losses. This results in errors and to overcome these, necessary correction has to be applied
(3) Personal errors: These errors are due to individuals performing the experiment, may be due to incorrect initial setting up of the experiment or carelessness of the individual making the observation due to improper precautions.
(4) Errors due to external causes: The change in the external conditions during an experiment can cause error in measurement. For example, changes in temperature, humidity, or pressure during measurements may affect the result of the measurement.
(5) Least count error: Least count is the smallest value that can be measured by the measuring instrument, and the error due to this measurement is least count error. The instrument’s
resolution hence is the cause of this error. Least count error can be reduced by using a high precision instrument for the measurement.
(ii) Random errors: Random errors may arise due to random and unpredictable variations in
experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer who performs the experiment. Random errors are sometimes called “chance error”. When different readings are obtained by a person every time he repeats the experiment, personal error occurs. For example, consider the case of the thickness of a wire measured using a screw gauge. The readings taken may be different for different trials. In this case, a large number of measurements are made and then the arithmetic mean is taken.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 26

Question 3.
What do you mean by propagation of errors? Explain the propagation of errors in addition and multiplication.
Answer:
A number of measured quantities may be involved in the final calculation of an experiment. Different types of instruments might have been used for taking readings. Then we may have to look at the errors in measuring various quantities, collectively.
The error in the final result depends on
(i) The errors in the individual measurements
(ii) On the nature of mathematical operations performed to get the final result. So we should know the rules to combine the errors.
The various possibilities of the propagation or combination of errors in different mathematical operations are discussed below:
(i) Error in the sum of two quantities
Let ∆A and ∆B be the absolute errors in the two quantities A and B respectively. Then, Measured value of A = A ± ∆A
Measured value of B = B ± ∆B
Consider the sum, Z = A + B
The error ∆Z in Z is then given by
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 27
The maximum possible error in the sum of two quantities is equal to the sum of the absolute errors in the individual quantities.
Error in the product of two quantities: Let ∆A and ∆B be the absolute errors in the two quantities A, and B, respectively. Consider the product Z = AB
The error ∆Z in Z is given by Z ± ∆Z = (A ± ∆A) (B ± ∆B)
= (AB) ± (A ∆ B) ± (B ∆ A) ± (∆A • ∆B)
Dividing L.H.S by Z and R.H.S by AB, we get,
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 28
As ∆A/A, ∆B/B are both small quantities, their product term Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 1215 can be neglected.
The maximum fractional error in Z is
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 1216

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 4.
Write short note on the following:
(a) Unit
(b) Rounding – off
(c) Dimensionless quantities
Answer:
(a) Unit: An arbitrarily chosen standard of measurement of a quantity, which is accepted internationally is called unit of the quantity.
The units in which the fundamental quantities are measured are called fundamental or base units and the units of measurement of all other physical quantities, which can be obtained by a suitable multiplication or division of powers of fundamental units, are called derived units.
(b) Rounding – off: In no case should the result have more significant figures than die figures involved in the data used for calculation. The result of calculation with numbers containing more than one uncertain digit should be rounded off. The rules for rounding off are given below.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 30
(c) Dimensionless quantities: On the basis of dimension, dimensionless quantities are classified into two categories.
(i) Dimensionless variables:
Physical quantities which have no dimensions, but have variable values are called dimensionless variables. Examples are specific gravity, strain, refractive index etc.
(ii) Dimensionless Constant:
Quantities which have constant values and also have no dimensions are called dimensionless constants. Examples are π, e, numbers etc.

Question 5.
Explain the principle of homogeniety of dimensions. What are its uses? Give example.
Answer:
The principle of homogeneity of dimensions states that the dimensions of all the terms in a physical expression should be the same. For example, in the physical expression v2 = u2 + 2as, the dimensions of v2, u2 and 2 as are the same and equal to [L2T-2].
(i) To convert a physical quantity from one system of units to another: This is based on the fact that the product of the numerical values (n) and its corresponding unit (u) is a constant, i.e, = constant n1[u1] = constant (or) n1[u1] = n2[u2].
Consider a physical quantity which has dimension ‘a’ in mass, ‘b’ in length and ‘c’ in time. If the fundamental units in one system are M1, L1 and T1 and the other system are M2, L2 and T2 respectively, then we can write, Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 31
We have thus converted the numerical value of physical quantity from one system of units into the other system.
Example: Convert 76 cm of mercury pressure into Nm-2 using the method of dimensions. Solution: In cgs system 76 cm of mercury pressure = 76 × 13.6 × 980 dyne cm-2
The dimensional formula of pressure P is [ML-1T-2]

(ii) To check the dimensional correctness of a given physical equation:
Example: The equation \(\frac{1}{2} m v^{2}\) = mgh can be checked by using this method as follows.
Solution: Dimensional formula for
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 311
Dimensional formula for
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 32
Both sides are dimensionally the same, hence the equations Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 300 is dimensionally correct.
(iii) To establish the relation among various physical quantities:
Example: An expression for the time period T of a simple pendulum can be obtained by using this method as follows.
Let true period T depend upon
(i) mass m of the bob
(ii) length l of the pendulum and
(iii) acceleration due to gravity g at the place where the pendulum is suspended. Let the constant involved is K = 2π.
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 312
Here k is the dimensionless constant. Rewriting the above equation with dimensions.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 321
Comparing the powers of M, L and T on both sides, a – 0, b + c = 0, -2c = 1
Solving for a, b and c a = 0, b = 1/2, and c = -1/2
From the above equation
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 33

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Numerical Problems

Question 1.
In a submarine equipped with sonar, the time delay between the generation of a pulse and its echo after reflection from an enemy submarine is observed to be 80 sec. If the speed of sound in water is 1460 ms-1. What is the distance of enemy submarine?
Answer:
Given:
Speed of sound in water = 1460 ms-1
Time delay = 80s
Distance of enemy ship = ?
Solution:
Total distance covered = speed × time
= 1460 ms-1 × 80s = 116800 m
Time taken is for forward and backward path of sound waves.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 34
= 58400 m (or) 58.4 km

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 2.
The radius of the circle is 3.12 m. Calculate the area of the circle with regard to significant figures.
Answer:
Given: radius : 3.12 m (Three significant figures)
Solution:
Area of the circle = πr2 = 3.14 × (3.12 m)2 = 30.566
If the result is rounded off into three significant figure, area of the circle = 30.6 m2

Question 3.
Assuming that the frequency v of a vibrating string may depend upon
(i) applied force (F)
(ii) length (l)
(iii) mass per unit length (m), prove that Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 315 using dimensional analysis. [Related to JIPMER 2001]
Answer:
Given: The frequency v of a vibrating string depends
(i) applied force (F)
(ii) length (l)
(iii) mass per unit length (m)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 351
Substitute the dimensional formulae of the above quantities
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 36
Comparing the powers of M, L, T on both sides,
x + z = 0, x + y – z = 0, -2x = -1
Solving for x, y, z, we get
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 37
Substitute x, y, z values in equ(1)
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 38

Question 4.
Jupiter is at a distance of 824.7 million km from the Earth. Its angular diameter is measured to be 35.72″. Calculate the diameter of Jupiter.
Answer:
Given,
Given Distance of Jupiter = 824.7 × 106 km = 8.247 × 1011 m
angular diameter = 35.72 × 4.85 × 10-6rad = 173.242 × 10-6 rad
= 1.73 × 10-4 rad
∴ Diameter of Jupiter D = D × d = 1.73 × 10-4 rad × 8.247 × 1011 m
= 14.267 × 1o7 m = 1.427 × 108 m (or) 1.427 × 105</sup km

Question 5.
The measurement value of length of a simple pendulum is 20 cm known with 2 mm accuracy. The time for 50 oscillations was measured to be 40 s within 1 s resolution. Calculate the percentage of accuracy in the determination of acceleration due to gravity ‘g’ from the above measurement.
Answer:
Given,
Length of simple pendulum (l) = 20 cm
absolute error in length (∆l) = 2 mm = 0.2 cm
Time taken for 50 oscillation (t) = 40 s
error in time ∆T = 1 s
Solution: Time period for one oscillation (T)
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 39
Hence, the percentage error in g is
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 40

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Conceptual Questions

Question 1.
Why is it convenient to express the distance of stars in terms of light year (or) parsec rather than in km?
Answer:
A parsec is 206, 265 AU and is roughly the distance to the nearest stars. If we were to view a giant star with a diameter of 1 AU at a distance of one parsec, it would appear to be just 1/3600th of a degree in angular size. For comparison, the sun and moon are both half a degree in angular size when viewed from Earth.

Question 2.
Show that a screw gauge of pitch 1 mm and 100 divisions is more precise than a vernier caliper with 20 divisions on the sliding scale.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 41
As shown, the least count of screw gauge is lesser then vernier caliper, hence screw gauge is more precise.

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 4.
Having all units in atomic standards is more useful. Explain.
Answer:
An atomic mass unit (symbolized AMU or amu) is defined as precisely 1/12 the mass of an atom of carbon-12. The carbon-12 (C-12) atom has six protons and six neutrons in its nucleus.
In imprecise terms, one AMU is the average of the proton rest mass and the neutron rest mass. This is approximately 1.67377 × 10-27 kilogram (kg), or 1.67377 × 10-24 gram (g). The mass of an atom in AMU is roughly equal to the sum of the number of protons and neutrons in the nucleus.
The AMU is used to express the relative masses of, and thereby differentiate between, various isotopes of elements. Thus, for example, uranium-235 (U-235) has an AMU of approximately 235, while uranium-238 (U-238) is slightly more massive. The difference results from the fact that U-238, the most abundant naturally occurring isotope of uranium, has three more neutrons than U-235, an isotope that has been used in nuclear reactors and atomic bombs.

Question 5.
Why dimensional methods are applicable only up to three quantities?
Answer:
Understanding dimensions is of utmost importance as it helps us in studying the nature of physical quantities mathematically. The basic concept of dimensions is that we can add or subtract only those quantities which have same dimensions. Also, two physical quantities are equal if they have same dimensions. these basic ideas help us in deriving the new relation between physical quantities, it is just like units.

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Additional Questions Solved

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Multiple Choose Questions

Question 1.
The unit of surface tension ……
(a) MT-2
(b) Nm-2
(c) Nm
(d) Nm-1
Answer:
(d) Nm-1

Question 2.
One atomus equal to ……
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 401
Answer:
(c) 160 ms

Question 3.
One light year is ……
(a) 3.153 × 107 m
(b) 1.496 × 107 m
(c) 9.46 × 1012 km
(d) 3.26 × 1015 km
Answer:
(c) 9.46 × 1012 km

Question 4.
One Astronomical unit is
(a) 3.153 × 107 m
(b) 1.496 × 107 m
(c) 9.46 × 1012 m
(d) 3.26 × 1015 m
Answer:
(b) 1.496 × 107 m

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 5.
One parsec is …..
(a) 3.153 × 107 m
(b) 3.26 × 1015 m
(c) 30.84 × 1015 m
(d) 9.46 × 1015 m
Answer:
(c) 30.84 × 1015 m

Question 6.
One Fermi is …..
(a) 10-9 m
(b) 10-10 m
(c) 10-12 m
(d) 10-15 m
Answer:
(d) 10-15 m

Question 7.
One Angstrom is ………
(a) 10-9 m
(b) 10-10m
(c) 10-12 m
(d) 10-15 m
Answer:
(b) 10-10 m

Question 8.
One solar mass is ….
(a) 2 × 1030 kg
(b) 2 × 1030 g
(c) 2 × 1030 mg
(d) 2 × 1030 tonne
Answer:
(d) 2 × 1030 tonne

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 9.
\(\frac{1}{12}\) of the mass of carbon 12 atom is …..
(a) 1 TMC
(b) mass of neutron
(c) 1 amu
(d) mass of hydrogen
Answer:
(d) mass of hydrogen

Question 10.
The word physics is derived from the word …..
(a) scientist
(b) fusis
(c) fission
(d) fusion
Answer:
(b) fusis

Question 11.
The study of forces acting on bodies whether at rest or in motion is …..
(a) classical mechanics
(b) quantum mechanics
(c) thermodynamics
(d) condensed matter physics
Answer:
(a) classical mechanics

Question 12.
Mass of observable universe …..
(a) 1031 kg
(b) 1041 kg
(c) 1055 kg
(d) 9.11 × 1031 kg
Answer:
(c) 1055 kg

Question 13.
Mass of an electron …….
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 301
Answer:
(b) 9.11 × 10-31 kg

Question 14.
The study of production and propagation of sound waves …..
(a) Astrophysics
(b) Acoustics
(c) Relativity
(d) Atomic physics
Answer:
(b) Acoustics

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 15.
The study of the discrete nature of phenomena at the atomic and subatomic levels.
(a) Quantum mechanics
(b) High energy physics
(c) Acoustics
(d) Classical mechanics
Answer:
(a) Quantum mechanics

Question 16.
The techniQuestion used to study the crystal structure of various rocks are …….
(a) diffraction
(b) interference
(c) total internal reflection
(d) refraction
Answer:
(a) diffraction

Question 17.
The astronomers used to observe distant points of the universe by …….
(a) Electron telescope
(b) Astronomical telescope
(c) Radio telescope
(d) Radar
Answer:
(c) Radio telescope

Question 18.
The comparison of any physical quantity with its standard unit is known as ……..
(a) fundamental quantities
(b) measurement
(c) dualism
(d) derived quantities
Answer:
(b) measurement

Question 19.
Fundamental quantities can also be known as …… quantities.
(a) original
(b) physical
(c) negative
(d) base
Answer:
(d) base

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 20.
Which one of the following is not a fundamental quantity?
(a) length
(b) luminous intensity
(c) temperature
(d) water current
Answer:
(d) water current

Question 21.
The system of unit not only based on length, mass, and time is
(a) FPS
(b) CGS
(c) MKS
(d) SI
Answer:
(d) SI

Question 22.
The coherent system of units …..
(a) CGS
(b) SI
(c) FPS
(d) MKS
Answer
(b) SI

Question 23.
The triple point temperature of water is ……
(a) -273.16 K
(b) 0K
(c) 273.16 K
(d) 100 K
Answer:
(d) 100 K

Question 24.
Which of the following is a unit of distance?
(a) Light year
(b) Leap year
(c) Dyne-sec
(d) Pauli
Answer:
(a) Light year

Question 25.
The unit of moment of force ……
(a) Nm2
(b) Nm
(c) N
(d) NJ rad
Answer:
(b) Nm

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 26.
1 radian is ……
(a) 2.91 × 10-4 m
(b) 57.27°
(c) 180°
(d) \(\frac{\pi}{180}\)
Answer:
(b) 57.27°

Question 27.
One degree of arc is …….
(a) 1″
(b) 60″
(c) 60′
(d) 60°
Answer:
(c) 60′

Question 28.
One degree of arc is equal to …….
(a) 1.457 × 102 rad
(b) 1.457 × 10-2 rad
(c) 1.745 × 102 rad
(d) 1.745 × 10-2 rad
Answer:
(b) 1.457 × 10-2 rad

Question 29.
1 minute of arc is equal to …….
(a) 1.745 × 10-2 rad
(b) 2.91 × 10-4 rad
(c) 2.91 × 104 rad
(d) 4.85 × 10-6 rad
Answer:
(b) 2.91 × 10-4 rad

Question 30.
1 second of arc is equal to ………
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 35
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 361

Question 31.
1 second of arc is equal to ….
(a) 0.00027°
(b) 1.745 × 10-2 rad
(c) 2.91 × 10-4 rad
(d) 4.85 × 10-6 rad
Answer:
(a) 0.00027°

Question 32.
Unit of impulse ….
(a) NS2
(b) NS
(c) Nm
(d) Kgms-2
Answer:
(b) NS

Question 33.
The ratio of energy and temperature is known as ……
(a) Stefen’s constant
(b) Boltzmann constant
(c) Plank’s constant
(d) Kinetic constant
Answer:
(b) Boltzmann constant

Question 34.
The range of distance can be measured by using direct methods is …..
(a) 10-2 to 10-5 m
(b) 10-2 to 102 m
(c) 102 to 1(T5 m {d) 10″2 to 105 m
Answer:
(b) 10-2 to 102 m

Question 35.
Which of the following is in increased order?
(a) exa, tera, hecto
(b) tera, exa, hecto
(c) giga, tera, exa
(d) hecto, exa, giga
Answer:
(c) giga, tera, exa

Question 36.
10-18 is called as ……
(a) nano
(b) pico
(c) femto
(d) atto
Answer:
(d) atto

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 37.
A radio signal sent towards the distant planet, returns after “t” s. If “c” is the speed of radio
waves then the distance of the planet and from the earth is …….
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 50
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 51

Question 38.
Find odd one out ….
(a) Newton
(b) metre
(c) candela
(d) Kelvin
Answer:
(a) Newton

Question 39.
The shift in the position of an object when viewed with two eyes, keeping one eye closed at a
time is known as …
(a) basis
(b) fundamental
(c) parallax
(d) pendulum
Answer:
(c) parallax

Question 40.
Chandrasekar limit is ….. times the mass of the sun.
(a) 1.2
(b) 1.4
(c) 1.6
(d) 1.8
Answer:
(b) 1.4

Question 41.
The smallest physical unit of time is
(a) second
(b) minute
(c) microsecond
(d) shake
Answer:
(d) shake

Question 42.
Size of atomic nucleus is …..
(a) 10-10 m
(b) 10-12 m
(c) 10-14 m
(d) 10-18 m
Answer:
(c) 10-14 m

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 43.
Time interval between two successive heart beat is in the order of …….
(a) 10° s
(b) 10 s
(c) 102 s
(d) 10-3 s
Answer:
(a) 10° s

Question 44.
Half life time of a free neutron is in the order of ……
(a) 10°
(b) 101 s
(c) 102 s
(d) 103 s
Answer:
(d) 103 s

Question 45.
The uncertainty contained in any measurement is ……
(a) rounding off
(b) error
(c) parallax
(d) gross
Answer:
(b) error

Question 46.
Zero error of an instrument is a ……
(a) Systematic error
(b) Random error
(c) Gross error
(d) Both (a) and (b)
Answer:
(a) Systematic error

Question 47.
Error in the measurement of radius of a sphere is 2%. Then error in the measurement of surface
area is ….
(a) 1%
(b) 2%
(c) 3%
(d) 4%
Answer:
(d) 4%

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 48.
Imperfections in experimental procedure gives ….. error.
(a) random
(b) gross
(c) systematic
(d) personal
Answer:
(c) Systematic

Question 49.
Random error can also be called as ….
(a) personal error
(b) chance error
(c) gross error
(d) system error
Answer:
(b) chance error

Question 50.
To get the best possible true value of the quantity has to be taken.
(a) rms value
(b) net value
(c) arithmetic mean
(d) mode
Answer:
(c) arithmetic mean

Question 51.
The error caused due to the shear carelessness of an observer is called as …… error.
(a) Systematise
(b) Gross
(c) Random
(d) Personal
Answer:
(b) Gross

Question 52.
The uncertainty in a measurement is called as ….
(a) error
(b) systematic
(c) random error
(d) gross error
Answer:
(a) error

Question 53.
The difference between the true value and the measured value of a quantity is known as …..
(a) Absolute error
(b) Relative error
(c) Percentage error
(d) Systemmatic error
Answer:
(a) Absolute error

Question 54.
If a1, a2, a3 …. an are the measured value of a physical quantity “a” and am is the true value then absolute error …..
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 60
Answer:
(d) \(\Delta a_{n}=a_{m}-a_{n}\)

Question 55.
If ‘am‘ and ‘∆am ‘ are true value and mean absolute error respectively, then the magnitude of the quantity may lie between …..
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 61
Answer:
(b) \(a_{m}-\Delta a_{m} \text { to } a_{m}+\Delta a_{m}\)

Question 56.
The ratio of the mean absolute error to the mean value is called as ……
(a) absolute error
(b) random error
(c) relative error
(d) percentage error
Answer:
(c) Relative error

Question 57.
Relative error can also be called as ……
(a) fractional error
(b) absolute error
(c) percentage error
(d) systematic error
Answer:
(a) fractional error

Question 58.
A measured value to be close to targeted value, percentage error must be close to
(a) 0
(b) 10
(c) 100
(d) ∝
Answer:
(a) 0

Question 59.
The maximum possible error in the sum of two quantities is equal to …….
(a) Z = A + B
(b) ∆Z = ∆A + ∆B
(c) ∆Z = ∆A/∆B
(d) ∆Z = ∆A – ∆B
Answer:
(b) ∆Z = ∆A + ∆B

Question 60.
The maximum possible error in the difference of two quantities is ……
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 62
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 63

Question 61.
The maximum fractional error in the division of two quantities is ….
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 64
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 65

Question 62.
The fractional error in the nth power of a quantity is …..
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 66
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 67

Question 63.
A physical quantity is given as y = \(\frac{a b^{3}}{c^{2}}\). If ∆a, ∆b, ∆c are absolute errors, the possible fractional error in y is …..
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 68
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 69

Question 64.
Number of significant digits in 3256 …
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 65.
Number of significant digits in 32005 ……
(a) 1
(b) 2
(c) 5
(d) 2
Answer:
(c) 5

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 66.
Number of significant digits in 2030 …
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 67.
Number of significant digits in 2030N …..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 68.
Number of significant digits in 0.0342 …..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c)3

Question 69.
Number of significant digit in 20.00 …..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 70.
Number of significant digit in 0.030400
(a) 6
(b) 5
(c) 4
(d) 3
Answer:
(b) 5

Question 71.
The force acting on a body is measured as 4.25 N. Round it off with two significant figure ..
(a) 4.3
(b) 4.2
(c) both
(a) or (b)
(d) 4.25
Answer:
(b) 4.2

Question 72.
The quantities a, b, c are measured as 3.21, 4.253, 7.2346. The sum (a + b + c) with proper
significant digits is ……
(a) 14.6976
(b) 14.697
(c) 14.69
(d) 14.6
Answer:
(c) 14.69

Question 73.
The dimensions of gravitational constant G are …
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 80
Answer:
(b) \(\mathbf{M}^{-1} \mathbf{L}^{3} \mathbf{T}^{-2}\)

Question 74.
The ratio of one nanometer to one micron is
(a) 10-3
(b) 103
(c) 10-9
(d) 10-6
Answer:
(b) 103

Question 75.
Which of the following pairs does not have same dimension?
(a) Moment of inertia and moment of force
(b) Work and torque
(c) Impulse and momentum
(d) Angular momentum and Plank’s constant
Answer:
(a) Moment of inertia and moment of force

Question 76.
Two quantities A and B have different dimensions. Which of the following is physically meaningful?
(a) A + B
(b) A – B
(c) A /B
(d) None
Answer:
(c) A /B

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 77.
The dimensional formula for moment of inertia ……
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 81
Answer:
(d) \(\mathbf{M} \mathbf{L}^{2} \mathbf{T}^{\mathbf{0}}\)

Question 78.
Which of the following is having same dimensional formula?
(a) Work and power
(b) Radius of gyration and displacement
(c) Impulse and force
(d) Frequencies and wavelength
Answer:
(b) Radius of gyration and displacement

Question 79.
Which of the following quantities is expressed as force per unit area?
(a) Pressure
(b) Stress
(c) Both (a) and (b)
(d) None
Answer:
(c) Both (a) and (b)

Question 80.
In equation of motion Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 85 the dimensional formula for K is …..
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 86
Answer:
(b) \(\left[\mathbf{L} \mathbf{T}^{-2}\right]\)

Question 81.
The dimensional formula for heat capacity ……
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 87
Answer:
(d) \(\left[\mathbf{M} \mathbf{L}^{2} \mathbf{T}^{2} \mathbf{K}^{-1}\right]\)

Question 82.
The product of Avogadro constant and elementary charge is known as …… constant.
(a) Planck’s
(b) Avagadro
(c) Boltzmann
(d) Faraday
Answer:
(d) Faraday

Question 83.
The force F is given by F = at + bt2 where t is time. The dimensions of ‘a’ and ‘b’ respectively are
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 861
Answer:
(d) \(\left[\mathrm{MLT}^{-2}\right]\) and \(\left[\mathrm{MLT}^{-0}\right]\)

Question 84.
Dimensions of impulse are …..
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 871
Answer:
(c) \(\left[\mathrm{MLT}^{-1}\right]\)

Question 85.
If speed of light (c), acceleration due to gravity (g) and pressure (P) are taken as fundamental
units, the possible relation to gravitational constant (G) is ….
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 88
Answer:
(c) \(c^{0} g^{2} p^{-1}\)

Question 86.
Equivalent of one joule is ……
(a) Nm2
(b) kg m2 s-2
(c) kg m s-1
(d) N kg m2
Answer:
(b) kg m2 s-2

Question 87.
Pick out the dimensionless quantity …..
(a) force
(b) specific gravity
(c) planck’s constant
(d) velocity
Answer:
(b) specific gravity

Question 88.
Odd one out ………
(a) strain
(b) refractive index
(c) numbers
(d) stress
Answer:
(d) stress

Question 89.
A wire has a mass 0.3 ± 0.003g, radius 0.5 ± 0.005 mm and length 6 + 0.06 cm. The maximum percentage error in the measurement of its density is …….
(a) 1%
(b) 2%
(c) 3%
(d) 4%
Answer:
(d) 4%

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 90.
The dimensions of planck constant equals to that of …..
(a) energy
(b) momentum
(c) angular momentum
(d) power
Answer:
(c) angular momentum

Question 91
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 90
(a) The unit of λ is same as that of x and A
(b) The unit of λ is same as that of x but not of A
(c) The unit of c is same as that of 2π/λ
(d) The unit of (ct – x)is same as that 2π/λ
Answer:
(a) The unit of λ is same as that of x and A

Question 92.
The number of significant figures in 0.06900 is …….
(a) 2
(b) 2
(c) 4
(d) 5
Answer:
(c) 4

Question 93.
The numbers 3.665 and 3.635 on rounding off to 3 significant figures will give
(a) 3.66 and 3.63
(b) 3.66 and 3.64
(c) 3.67 and 3.63
(d) 3.67 and 3.64
Answer:
(b) 3.66 and 3.64

Question 94.
Which of the following measurements is most precise?
(a) 4.00 mm
(b) 4.00 cm
(c) 4.00 m
(d) 4.00 km
Answer:
(a) 4.00 mm

Question 95.
The mean radius of a wire is 2 mm. Which of the following measurements is most accurate? (a) 1.9 mm
(b) 2.25 mm
(c) 2.3 mm
(d) 1.83 mm
Answer:
(a) 1.9 mm

Question 96.
If error in measurement of radius of sphere is 1%. What will be the error in measurement of volume?
(a) 1%
(b) \(\frac{1}{3}\)%
(c) 3%
(d) 10%
Answer:
(c) 3%

Question 97.
Dimensions [M L-1 T-1] are related to …….
(a) torque
(b) work
(c) energy
(d) Coefficient of viscosity
Answer:
(d) Coefficient of viscosity

Question 98.
Heat produced by a current is obtained a relation H = I2RT. If the errors in measuring these quantities current, resistance, time are 1%, 2%, 1% respectively then total error in calculating the energy produced is
(a) 2%
(b) 4%
(c) 5%
(d) 6%
Answer:
(c) 5%

Question 99.
Length cannot be measured by ….
(a) fermi
(b) angstrom
(c) parsec
(d) debye
Answer:
(d) debye

Question 100.
The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate by using the formula p = \([/late\frac{\mathrm{F}}{l^{2}}x]. If the maximum errors in the measurement
of force and length are 4% and 2% respectively, then the maximum error in the measurement of pressure is ……..
(a) 1%
(b) 2%
(c) 8%
(d) 10%
Aswer:
(c) 8%

Question 101.
Which of the following cannot be verified by using dimensional analysis?
1 mv2
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 100
Answer:
(b) y = a sin wt

Question 102.
Percentage errors in the measurement of mass and speed are 3% and 2% respectively. The error in the calculation of kinetic energy is …….
(a) 2%
(b) 3%
(c) 5%
(d) 7%
Answer:
(d) 7%

Question 103.
More number of readings will reduce …….
(a) random error
(b) systematic error
(c) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(a) random error

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 104.
If the percentage error in the measurement of mass and momentum of a body are 3% and
2%respectively, then maximum possible error in kinetic energy is
(a) 2%
(b) 3%
(c) 5%
(d) 7%
Answer:
(d) 7%

Question 105.
In a vernier caliper, n divisions of vernier scale coincides with (n – 1) divisions of main scale. The least count of the instrument is ………
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 121
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 103

Question 106.
The period of a simple pendulum is recorded as 2.56s, 2.42s, 2.71s and 2.80s respectively.
The average absolute error is
(a) 0.1s
(b) 0.2s
(c) 1.0s
(d) 0.11s
Answer:
(d) 0.11s

Question 107.
In a system of units, if force (F), acceleration (A) and time (T) are taken as fundamental units
then the dimensional formula of energy is
(a) [FA2T]
(b) [FAT2]
(c) [F2AT]
(d) [FAT]
Answer:
(b) [FAT2]

Question 108.
The random error in the arithmetic mean of 50 observations is ‘a’, then the random error in the
arithmetic mean of 200 observations a would be
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 104
Answer:
(c) [latex]\frac{a}{4}\)

Question 109.
Which of the following is not dimensionless?
(a) Relative permittivity
(b) Refractive index
(c) Relative density
(d) Relative velocity
Answer:
(d) Relative velocity

Question 110.
If V-velocity, K – kinetic energy and T – time are chosen as the fundamental units, then what is the dimensional formula for surface tension?
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 110
Answer:
(a) [K V-2 T-2]

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Short Answer Questions (1 Mark)

Question 1.
A new unit of length is chosen such that the speed of light in vaccum is unity. What is the distance between the sun and the earth in terms of the new unit if light takes 8 min and 20 s to cover this distance.
Answer:
Speed of light in vacuum, c = 1 new unit of length s-1
t = 8 min. 20 sec, = 500 s
x = ct= 1 new unit of length s-1 × 500s
x = 500 new unit of length

Question 2.
If x = a + bt + ct2, where x is in metre and t in seconds, what is the unit of c ?
Answer:
The unit of left hand side is metre so the units of ct2 should also be metre.
Since t2 has unit of s2, so the unit of c is m/s2.

Question 3.
What is the difference between mN, Nm and nm ?
Answer:
mN means milli newton, 1 mN = 10-3 N, Nm means Newton meter, nm means nano meter.

Question 4.
The radius of atom is of the order of 1 A° & radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of the atom as compared to the volume of nucleus ?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 152

Question 5.
How many kg make 1 unified atomic mass unit ?
Answer:
1u = 1.66 × 10-27 kg

Question 6.
Name some physical quantities that have same dimension.
Answer:
Work, energy and torque.

Question 7.
Name the physical quantities that have dimensional formula [ML -1T-2].
Answer:
Stress, pressure, modulus of elasticity.

Question 8.
Give two examples of dimensionless variables.
Answer:
Strain, refractive index.

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 9.
State the number of significant figures in
(i) 0.007 m2
(ii) 2.64 × 1024 kg
(iii) 0.2370 g cm-3
(iv) 0.2300m
(v) 86400
(vi) 86400 m
Answer:
(i) 1,
(ii) 3,
(iii) 4,
(iv) 4,
(v) 3,
(vi) 5 since it comes from a measurement the last two zeros become significant.

Question 10.
Given relative error in the measurement of length is 0.02, what is the percentage error ?
Answer:
2%.

Question 11.
A physical quantity P is related to four observables a, b, c and d as follows :
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 1101
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. What is the percentage error in the quantity P?
Answer:
Relative error in P is given by
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 120

Question 12.
A boy recalls the relation for relativistic mass (m) in terms of rest mass (m0) velocity of particle V, but forgets to put the constant c (velocity of light). He writes Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 125 correct the equation by putting the missing ‘c’.
Answer:
Since quantities of similar nature can only be added or subtracted, v2 cannot be subtracted from 1 but v2/c2 can be subtracted from 1.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 1212

Question 13.
Name the technique used in locating.
(a) an under water obstacle
(b) position of an aeroplane in space.
Answer:
(a) SONAR ➝ Sound Navigation and Ranging.
(b) RADAR ➝ Radio Detection and Ranging.

Question 14.
Deduce dimensional formulae of—
(i) Boltzmann’s constant
(ii) mechanical equivalent of heat.
Answer:
(i) Boltzmann Constant:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 136

Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement

Question 15.
Give examples of dimensional constants and dimensionless constants.
Answer:
Dimensional Constants : Gravitational constant, Plank’s constant. Dimensionless Constants : it, e.

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Short Answer Questions (2 Marks)

Question 16.
The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm. Calculate the minimum inaccuracy in the measurement of distance.
Answer:
Minimum inaccuracy = Vernier constant
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 137

Question 17.
If the unit of force is 100N, unit of length is 10m and unit of time is 100s. What is the unit of Mass in this system of units ?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 138

Question 18.
State the principle of homogeneity. Test the dimensional homogeneity of equations
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 139
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 140
as Dimensions of L.H.S. = Dimensions of R.H.S.
∴ The equation to dimensionally homogeneous.

(ii) Sn = Distance travelled in nth sec that is (Sn – Sn – 1)
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 141
Hence this is dimensionally correct.

Question 19.
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 142
Answer:
Since dimensionally similar quantities can only be added
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 143

Question 20.
Magnitude of force experienced by an object moving with speed v is given by F = kv2. Find dimensions of k.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 144

Question 21.
A book with printing error contains four different formulae for displacement. Choose the
correct formula/formulae
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 145
Answer:
The arguments of sine and cosine function must be dimensionless so (a) is the probable correct formulae. Since
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 146

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Numericals Questions

Question 22.
Determine the number of light years in one metre.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 240

Question 23.
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces 20.15 g and 20.17 g are added to the box.
(i) What is the total mass of the box ?
(ii) The difference in masses of the pieces to correct significant figures.
Answer:
(i) Mass of box = 2.3 kg
Mass of gold pieces = 20.15 + 20.17 = 40.32 g = 0.04032 kg.
Total mass = 2.3 + 0.04032 = 2.34032 kg
In correct significant figure mass = 2.3 kg (as least decimal)
(ii) Difference in mass of gold pieces = 0.02 g
In correct significant figure (2 significant fig. minimum decimal) will be 0.02 g.

Question 24.
5.74 g of a substance occupies 1.2 cm3. Express its density to correct significant figures.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 245
Here least significant figure is 2, so density = 4.8 g/cm3.

Question 25.
If displacement of a body s = (200 ± 5) m and time taken by it t = (20 + 0.2) s, then find the percentage error in the calculation of velocity.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 246

Question 26.
If the error in measurement of mass of a body be 3% and in the measurement of velocity be 2%. What will be maximum possible error in calculation of kinetic energy.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 247

Question 27.
The length of a rod as measured in an experiment was found to be 2.48 m, 2.46 m, 2.49 m, 2.50 m and 2.48 m. Find the average length, absolute error and percentage error. Express the result with error limit.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 248

Question 28.
A physical quantity is measured as Q = (2.1 ± 0.5) units. Calculate the percentage error in (1) Q2 (2) 2Q.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 249

Question 29.
When the planet Jupiter is at a distance of 824.7 million km from the earth, its angular diameter is measured to be 35.72″ of arc. Calculate diameter of Jupiter.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 250

Question 30.
A laser light beamed at the moon takes 2.56s and to return after reflection at the moon’s surface. What will be the radius of lunar orbit?
Answer:
t = 2.54 s
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 251

Question 31.
Convert
(i) 3 m.s-2 to km h-2
(ii) G = 6.67 × 10-11 N m2 kg-2 to cm3 g-1 s-2
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 252
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 253

Question 32.
A calorie is a unit of heat or energy and it equals 4.2 J where 1J = 1 kg m2s-2. Suppose we employ a system of units in which unit of mass is α kg, unit of length is β m, unit of time γs. What will be magnitude of calorie in terms of this new system.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 254

Question 33.
The escape velocity v of a body depends on—
(i) the acceleration due to gravity ‘g’ of the planet,
(ii) the radius R of the planet. Establish dimensionally the relation for the escape velocity.
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 255

Question 34.
The frequency of vibration of a string depends of on,
(i) tension in the string
(ii) mass per unit length of string,
(iii) vibrating length of the string. Establish dimensionally the relation for frequency.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 256
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 257

Question 35.
One mole of an ideal gas at STP occupies 22.4 L. What is the ratio of molar volume to atomic volume of a mole of hydrogen? Why is the ratio so large? Take radius of hydrogen molecule to be 1°A.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 1 Nature of Physical World and Measurement 258
This ratio is large because actual size of gas molecule is negligible in comparison to the inter molecular separation.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

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Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

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Samacheer Kalvi 11th Physics Waves Multiple Choice Questions
Question 1.
A student tunes his guitar by striking a 120 Hertz with a tuning fork, and simultaneously plays the 4th string on his guitar. By keen observation, he hears the amplitude of the combined sound oscillating thrice per second. Which of the following frequencies is the most likely the frequency of the 4th string on his guitar?
(a) 130
(b) 117
(c) 110
(d) 120
Answer:
(b) 117

Question 2.
A transverse wave moves from a medium A to a medium B. In medium A, the velocity of the transverse wave is 500 ms-1 and the wavelength is 5 m. The frequency and the wavelength of the wave in medium B when its velocity is 600 ms-1, respectively are
(a) 120 Hz and 5 m
(b) 100 Hz and 5 m
(c) 120 Hz and 6 m
(d) 100 Hz and 6 m
Answer:
(d) 100 Hz and 6 m

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 3.
For a particular tube, among six harmonic frequencies below 1000 Hz, only four harmonic frequencies are given : 300 Hz, 600 Hz, 750 Hz and 900 Hz. What are the two other frequencies missing from this list?
(a) 100 Hz, 150 Hz
(b) 150 Hz, 450 Hz
(c) 450 Hz, 700 Hz
(d) 700 Hz, 800 Hz
Answer:
(b) 150 Hz, 450 Hz
Hint:
If the tube is open at both ends so the harmonic frequencies are based on 150 Hz.
1st = 150 Hz ; 2nd = 300 Hz ; 3rd = 450 Hz ; 4th = 600 Hz ; 5th = 750 Hz ; 6th = 900 Hz
The above frequencies the missing frequency in the list 150 Hz, 450 Hz

Question 4.
Which of the following options is correct?
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1
Options for (1), (2) and (3), respectively are
(a) (B), (C) and (A)
(b) (C), (A) and (B)
(c) (A), (B) and (C)
(d) (B), (A) and (C)
Answer:
(a) (B), (C) and (A)

Question 5.
Compare the velocities of the wave forms given below, and choose the correct option.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 2
where, vA, vB, vC and vD are velocities given in (A), (B), (C) and (D), respectively.
(a) VA > VB > VD > VC
(b) VA < VB < VD < VC
(c) VA = VB = VD = VC
(d) VA > VB = VD > VC
Answer:
(c) VA = VB = VD = VC

Question 6.
A sound wave whose frequency is 5000 Hz travels in air and then hits the water surface. The ratio of its wavelengths in water and air is …….
(a) 4.30
(b) 0.23
(c) 5.30
(d) 1.23
Answer:
(a) 4.30
Hint.
Frequency of sound, f = 5000 Hz
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 5

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 7.
A person standing between two parallel hills fires a gun and hears the first echo after t1 sec and the second echo after t2 sec. The distance between the two hills is …..
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 6
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 7
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 8

Question 8.
An air column in a pipe which is closed at one end, will be in resonance with the vibrating body of frequency 83 Hz. Then the length of the air column is ………
(a) 1.5 m
(b) 0.5 mSamacheer Kalvi 11th Physics Solutions Chapter 11 Waves
(c) 1.0 m
(d) 2.0 m
Answer:
(c) 1.0 m
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9
Velocity of sound in air y = 343 ms-1
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 10

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 9.
The displacement y of a wave travelling in the x direction is given by Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 11 where x and y are measured in metres and t in second. The speed of the wave is ………
(a) 150 ms-1
(b) 300 ms-1
(c) 450 ms-1
(d) 600 ms-1
Answer:
(a) 150 ms-1
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 111

Question 10.
Consider two uniform wires vibrating simultaneously in their fundamental notes. The tensions, densities, lengths and diameter of the two wires are in the ratio 8 : 1, 1 : 2, x : y and 4 : 1 respectively. If the note of the higher pitch has a frequency of 360 Hz and the number of beats produced per second is 10, then the value of x : y is ……….
(a) 36 : 35
(b) 35 : 36
(c) 1 : 1
(d) 1 : 2
Answer:
(a) 36 : 35

Question 11.
Which of the following represents a wave?
(a) (x – vt)3
(b) x(x + vt)
(c) \(\frac{1}{(x+v t)}\)
(d) sin (x + vt)
Answer:
(d) sin (x + vt)

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 12.
A man sitting on a swing which is moving to an angle of 60° from the vertical is blowing a whistle which has a frequency of 2.0 k Hz. The whistle is 2.0 m from the fixed support point of the swing. A sound detector which detects the whistle sound is kept in front of the swing. The maximum frequency the sound detector detected is …….
(a) 2.027 kHz
(b) 1.947 kHz
(c) 9.74 kHz
(d) 1.011 kHz
Answer:
(a) 2.027 kHz

Question 13.
Let. y = \(\frac{1}{1+x^{2}}\) at t = 0s be the amplitude of the wave propagating in the positive x-direction. At t = 2s, the amplitude of the wave propagating becomes Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 13. Assume that the shape of the wave does not change during propagation. The velocity of the wave is …..
(a) 0.5 ms-1
(b) 1.0 ms-1
(c) 1.5 ms-1
(d) 2.0 ms-1
Answer:
(b) 1.0 ms-1
Hint.
The general expression y in terms of x
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 131
The shape of wave does not change, also wave move in 2 sec, 2m in positive ‘x’ direction. So, wave moves 2m in 2 sec.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 14

Question 14.
A uniform rope having mass m hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. Which of the following plots shows the correct variation of speed v with height h from the lower end?
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 15
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 16

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 15.
An organ pipe A closed at one end is allowed to vibrate in its first harmonic and another pipe B open at both ends is allowed to vibrate in its third harmonic. Both A and B are in resonance with a given tuning fork. The ratio of the length of A and B is …….
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 17
Answer:
(c) \(\frac{1}{6}\)
Hint:
(c) \(\frac{1}{6}\)

Samacheer Kalvi 11th Physics Waves Short Answer Questions

Question 1.
What is meant by waves?
Answer:
The disturbance which carries energy and momentum from one point in space to another point in space without the transfer of the medium is known as a wave.

Question 2.
Write down the types of waves.
Answer:
Waves can be classified into two types:
(a) Transverse waves
(b) Longitudinal waves

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 3.
What are transverse waves? Give one example.
Answer:
In transverse wave motion, the constituents of the medium oscillate or vibrate about their mean positions in a direction perpendicular to the direction of propagation (direction of energy transfer) of waves.
Example: light (electromagnetic waves)

Question 4.
What are longitudinal waves? Give one example.
Answer:
In longitudinal wave motion, the constituent of the medium oscillate or vibrate about their mean positions in a direction parallel to the direction of propagation (direction of energy transfer) of waves.
Example: Sound waves travelling in air.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 5.
Define wavelength.
Answer:
For transverse waves, the distance between two neighbouring crests or troughs is known as the wavelength. For longitudinal waves, the distance between two neighbouring compressions or rarefactions is known as the wavelength. The SI unit of wavelength is meter.

Question 6.
Write down the relation between frequency, wavelength and velocity of a wave.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 19

Question 7.
What is meant by interference of waves?
Answer:
Interference is a phenomenon in which two waves superimpose to form a resultant wave of greater, lower or the same amplitude.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 8.
Explain the beat phenomenon.
Answer:
When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two. sources, then their difference in frequency gives the beat frequency. Number of beats per second n = | f1 – f2| per second

Question 9.
Define intensity of sound and loudness of sound.
Answer:

  1. The loudness of sound is defined as “the degree of sensation of sound produced in the ear or the perception of sound by the listener”.
  2. The intensity of sound is defined as “the sound power transmitted per unit area taken normal to the propagation of the sound wave”.

Question 10.
Explain Doppler Effect.
Answer:
When the source and the observer are in relative motion with respect to each other and to the medium in which sound propagates, the frequency of the sound wave observed is different from the frequency of the source. This phenomenon is called Doppler Effect.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 11.
Explain red shift and blue shift in Doppler Effect.
Answer:
If the spectral lines of the star are found to shift towards red end of the spectrum (called as red shift) then the star is receding away from the Earth. Similarly, if the spectral lines of the star are found to shift towards the blue end of the spectrum (called as blue shift) then the star is approaching Earth.

Question 12.
What is meant by end correction in resonance air column apparatus?
Answer:
The antinodes are not exactly formed at the open end, we have to include a correction, called end correction e, by assuming that the antinode is formed at some small distance above the open end. Including this end correction, the first resonance is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 20
Again taking end correction into account, we have
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 21

Question 13.
Sketch the function Y = x + a. Explain your sketch
Answer:
When a = 0, y = x
when a = 1; x = 1: y = 1 + 1 = 2
when a = 2; x = 2; y = 2 + 2 = 4
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 22
Explanation: This implies, when increasing the value of a, the line shifts towards right side at a = 0, and line shifts towards left side at a = 1, 2, ….For a = vt, y = x – vt satisfies the differential equation. Though this function satisfies the differential equation, it is not finite for all values of x and t. Hence it does not represent a waves.

Question 14.
Write down the factors affecting velocity of sound in gases.
Answer:
(a) Effect of pressure
(b) Effect of temperature
(c) Effect of density
(e) Effect of wind

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 15.
What is meant by an echo? Explain.
Answer:
Echo: An echo is a repetition of sound produced by the reflection of sound waves from a wall, mountain or other obstructing surfaces.
Explanation: The speed of sound in air at 20°C is 344 m s-1. If we shout at a wall which is at 344 m away, then the sound will take 1 second to reach the wall. After reflection, the sound will take one more second to reach us. Therefore, we hear the echo after two seconds. Scientists have estimated that we can hear two sounds properly if the time gap or time interval between each sound is \(\left(\frac{1}{10}\right)^{\text {th }}\) of a second (persistence of hearing) i.e., 0.1 s. Then,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 221
2d= 344 × 0.1 = 34.4m ;d= 17.2m
The minimum distance from a sound reflecting wall to hear an echo at 20°C is 17.2 meter.

Samacheer Kalvi 11th Physics Waves Long Answer Questions

Question 1.
Discuss how ripples are formed in still water.
Answer:
Suppose we drop a stone in a trough of still water, we can see a disturbance produced at the place where the stone strikes the water surface. We find that this disturbance spreads out (diverges out) in the form of concentric circles of ever increasing radii (ripples) and strike the boundary of the trough. This is because some of the kinetic energy of the stone is transmitted to the water molecules on the surface. Actually the particles of the water (medium) themselves do not move outward with the disturbance. This can be observed by keeping a paper strip on the water surface. The strip moves up and down when the disturbance (wave) passes on the water surface. This shows that the water molecules only undergo vibratory motion about their – mean positions.

Question 2.
Briefly explain the difference between travelling waves and standing waves.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 23

Question 3.
Show that the velocity of a travelling wave produced in a string is v =\(\sqrt{\frac{\mathrm{T}}{\mu}}\)
Answer:
Velocity of transverse waves in a stretched string: Let us compute the velocity of transverse travelling waves on a string. When a jerk is given at one end (left end) of the rope, the wave pulses move towards right end with a velocity v. This means that the pulses move with a velocity v with respect to an observer who is at rest frame. Suppose an observer also moves with same velocity v in the direction of motion of the wave pulse, then that observer will notice that the wave pulse is stationary and the rope is moving with pulse with the same velocity v. Consider an elemental segment in the string. Let A and B be two points on the string at an instant of time. Let dl and dm be the length and mass of the elemental string, respectively. By definition, linear mass density, μ is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 24
The elemental string AB has a curvature which looks like an arc of a circle with centre at O, radius R and the arc subtending an angle θ at the origin O. The angle θ can be written in terms of arc length and radius as θ = \(\frac{d l}{R}\). The centripetal acceleration supplied by the tension in the string is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 25
Then, centripetal force can be obtained when mass of the string (dm) is included in equation (3)
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 26
The centripetal force experienced by elemental string can be calculated by substituting equation (2) in equation (4) we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 27
The tension T acts along the tangent of the elemental segment of the string at A and B. Since the arc length is very small, variation in the tension force can be ignored. We can resolve T into horizontal component T cos \(\left(\frac{\theta}{2}\right)\) and vertical component T sin \(\left(\frac{\theta}{2}\right)\) The horizontal component at A and B are equal in magnitude but opposite in direction; therefore, they cancel each other. Since the elemental arc length AB is taken to be very small, the vertical components at A and B appears to acts Vertical towards the centre of the arc and hence, they add up. The net radial force Fr is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 28
Since the amplitude of the wave is very small when it is compared with the length of the spring, the sine of small angle is approximated as sin \(\left(\frac{\theta}{2}\right) \approx \frac{\theta}{2}\). Hence equation (6) can be written as
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 29
Applying Newton’s second law to the elemental string in the radial direction, under equilibrium, the radial component of the force is equal to the centripetal force. Hence equating equation (5)
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 30

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 4.
Describe Newton’s formula for velocity of sound waves in air and also discuss the Laplace’s correction.
Answer:
Newton’s formula for speed of sound waves in air: Sir Isaac Newton assumed that when sound propagates in air, the formation of compression and rarefaction takes place in a very slow manner so that the process is isothermal in nature. That is, the heat produced during compression (pressure increases, volume decreases), and heat lost during rarefaction (pressure decreases, volume increases) occur over a period of time such that the temperature of the medium remains constant. Therefore, by treating the air molecules to form an ideal gas, the changes in pressure and volume obey Boyle’s law, Mathematically
PV = constant …(1)
Differentiating equation (1), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 31
where, BT is an isothermal bulk modulus of air. Substituting equation (2) in equation the speed of sound in air is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 32
Since P is the pressure of air whose value at NTP (Normal Temperature and Pressure) is 76 cm of mercury, we have
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 33
Here ρ is density of air, then the speed of sound in air at Normal Temperature and Pressure (NTP) is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 34
But the speed of sound in air at 0°C is experimentally observed as 332 m s-1 which is close upto 16% more than theoretical value (Percentage error is Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 35). This error is not small.
Laplace’s correction: In 1816, Laplace satisfactorily corrected this discrepancy by assuming that when the sound propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast. Hence the exchange of heat produced due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a bad conductor of heat. Since, temperature is no longer considered as a constant here, sound propagation is an adiabatic process. By adiabatic considerations, the gas obeys Poisson’s law (not Boyle’s law as Newton assumed), which is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 351
Since air contains mainly, nitrogen, oxygen, hydrogen etc, (diatomic gas), we take γ = 1.47.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 36

Question 5.
Write short notes on reflection of sound waves from plane and curved surfaces. Reflection of sound through the plane surface
Answer:
When the sound waves hit the plane wall, they bounce off in a manner similar to that of light. Suppose a loudspeaker is kept at an angle with respect to a wall (plane surface), then the waves coming from the source (assumed to be a point source) can be treated as spherical wave fronts (say, compressions moving like a spherical wave front). Therefore, the reflected wave front on the plane surface is also spherical, such that its centre of curvature (which lies on the other side of plane surface) can be treated as the image of the sound source (virtual or imaginary loud speaker) which can be assumed to be at a position behind the plane surface.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 37
Reflection of sound through the curved surface: The behaviour of sound is different when – it is reflected from different surfaces-convex or concave or plane. The sound reflected from a convex surface is spread out and so it is easily attenuated and weakened. Whereas, if it is reflected from the concave surface it will converge at a point and this can be easily amplified. The parabolic reflector (curved reflector) which is used to focus the sound precisely to a point is used in designing the parabolic mics which are known as high directional microphones.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 38
We know that any surface (smooth or rough) can absorb sound. For example, the sound produced in a big hall or auditorium or theatre is absorbed by the walls, ceilings, floor, seats etc. To avoid such losses, a curved sound board (concave board) is kept in front of the speaker, so that the board reflects the sound waves of the speaker towards the audience. This method will minimize the spreading of sound waves in all possible direction in that hall and also enhances the uniform distribution of sound throughout the hall. That is why a person sitting at any position in that hall can hear the sound without any disturbance.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 6.
Briefly explain the concept of superposition principle.
Answer:
Superposition Principle: When a jerk is given to a stretched string which is tied at one end, a wave pulse is produced and the pulse travels along the string. Suppose two persons holding the stretched string on either side give a jerk simultaneously, then these two wave pulses move towards each other, meet at some point and move away from each other with their original identity. Their behaviour is very different only at the crossing/meeting points; this behaviour depends on whether the two pulses have the same or different shape.
When the pulses have the same shape, at the crossing, the total displacement is the algebraic sum of their individual displacements and hence its net amplitude is higher than the amplitudes of the individual pulses. Whereas, if the two pulses have same amplitude but shapes are 180° out of phase at the crossing point, the net amplitude vanishes at that point and the pulses will recover their identities after crossing.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 39
Only waves can possess such a peculiar property and it is called superposition of waves. This means that the principle of superposition explains the net behaviour of the waves when they overlap. Generalizing to any number of waves i.e., if two are more waves in a medium move simultaneously, when they overlap, their total displacement is the vector sum of the individual displacements. We know that the waves satisfy the wave equation which is a linear second order homogeneous partial differential equation in both space coordinates and time. Hence, their linear combination (often called as linear superposition of waves) will also satisfy the same differential equation. To understand mathematically, let us consider two functions which characterize the displacement of the waves, for example,
y1 = A1 sin (kx – ωt) and y2= A2 cos (kx – ωt)
Since, both y1 and y2 satisfy the wave equation (solutions of wave equation) then their algebraic sum
y = y1 + y2
also satisfies the wave equation. This means, the displacements are additive. Suppose we multiply y1 and y2 with some constant then their amplitude is scaled by that constant Further, if C1 and C2 are used to multiply the displacements y1 and y2, respectively, then, their net displacement y is
C = C1y1 + C2y2
This can be generalized to any number of waves. In the case of n such waves in more than one dimension the displacements are written using vector notation. Here, the net displacement \(\vec{y}\) is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 42
The principle of superposition can explain the following :
(a) Space (or spatial) Interference (also known as Interference)
(b) Time (or Temporal) Interference (also known as Beats)
(c) Concept of stationary waves
Waves that obey principle of superposition are called linear waves (amplitude is much smaller than their wavelengths). In general, if the amplitude of the wave is not small then they are called non-linear waves. These violate the linear superposition principle, e.g., laser. In this chapter, we will focus our attention only on linear waves.

Question 7.
Explain how the interference of waves is formed.
Answer:
Consider two harmonic waves having identical frequencies, constant phase difference cp and same wave form (can be treated as coherent source), hut having amplitudes A1 and A2, then
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 43
Suppose they move simultaneously in a particular direction, then interference occurs (i.e., overlap of these two waves). Mathematically
y = y1 + y2 …. (3)
Therefore, substituting equation (1) and equation (3) in equation (3), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 46
By squaring and adding equation (5) and equation (6), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 461
Since, intensity is square of the amplitude (I = A2), we have
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 462
This means the resultant intensity at any point depends on the phase difference at that point.

(a) For constructive interference:
When crests of one wave overlap with crests of another wave, their amplitudes will add up and we get constructive interference. The resultant wave has a larger amplitude than the individual waves as shown in figure (a).
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 4611
The constructive interference at a point occurs if there is maximum intensity at that point, which means that
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 47
This is the phase difference in which two waves overlap to give constructive interference. Therefore, for this resultant wave,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 48
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 49

(b) For destructive interference: When the trough of one wave overlaps with the crest of another wave, their amplitudes “cancel” each other and we get destructive interference as shown in figure (b). The resultant amplitude is nearly zero. The destructive interference occurs if there is minimum intensity at that point, which means cos φ = – 1 ⇒ φ = π, 3π, 5π,… = (2n – 1) K, where n = 0, 1, 2, …. i.e. This is the phase difference in which two waves overlap to give destructive interference. Therefore,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 50
Hence, the resultant amplitude A = |A1 – A2|

Question 8.
Describe the formation of beats.
Answer:
Formation of beats: When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two sources, then their difference in frequency gives the beat frequency. Number of beats per second
n = |f1 – f2|

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 9.
What are stationary waves?. Explain the formation of stationary waves and also write down the characteristics of stationary waves.
Answer:
Explanation of stationary waves: When the wave hits the rigid boundary it bounces back to the original medium and can interfere with the original waves. A pattern is formed, which are known as standing waves or stationary waves.
Explanation: Consider two harmonic progressive waves (formed by strings) that have the same amplitude and same velocity but move in opposite directions. Then the displacement of the first wave (incident wave) is
y1 = A sin (kx – ωt) (waves move toward right) …(1)
and the displacement of the second wave (reflected wave) is
y2 = A sin (kx + ωt) (waves move toward left) …(2)
both will interfere with each other by the principle of superposition, the net displacement is
y = y1 + y2 …… (3)
Substituting equation (1) and equation (2) in equation (3), we get
y = A sin (kx – ωt) + A sin (kx + ωt) …(4)
Using trigonometric identity, we rewrite equation (4) as
y(x, t) = 2A cos (ωt) sin (kx) …(5)
This represents a stationary wave or standing wave, which means that this wave does not move either forward or backward, whereas progressive or travelling waves will move forward or backward. Further, the displacement of the particle in equation (5) can be written in more compact form,
y(x, t) = A’ cos (ωt)
where, A’ = 2A sin (foe), implying that the particular element of the string executes simple harmonic motion with amplitude equals to A’. The maximum of this amplitude occurs at positions for which
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 52
where m takes half integer or half integral values. The position of maximum amplitude is known as antinode. Expressing wave number in terms of wavelength, we can represent the anti-nodal positions as
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 53
The distance between two successive antinodes can be computed by
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 54
Similarly, the minimum of the amplitude A’ also occurs at some points in the space, and these points can be determined by setting
sin (kx) = 0 ⇒ kx = 0, π, 2π, 3π, … = nπ
where n takes integer or integral values. Note that the elements at these points do not vibrate (not move), and the points are called nodes. The nth nodal positions is given by,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 55

Characteristics of stationary waves:

  1. Stationary waves are characterised by the confinement of a wave disturbance between two rigid boundaries. This means, the wave does not move forward or backward in a medium (does not advance), it remains steady at its place. Therefore, they are called “stationary waves or standing waves”.
  2. Certain points in the region in which the wave exists have maximum amplitude, called as anti-nodes and at certain points the amplitude is minimum or zero, called as nodes.
  3. The distance between two consecutive nodes (or) anti-nodes is \(\frac{\lambda}{2}\)
  4. The distance between a node and its neighbouring anti-node is \(\frac{\lambda}{4}\)
  5. The transfer of energy along the standing wave is zero.

Question 10.
Discuss the law of transverse vibrations in stretched strings.
Answer:
Laws of transverse vibrations in stretched strings: There are three laws of transverse vibrations of stretched strings which are given as follows:
(i) The law of length: For a given wire with tension T (which is fixed) and mass per unit length µ (fixed) the frequency varies inversely with the vibrating length. Therefore,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 58
⇒ l × f = C, where C is a constant

(ii) The law of tension: For a given vibrating length l (fixed) and mass per unit length p , (fixed) the frequency varies directly with the square root of the tension T,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 59

(iii) The law of mass: For a given vibrating length l (fixed) and tension T (fixed) the frequency varies inversely with the square root of the mass per unit length µ,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 60

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 11.
Explain the concepts of fundamental frequency, harmonics and overtones in detail. Fundamental frequency and overtones in detail.
Answer:
Fundamental frequency and overtones: Let us now keep the rigid boundaries at x = 0 and x = L and produce a standing waves by wiggling the string (as in plucking strings in a guitar). Standing waves with a specific wavelength are produced. Since, the amplitude must vanish at the boundaries, therefore, the displacement at the boundary must satisfy the following conditions
x(x = 0, t) = 0 and y(x = L, t) = 0
Since, the nodes formed at a distance \(\frac{\lambda_{n}}{2}\) apart, we have \(n\left(\frac{\lambda_{n}}{2}\right)=\mathrm{L}\), where n is an integer, L is the length between the two boundaries and λn is the specific wavelength that satisfy the specified boundary conditions. Hence,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 63
Therefore, not all wavelengths are allowed. The (allowed) wavelengths should fit with the specified boundary conditions, i.e., for n = 1, the first mode of vibration has specific wavelength λ1 = 2L. Similarly for n = 2, the second mode of vibration has specific wavelength
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 64
For n = 3, the third mode of vibration has specific wavelength
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 65
The frequency of each mode of vibration (called natural frequency) can be calculated.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 66
The lowest natural frequency is called the fundamental frequency.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 67
The second natural frequency is called the first over tone.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 68
The third natural frequency is called the second over tone.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 69
Therefore, the nth natural frequency can be computed as integral (or integer ) multiple of fundamental frequency, i.e.,
fn = nf1 where n is an integer …(5)
If natural frequencies are written as integral multiple of fundamental frequencies, then the frequencies are called harmonics. Thus, the first harmonic is f1 = f1 (the fundamental frequency is called first harmonic), the second harmonic is f2 = 2f1, the third harmonic is f3 = 3f1 etc.

Question 12.
What is a sonometer? Give its construction and working. Explain how to determine the frequency of tuning fork using sonometer.
Answer:
Stationary waves in sonometer: Sono means sound related, and sonometer implies sound-related measurements.
It is a device for demonstrating the relationship between the frequency of the sound produced in the transverse standing wave in a string, and the tension, length and mass per unit length of the string. Therefore, using this device, we can determine the following quantities:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 70
(a) the frequency of the tuning fork or frequency of alternating current
(b) the tension in the string
(c) the unknown hanging mass
Construction: The sonometer is made up of a hollow box which is one meter long with a uniform metallic thin string attached to it. One end of the string is connected to a hook and the other end is connected to a weight hanger through a pulley as shown in figure. Since only one string is used, it is also known as monochord. The weights are added to the free end of the wire to increase the tension of the wire. Two adjustable wooden knives are put over the board, and their positions are adjusted to change the vibrating length of the stretched wire.
Working: A transverse stationary or standing wave is produced and hence, at the knife edges P and Q, nodes are formed. In between the knife edges, anti-nodes are formed. If the length of the vibrating element is then
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 71
Let f be the frequency of the vibrating element, T the tension of in the string and p the mass per unit length of the string. Then using equation, we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 72
Let ρ be the density of the material of the string and d be the diameter of the string. Then the mass per unit length μ
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 73

Question 13.
Write short notes on intensity and loudness.
Answer:
Intensity and loudness: Consider a source and two observers (listeners). The source emits sound waves which carry energy. The sound energy emitted by the source is same regardless of whoever measures it, i.e., it is independent of any observers standing in that region. But the sound received by the two observers may be different; this is due to some factors like sensitivity of ears, etc. To quantify such thing, we define two different quantities known as intensity and loudness of sound.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 74
Intensity of sound: When a sound wave is emitted by a source, the energy is carried to all possible surrounding points. The average sound energy emitted or transmitted per unit time or per second is called sound power.
Therefore, the intensity of sound is defined as “the sound power transmitted per unit area taken normal to the propagation of the sound wave ”.
For a particular source (fixed source), the sound intensity is inversely proportional to the square of the distance from the source.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 75
This is known as inverse square law of sound intensity.
the power output does not depend on the observer and depends on the baby. Therefore, Loudness of sound: Two sounds with same intensities need not have the same loudness. For example, the sound heard during the explosion of balloons in a silent closed room is very loud when compared to the same explosion happening in a noisy market. Though the intensity of the sound is the same, the loudness is not. If the intensity of sound is increased then loudness also increases. But additionally, not only does intensity matter, the internal and subjective experience of “how loud a sound is” i.e., the sensitivity of the listener also matters here. This is often called loudness. That is, loudness depends on both intensity of sound wave and sensitivity of the ear (It is purely observer dependent quantity which varies from person to person) whereas the intensity of sound does not depend on the observer. The loudness of sound is defined as “the degree of sensation of sound produced in the ear or the perception of sound by the listener”.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 14.
Explain how overtones are produced in a:
(a) Closed organ pipe
(b) Open organ pipe
Answer:
(a) Closed organ pipes: Clarinet is an example of a closed organ pipe. It is a pipe with one end closed and the other end open. If one end of a pipe is closed, the wave reflected at this closed end is 180° out of phase with the incoming wave. Thus there is no displacement of the particles at the closed end. Therefore, nodes are formed at the closed end and anti-nodes are formed at open end.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 76
Let us consider the simplest mode of vibration of the air column called the fundamental mode. Anti-node is formed at the
open end and node at closed end. From the figure, let L be the length of the tube and the wavelength of the wave produced. For the fundamental mode of vibration, we have,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 77
which is called the fundamental note
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 771
The frequencies higher than fundamental frequency can be produced by blowing air strongly at open end. Such frequencies are called overtones.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 78
The figure (b) shows the second mode of vibration having two nodes and two antinodes,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 79
is called first over tone, since here, the frequency is three times the fundamental frequency it is called third harmonic.
The figure (c) shows third mode of vibration having three nodes and three anti-nodes.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 80
is called second over tone, and since n = 5 here, this is called fifth harmonic. Hence, the closed organ pipe has only odd harmonics and frequency of the nth harmonic is fn = (2n + 1)f1. Therefore, the frequencies of harmonics are in the ratio
f1 : f2 : f3 : f4 …… = 1 : 3 : 5 : 7 : …… ……… (3)

(b) Open organ pipes: Flute is an example of open organ pipe. It is a pipe with both the ends open. At both open ends, anti-nodes are formed. Let us consider the simplest mode of vibration of the air column called fundamental mode. Since anti-nodes are formed at the open end, a node is formed at the mid-point of the pipe.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 81
From figure (d), if L be the length of the tube, the wavelength of the wave produced is given by
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 82
The frequency of the note emitted is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 83
which is called the fundamental note. The frequencies higher than fundamental frequency can be produced by blowing air strongly at one of the open ends. Such frequencies are called overtones.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 84
The Figure (e) shows the second mode of vibration in open pipes. It has two nodes and three anti-nodes, and therefore,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 85
is called first over tone. Since n = 2 here, it is called second harmonic.
The Figure (f) above shows the third mode of vibration having three nodes and four anti-nodes.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 86
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 87
is called second over tone. Since n = 3 here, it is called the third harmonic.
Hence, the open organ pipe has all the – harmonics and frequency of nth harmonic is fn = nf1. Therefore, the frequencies of harmonics are in the ratio
f1 : f2 : f3 : f4 … = 1 : 2 : 3 : 4 : … …(6)

Question 15.
How will you determine the velocity of sound using resonance air column apparatus?
Answer:
Resonance air column apparatus:
The resonance air column apparatus and first, second and third resonance The resonance air column apparatus is one of the simplest techniques to measure the speed of sound in air at room temperature. It consists of a cylindrical glass tube of one meter length whose one end A is open and another end B is connected to the water reservoir R through a rubber tube as shown in figure. This cylindrical glass tube is mounted on a vertical stand with a scale attached to it. The tube is partially filled with water and the water level can be adjusted by raising or lowering the water in the reservoir R. The surface of the water will act as a closed erid and other as the open end. Therefore, it behaves like a closed organ pipe, forming nodes at the surface of water and antinodes at the closed end. When a vibrating tuning fork is brought near the open end of the tube, longitudinal waves are formed inside the air column. These waves move downward as shown in Figure, and reach the surfaces of water and get reflected and produce standing waves. The length of the air column is varied by changing the water level until a loud sound is produced in the air column. At this particular length the frequency of Waves in the air column resonates with the frequency of the tuning fork (natural frequency of the tuning fork). At resonance, the frequency of sound waves produced is equal to the frequency of the tuning fork. This will occur only when the length of air column is proportional to \(\left(\frac{1}{4}\right)^{t h}\) of the wavelength of the sound waves produced. Let the first resonance occur at length L1, then
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 88
But since the antinodes are not exactly formed at the open end, we have to include a correction, called end correction e, by assuming that the antinode is formed at some small distance above the open end. Including this end correction, the first resonance is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 881
Now the length of the air column is increased to get the second resonance. Let L2 be the length at which the second resonance occurs. Again taking end correction into account, we have
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 90
In order to avoid end correction, let us take the difference of equation (3) and equation (2)
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 91
The speed of the sound in air at room temperature can be computed by using the formula
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 92
Further, to compute the end correction, we use equation (2) and equation (3), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 93

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 16.
What is meant by Doppler effect? Discuss the following cases
(1) Source in motion and Observer at rest
(a) Source moves towards observer
(b) Source moves away from the observer
(2) Observer in motion and Source at rest.
(a) Observer moves towards Source
(b) Observer resides away from the Source
(3) Both are in motion
(a) Source and Observer approach each other
(b) Source and Observer resides from each other
(c) Source chases Observer
(d) Observer chases Source
Answer:
Doppler Effect: When the source and the observer are in relative motion with respect to each other and to the medium in which sound propagates, the frequency of the sound wave observed is different from the frequency of the source. This phenomenon is called Doppler Effect.
1. Source in motion and the observer at rest
(a) Source moves towards the observer: Suppose a source S moves to the right (as shown in figure) with a velocity vs and let the frequency of the sound waves produced by the source be fs. We assume the velocity of sound in a medium is v.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 94
The compression (sound wave front) produced by the source S at three successive instants of time are shown in the figure. When S is at position x1 the compression is at C1. When S is at position x2, the compression is at C2 and similarly for x3 and C3. Assume that if reaches the observer’s position A then at that instant C2 reaches the point B and C3 reaches the point C as shown in the figure. It is obvious to see that the distance between compressions C2 and C3 is shorter than distance between C1 and C2. This means the wavelength decreases when the source S moves towards the observer O (since sound travels longitudinally and wavelength is the distance between two consecutive compressions). But frequency is inversely related to wavelength and therefore, frequency increases.
Let λ be the wavelength of the source S as measured by the observer when S is at position x1 and λ’ be wavelength of the source observed by the observer when S moves to position x2. Then the change in wavelength is ∆λ = λ – λ’ = vs t, where t is the time taken by the source to travel between x1 and x2. Therefore,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 95
On substituting equation (2) in equation (1), we get Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 96
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 961

(b) Source moves away from the observer: Since the velocity here of the source is opposite in direction when compared to case (a), therefore, changing the sign of the velocity of the source in the above case i.e, by substituting (vs ➝ – vs) in equation (1), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 962
Using binomial expansion again, we get,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 963

2. Observer in motion and source at rest:
(a) Observer moves towards Source:
Let us assume that the observer O moves towards the source S with velocity vo. The source S is at rest and the velocity of sound waves (with respect to the medium) produced by the source is v. From the figure, we observe that both vo and v are in opposite direction. Then, their relative velocity is vr = v + vo. The wavelength of the sound wave is \(\lambda=\frac{v}{f}\), which means the frequency observed by the observer O is f’ = \(\). Then
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 964
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9641

(b) Observer recedes away from the Source: If the observer O is moving away (receding away) from the source S, then velocity v0 and v moves in the same direction. Therefore, their relative velocity is vr = v – vr. Hence, the frequency observed by the observer O is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 965

3. Both are in motion:
(a) Source and observer approach each other:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9651
Let vs and vo be the respective velocities of source and observer approaching each other as shown in figure. In order to calculate the apparent frequency observed by the observer, as a simple calculation, let us have a dummy (behaving as observer or source) in between the source and observer. Since the dummy is at rest, the dummy (observer) observes the apparent frequency due to approaching source as given in equation (3) as
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 966
At that instant of time, the true observer approaches the dummy from the other side. Since the source (true source) comes in a direction opposite to true observer, the dummy (source) is treated as stationary source for the true observer at that instant. Hence, apparent frequency when the true observer approaches the stationary source (dummy source), from equation (7) is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 967
Since this is true for any arbitrary time, therefore, comparing equation (9) and equation (10), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 958

(b) Source and observer resides from each other
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 969
Here, we can derive the result as in the previous case. Instead of a detailed calculation, by inspection from figure, we notice that the velocity of the source and the observer each point in opposite directions with respect to the case in (a) and hence, we substitute (vs ➝ – vs) and (v0 ➝ – vo) in equation (11), and therefore, the apparent frequency observed by the observer when the source and observer recede from each other is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 970

(c) Source chases the observer
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 971
Only the observer’s velocity is oppositely directed when compared to case (a).
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 972

(d) Observer chases the source
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 973
Only the source velocity is oppositely directed when compared to case (a). Therefore, substituting vs ➝ – vs in equation (12), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9731

Samacheer Kalvi 11th Physics Waves Numerical Problems

Question 1.
The speed of a wave in a certain medium is 900 m/s. If 3000 waves passes over a certain point of the medium in 2 minutes, then compute its wavelength?
Answer:
Speed of the wave in medium v = 900 ms-1
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 974

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 2.
Consider a mixture of 2 mol of helium and 4 mol of oxygen. Compute the speed of sound in this gas mixture at 300 K.
Answer:
The mixture of helium and oxygen.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 975
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 976

Question 3.
A ship in a sea sends SONAR waves straight down into the seawater from the bottom of the ship. The signal reflects from the deep bottom bed rock and returns to the ship after 3.5 s. After the ship moves to 100 km it sends another signal which returns back after 2s. Calculate the depth of the sea in each case and also compute the difference in height between two cases.
Answer:
Speed of SONAR waves in water c = 1500 ms-1
Time taken to reflect from the bottom of the sea, 2t = 3.5 sec
∴ t = 1.75 sec
Distance covered in forward and reflected backward (d1) = c × t
d2 = 1500 × 1.75 = 2625 m
After ship moves in a distance = 150 km
Time taken to reflect by the waves 2t = 2s
t = 1s
Distance covered by the waves (d2) = c × t = 1500 × 1 = 1500 m
The different between the height of two cases = 2625 – 1500
hdifference = 1124 m

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 4.
A sound wave is transmitted into a tube as shown in figure. The sound wave splits into two waves at the point A which recombine at point B. Let R be the radius of the semicircle which is varied until the first minimum. Calculate the radius of the semi-circle if the wavelength of the sound is 50.0 m
Answer:
The sound travelling in the curved path distance = πR
L1 = πR
The sound travelling in the straight path distance = 2R
L2 = 2R
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 977
The path distance of straight and curved path AP = L1 – L2
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 978

Question 5.
N tuning forks are arranged in order of increasing frequency and any two successive tuning forks give n beats per second when sounded together. If the last fork gives double the frequency of the first (called as octave), Show that the frequency of the first tuning fork is f = (N – 1)n.
Answer:
Total number of fork = N
The frequency of the 1st fork = f
The frequency of the last fork = 2f
∴ an = a + (n – 1)d
2f = f + (N – 1)n
2f – f = (N – 1)n
∴ f = (N – 1)n

Question 6.
Let the source propagate a sound wave whose intensity at a point (initially) be I. Suppose we consider a case when the amplitude of the sound wave is doubled and the frequency is reduced to one-fourth. Calculate now the new intensity of sound at the same point?
Answer:
Intensity of sound wave (old) = I1
Amplitude of sound wave (A2) = 2A1
Frequency of the sound wave I2 = ?
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 979
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9781

Question 7.
Consider two organ pipes of same length in which one organ pipe is closed and another organ pipe is open. If the fundamental frequency of closed pipe is 250 Hz. Calculate the fundamental frequency of the open pipe.
Answer:
Fundamental frequency of closed organ pipe
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9791
Fundamental frequency:f open organ pipe \(f_{o}=\frac{v}{2 l}=?\)
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 980

Question 8.
A police in a siren car moving with a velocity 20 ms chases a thief who is moving in a car with a velocity v0 ms-1. The police car sounds at frequency 300 Hz, and both of them move towards a stationary siren of frequency 400 Hz. Calculate the speed in which thief is moving.
(Assume the thief does not observe any beat)
Answer:
Velocity of sound v = 330 ms-1
Velocity of car (vs ) = 20 ms-1
Frequency of car (f1) = 300 Hz
Frequency of stationary siren (f2) = 400 Hz
The speed of the thief (vo ) = ?
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 981
330 – vo = 413.325 + 1.2525vo
2.2525vo = – 83. 325
vo = – 36.99
∴ speed of the thief in moving = 36.99 ms-1

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 9.
Consider the following function:
(a) y = x2 + 2 α tx
(b) y = (x + vt)2 which among the above function can be characterized as a wave?
Answer:
(a) y = x2 + 2 α tx. This expression is not a wave equation.
(b) y = (x + vt)2 . This expression is satisfies the wave equation.

Samacheer Kalvi 11th Physics Waves Conceptual Questions

Question 1.
Why is it that transverse waves cannot be produced in a gas? Can the transverse waves can be produced in solids and liquids?
Answer:
Transverse waves travel in the form of crests and through. They involve changes in the shape of the medium. As gas has no elasticity of shape, hence transverse waves cannot be produced in it. So, they can be transmitted through media which sustain shearing stress such as solids, strings and liquid surface.

Question 2.
Why is the roar of our national animal different from the sound of a mosquito?
Answer:
Both sounds travel at the speed of sound. The speed of sound varies according to the density and temperature of the air, but not according to the loudness of the sound,at least not for the levels of loudness we talking about here.
The roaring of the lion will be audible a lot further away, but that’s simply because it’s louder.

Question 3.
A sound source and listener are both stationary and a strong wind is blowing. Is there a Doppler effect?
Answer:
Yes, It does not matter whether the sound source or the transmission media are in motion, vibrations will be compressed in the direction of convergence and dilated in the direction of divergence.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 4.
In an empty room why is it that a tone sounds louder than in the room having things like furniture etc.
Answer:
Because in a furniture room will absorb the sound waves, hence there went be any echo. But in an empty room reflect the sound. Therefore there will be echo hence we hear sound louder.

Question 5.
How do animals sense impending danger of hurricane?
Answer:
Some animals are believed to be sensitive to be low frequency sound waves emitted by hurricanes, they can also detect the slight drops in air and water pressure that signal a storm’s approach.

Question 6.
Is it possible to realize whether a vessel kept under the tap is about to fill with water?
Answer:
The frequency of the note produced by an air column is inversely proportional to its length. As the level of water is the vessel rises, the length of the air column above it decreases. It produces sound of decreasing frequency, i.e., the sound becames shorter. From the shrillness of sound, it is possible to realize whether the vessel is filled which water.
vmin = 11.71 ms-1

Samacheer Kalvi 11th Physics Waves Textual Evaluation Solved Additional Questions Solved

I. Choose the correct answer from the following:

Question 1.
Mechanical Waves
(a) are longitudinal only
(b) are transverse only
(c) can be both longitudinal and transverse.
(d) are neither longitudinal for transverse waves.
Answer:
(c) can be both longitudinal and transverse.

Question 2.
Sound whose frequency is 50 Hz?
(a) has a relatively short wavelength.
(b) has a relatively long wavelength
(c) is very loud
(d) is very intense
Answer:
(a) has a relatively short wavelength.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 3.
Sound travels fastest in …….
(a) Steel
(b) air
(c) water
(d) vaccum
Answer:
(a) steel

Question 4.
A boat at anchor is rocked by waves of velocity 25m/s, having crests 100 m apart. The boat bounches up once in every
(a) 4.0s
(b) 2500s
(c) 0.25s
(d) 75s
Answer:
(a) 4.0s
Hint:
λ = distance between crests = 100 m frequency v = \(\frac{25}{100}=\frac{1}{4} \mathrm{s}^{-1}\)
Therefore, the crests reach the boat once every 4 seconds.

Question 5.
Choose the correct statement:
(a) sound waves are transverse waves
(b) sound travels fastest through vaccum.
(c) sound travels faster in solids than in gases.
(d) sound travels faster in gases than in liquids.
Answer:
(c) sound travels faster in solids than in gases.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 6.
Transverse waves can propagate ……
(a) both in a gas and in a metal
(b) in a gas but not in a metal
(c) not in a gas but in a metal
(d) neither in a gas nor in a metal
Answer:
(a) not in a gas but in a metal.

Question 7.
The speed of the wave represented by y = A sin(ωt – kx) is ……..
(a) k/ω
(b) ω/k
(c) ωk
(d) 1/ωk
Answer:
(b) ω/k

Question 8.
The equation of a wave travelling in a string can be written as y = 3 cos {π(100t – x)} where y and x are in cm and t is in seconds. Then the value of wavelength is …….
(a) 100 cm
(b) 2 cm
(c) 50 cm
(d) 4 cm
Answer:
(b) 2 cm
Hint:
On comparing given equation with y = A cos (kx – ωt), we get
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 985

Question 9.
A wave of frequency 500 Hzhas a velocity 300 m/s. The distance between two nearest points which are 60° out of phase, is ……
(a) 0.2 m
(b) 0.1 m
(c) 0.4 m
(d) 0.5 m
Answer:
(a) 0.1 cm
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 986

Question 10.
The equation of a wave travelling on a string is Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9792, where x, y are in cm and t in seconds. The velocity of the waves is ……..
(a) 64 cm/s in – x direction
(b) 32 cm/s in – x direction
(c) 32 cm/s in +x direction
(d) 64 cm/s in + x direction
Answer:
(d) 64 cm/s in + x direction to S
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9793

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 11.
The equation of a wave is Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 9794 where y, x are in cm and t in seconds. The amplitude wavelength, velocity and frequency of the wave are, respectively, …….
(a) 4 cm, 32 cm, 16 cm/s, 0.5 Hz
(b) 8 cm, 16 cm, 32 cm/s, 1.0 Hz
(c) 4 cm, 32 cm, 32 cm/s, 0.5 Hz
(d) 8 cm, 16 cm, 16 cm/s, 1.0 Hz
Answer:
(a) 4 cm, 32 cm, 16 cm/s, 0.5 Hz

Question 12.
The diagram shows the profile of a wave, which of the following pairs of points are in phase?
(a) A, B
(b) B, C
(c) B, D
(d) B, E
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 120
Answer:
(d) B, E

Question 13.
Ultrasonic waves are those waves which ………
(a) human beings cannot hear
(b) human beings can hear
(c) have high velocity
(d) have large amplitude
Answer:
(a) human beings cannot hear

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 14.
A transverse wave of amplitude 0.5m, wavelength 1 m and frequency 2Hz is propogating in a string in the negative x direction. The equation of this wave is ….
(a) y = 0.5 sin (2πx – 4πt)
(b) y = 0.5 sin (2πx + 4πt)
(c) y = 0.5 sin (πx – 2πt)
(d) y = 0.5 cos (kx – 2πt)
Answer:
(b) y = 0.5 sin (2πx + 4πt)
Hint:
y = A sin(kx + ωt)
Here A = 0.5 m
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 121

Question 15.
With the rise of temperature, the speed of sound in a gas ……..
(a) increases
(b) decreases
(c) remain the same
(d) may increase or decrease depending on the corresponding change in pressure.
Answer:
(a) increases

Question 16.
Speed of sound in a gas in proportional to …….
(a) square root of isothermal elasticity
(b) square root of adiabatic elasticity
(c) isothermal elasticity
(d) adiabatic elasticity
Answer:
(b) square root of adiabatic elasticity

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 17.
The velocity of sound in are is not affected by change in the …….
(a) atmospheric pressure
(b) moisture content of air
(c) temperature of air
(d) composition of air
Answer:
(a) atmospheric pressure

Question 18.
A longitudinal wave is described by the equation y = y0 sin 2π (ft – x/λ). The maximum particle velocity is equal to four times the wave velocity if ……..
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 122
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 123

Question 19.
If v0 and v denote the sound velocity and the rms velocity of the molecules in a gas, then ……..
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 124
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 125
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1256

Question 20.
With the propagation of a longitudinal wave through a material medium, the quantities transferred in the direction of propagation are ……..
(a) energy, momentum and mass
(b) energy and momentum
(c) energy and mass
(d) energy
Answer:
(b) energy and momentum

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 21.
If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity will ………
(a) increase by a factor of 2
(b) decrease by a factor of 2
(c) decrease by a factor of 4
(d) remain unchanged
Answer:
(c) decrease by a factor of 4
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1241

Question 22.
When a source of sound is in motion towards a stationary observer, the effect observed is
(a) increase in the velocity of sound only
(b) decrease in the velocity of sound only
(c) increase in frequency of sound only
(d) increase in both the velocity and the frequency of sound
Answer:
(c) increase in frequency of sound only

Question 23.
The apparent wavelength of the light from a star, moving away from the earth, is 0.01% more than its real wave length. The speed of the star with respect to the earth is ……
(a) 10 km/s
(b) 15 km/s
(c) 30 km/s
(d) 60 km/s
Answer:
(c) 30 km/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1253

Question 24.
The frequency of a radar is 780 MHz. When it is reflected from an approaching aeroplane the opponent frequency is more than the actual frequency by 2.6 kHz. The speed of the aeroplane is ……
(a) 0.25 km/s
(b) 0.5 km/s
(c) 1.0 km/s
(d) 2.0 km/s
Answer:
(b) 0.5 km/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 126

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 25.
The temperature at which the speed of sound in air becomes double its value at 27°C is …….
(a) 54°C
(b) 327°C
(c) 927°C
(d) -123°C
Answer:
(c) 927°C
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 127

Question 26.
The equation of a transverse wave is given by y = 10 sin{π(0.01x – 2t)} where y and x are in cm and t is in seconds. Its frequency is …….
(a) 10 s-1
(b) 2 s-1
(c) 1 s-1
(d) 0.01 s-1
Answer:
(c) 1 s-1
Hint:
Comparing with the standard equation y = A sin(kx – ωt),
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 128

Question 27.
When sound waves travel from air to water, which of the following remains constant?
(a) velocity
(b) frequency
(c) wavelength
(d) all of these
Answer:
(b) frequency

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 28.
The speed of sound in oxygen is 332 m/s at STP. The speed of sound in hydrogen at STP will be ……..
(a) 53/2 m/s
(b) 2546 m/s
(c) 1328 m/s
(d) 664 m/s
Answer:
(c) 1328 m/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 129

Question 29.
If va, vh, and vm are the speeds of sound in air, hydrogen and a metal at the same temperature, then ……..
(a) vh > va > vm
(b) vm > vh > va
(c) vh > vm > va
(d) va > vh > vm
Answer:
(b) vm > vh > va

Question 30.
Ultrasonic waves can be detected by ……
(a) telephone
(b) Hebb’s method
(c) Kundt’s tube
(d) Quincke’s tube
Answer:
(c) Kundt’s tube

Question 31.
The velocity of sound in a gas depends on ….
(a) Wavelength only
(b) density and elasticity of gas
(c) intensity only
(d) amplitude and frequency
Answer:
(b) density and elasticity of gas

Question 32.
When sound waves travel from air to water, which of these remains constant?
(a) velocity
(b) wavelength
(c) frequency
(d) all the above
Answer:
(c) frequency

Question 33.
When a wave goes from one medium to another, there is a change in
(a) velocity
(b) amplitude
(c) wavelength
(d) all the above
Answer:
(d) all the above

Question 34.
The equation of a sound wave is y = 0.0015 sin (62.8x + 316t). Find the wave length of the above ……
(a) 0.2 units
(b) 0.3 units
(c) 0.1 units
(d) 0.15 units
Answer:
(c) 0.1 units
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 130

Question 35.
Red shift is an illustration of ……….
(a) low temperature emission
(b) high frequency absorption
(c) Doppler effect
(d) Same unknown Phenomenon.
Answer:
(c) Doppler effect

Question 36.
The ratio of the velocity of sound in a monatomic gas to that in a triatomic gas having same molar mass, under similar conditions of temperature and pressure, is ………
(a) 1.12
(6) 1.25
(c) 1.50
(d) 1.6
Answer:
(a) 1.12
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1312

Question 37.
Doppler shift in frequency does not depend upon …….
(a) the actual frequency of the wave
(b) the velocity of the source from the listener.
(c) the velocity of the source.
(d) the velocity of the observer.
Answer:
(b) the velocity of the source from the listener.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 38.
If the density of oxygen is 16 times that of hydrogen, what will be the ratio of the velocities of sound in them?
(a) 1 : 4
(b) 4 : 1
(c) 2 : 1
(d) 1 : 16
Answer:
(a) 1 : 4
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 132

Question 39.
Pitch of sound depends on ……
(a) frequency
(b) wavelength
(c) amplitude
(d) speed
Answer:
(a) frequency

Question 40.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 133
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 134
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 135

Question 41.
Which of the following equations represents a wave?
(a) y = A(ωt – kx)
(b) y = A sin ωt
(c) y = A cos kx
(d) y = A sin (at – bx + c)
Answer:
(d) y = A sin (at -bx + c)

Question 42.
A wave travels in a medium according to the equation of displacement given by y(x, t) = 0.03 sin{π(2t – 0.01 x)} where y and x are in metres and t in seconds. The wave length of the wave is …..
(a) 200 m
(b) 100 m
(c) 20 m
(d) 10 m
Answer:
(a) 200 m
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 140

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 43.
The equation of a wave moving on string is y = 8 sin{π(0.002 x – 4t)} where x, y are in centimeter and t in seconds. The velocity of the wave is ……
(a) 100 cm/s
(b) 0.2π cm/s
(c) 4π cm/s
(d) 200 cm/s
Answer:
(d) 200 cm/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 141

Question 44.
If the velocity of sound in air is 340 ms-1, a person singing a note of frequency 250 cps is producing sound waves with a wavelength of ……..
(a) 0.7
(b) 1.36 cm
(c) 1.36 m
(d) 85 km
Answer:
(c) 1.36
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 142

Question 45.
Asa transverse wave strikes against a fixed end …….
(a) its phase changes by 180°, but velocity does not change.
(b) its phase does not change, but velocity changes
(c) its velocity changes and phase too changes by 180°
(d) nothing can be predicted about changes in its velocity and phase.
Answer:
(a) its phase changes by 180°, but velocity does not change

Question 46.
A source of sounds is travelling with a velocity of 40 km/hr towards an observer and emits sound of frequency 2000 Hz. If the velocity of sound is 1220 km/hr, then what is the apparent frequency heard by the observer?
(a) 2068 Hz
(b) 2180 Hz
(c) 2000 Hz
(d) 1980 Hz
Answer:
(a) 2068 Hz
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 143

Question 47.
A vehicle with a horn of frequency n is moving with a velocity of 30m/s in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency n + n1. Then (if the sound velocity in air is 300 m/s)
(a) n1 = 10 n
(b) n1 = 0
(c) n1 = – 0.1 n
(d) n1 = 0.1 n
Answer:
(b) n1 = 0
Hint:
No Doppler effect is observed if the source moves perpendicular to the line joining the source and the observer. Therefore, the correct choice is (b).

Question 48.
The Doppler effect is applicable for ……..
(a) light waves
(b) sound waves
(c) space waves
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 49.
The speed of a wave in a medium is 760 m/s. If 3600 waves are passing through a point in the medium in 2 minutes, then its wavelength is ……
(a) 13.8 m
(b) 25.3 m
(c) 41.5 m
(d) 57.2 m
Answer:
(b) 25.3 m
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 150

Question 50.
If a sound wave travels from air to water, the quantity that remain unchanged is …….
(a) velocity
(b) wavelength
(c) frequency
(d) amplitude
Answer:
(c) frequency

Question 51.
Asa spherical wave propagates, …….
(a) the wave intensity remains constant
(b) the wave intensity decrease as the inverse of the distance from the source
(c) the wave intensity decreases as the inverse square of the distance from the source.
(d) The wave intensity decreases as the inverse cube of the distance from the source.
Answer:
(c) The wave intensity decreases as the inverse square of the distance from the source.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 52.
A source of sound and a listener are approaching each other with a speed of 40ms-1.The apparent frequency of a note produced by the source is 400 Hz. Then its true frequency is (velocity of sound in air = 360 ms-1)
(a) 320 Hz
(b) 400 Hz
(c) 360 Hz
(d) 420 Hz
Answer:
(a) 320 Hz
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 160

Question 53.
Sound waves of wavelength greater than that of audible sound are called ……
(a) infrasonic waves
(b) ultrasonic waves
(c) sonic waves
(d) seismic waves
Answer:
(a) infrasonic waves

Question 54.
The frequency of a sound wave is/and its velocity is v. If the frequency is increased to 4 f the velocity of the wave will be:
(a) v
(b) 2v
(c) 4 v
(d) v/4
Answer:
(a) v
Hint:
The velocity is a characteristic of the medium and, therefore, it remains constant.

Question 55.
Which of the following statement is untrue? The velocity of sound in a gas …….
(a) is independent of pressure
(b) increases with increase in temperature
(c) is dependent on molecular weight
(d) is greater in dry air than in moist air
Answer:
(d) is greater in dry air than in moist air

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 56.
When a stone is dropped on the surface of still water, the waves produced are …….
(a) transverse
(b) longitudinal
(c) Stationary
(d) partly longitudinal and partly transverse.
Answer:
(d) Partly longitudinal and partly transverse.

Question 57.
The equation of a wave is y = 0.1 sin (100πt – kx) where x, y are in metres and t in seconds. If – the velocity of the wave is 100 m/s, then the value of k is
(a) 1 m-1
(b) 2m-1
(c) πm-1
(d) 2πm-1
Answer:
(c) πm-1

Question 58.
A transverse wave propagating on a stretched string of linear density 3 × 10-4 kg m-1 is represented by the equation, y = 0.2 sin (1.5x + 60t)
Where x is in metres and t is in seconds. The tension in the string (in newtons) is:
(a) 0.24
(b) 0.48
(c) 1.20
(d) 1.80
Answer:
(a) 0.48
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 500

Question 59.
A transverse wave propagating along x-axis is represented by
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 170
where x is in metres and t is in seconds. The speed of the wave is ……….
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 171
Answer:
(c) 8m/s
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 172

Question 60.
Two waves represented by the following equation are travelling in the same medium: y1 = 5 sin 2π (75t – 0.25 x) and y2 = 10 sin 2π (150 – 0.25x) The intensity ratio of the two waves is ……..
(a) 1 : 2
(b) 1 : 4
(c) 1 : 8
(d) 1 : 16
Answer:
(b) 1 : 4
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 173

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 61.
A point source emits sound equally in all direction is a non-absorbing medium. Two points P and Q are at distances of 2m and 3m, respectively, from the source. The ratio of the intensities of the waves at P and Q is …….
(a) 3 : 2
(b) 4 : 9
(c) 2 : 3
(d) 9 : 4
Answer:
(d) 9 : 4
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 174

Question 62.
The waves produced by a motor boat sailing in water are ……..
(a) transverse
(b) longitudinal
(c) longitudinal and transverse
(d) stationary
Answer:
(c) longitudinal and transverse

Question 63.
Doppler effect in sound is due to ………
(a) motion of source
(b) motion of observer
(c) relative motion of source and observer
(d) none of the above
Answer:
(c) relative motion of source and observer

Question 64.
The velocity of sound in air at NTP is 330m/s. What will be its value when temperature is doubled and pressure is halved?
(a) 165 m/s
(b) 330 m/s
(c) 330 /\(\sqrt{2}\)
(d) 300/\(\sqrt{2}\) m/s
Answer:
(c) 330 /\(\sqrt{2}\)
Hint:
There is no effect of change of pressure on the velocity of sound in air. Further, v ∝ \(\sqrt{\mathrm{T}}\)

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 65.
Sound waves travel at 350 m/s through warm air and at 3500 m/s through brass. The wavelength of a 700 Hz acoustic wave as it enters brass from warm air
(a) increases by a factor 20
(b) increases by a factor 10
(c) decreases by a factor 20
(d) decreases by a factor 10
Answer:
(b) increase by a factor 10
Hint:
Since the frequency remains the same, we have
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1701

Question 66.
A train moving at a speed of 220 m/s towards a stationary object, emits a sound of frequency 1000 Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is
(a) 3000 Hz
(b) 3500 Hz
(c) 4000 Hz
(d) 5000 Hz
Answer:
(d) 5000 Hz
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1711

Question 67.
A source of sounds emitting waves of frequency 100 Hz and an observer O are located at same distance from each other The source is moving with a speed of 19.4 ms-1 at an angle of 60° with the source-observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 ms-1) is ……
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1721
(a) 97 Hz
(b) 100 Hz
(c) 103 Hz
(d) 106 Hz
Answer:
(c) 103 Hz
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1723

Question 68.
Beats occur because of
(a) interference
(b) reflection
(c) refraction
(d) Doppler effect
Answer:
(a) interference

Question 69.
A vibrating stretched string resonates with a tuning fork of frequency 512 Hz when the length of the string is 0.5 m. The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz would be ……
(a) 0.25 m
(b) 0.75 m
(c) 1.0 m
(d) 2.0 m
Answer:
(c) 1.0 m
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 600

Question 70.
A cylindrical tube, open at both ends has a fundamental frequency f in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now …….
(a) f/2
(b) f
(c) 3f/4
(d) 2f
Answer:
(a) f
Hint:
When the tube is dipped in water, it become a closed pipe of length L/2. Its fundamental frequency is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1734

Question 71.
With the increase in temperature, the frequency of the found from an organ pipe
(a) decrease
(b) increase
(c) remains unchanged
(d) changes erractically
Answer:
(b) increase
Hint:
Frequency ∝ v/L. Now v and L both increase with temperature but increase of v is much more than the increase of L which is negligible. Thus frequency increases with temperature.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 72.
Two waves of the same frequency and amplitude super impose to produce a resultant disturbance of the same amplitude. The phase difference between the waves is ……
(a) zero
(b) π/3
(c) π/4
(d) 2π/3
Answer:
(d) 2π/3
Hint:
Let the amplitude of each wave be A and phase difference between them be φ. Then,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1742

Question 73.
A sonometer wire is vibrating in the second overtone. In the wire there are ……
(a) two nodes and two antinodes
(b) one node and two antinodes
(c) four nodes and three antinodes
(d) three nodes and three antinodes
Answer:
(c) four nodes and three antinodes

Question 74.
If a resonance tube is sounded with a tuning fork of frequency 256 Hz, resonance occurs at 35 cm and 105 cm. The velocity of sound is about ……
(a) 358 m/s
(b) 512 m/s
(c) 524 m/s
(d) none of these
Answer:
(a) 358 m/s
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 175

Question 75.
A wave of frequency 100 Hz is sent along a string towards a fixed end when this wave travels back after reflection, a node is formed at a distance of 10 cm from the fixed end of the string. The speed of the incident wave is ……
(a) 40 m/s
(b) 20 m/s
(c) 10 m/s
(d) 5 m/s
Answer:
(b) 20 m/s
Hint:
The fixed end is also a node distance between two nodes = \(\frac{\lambda}{2}\) = 10 cm
or λ = 20 cm = 0.2 cm
Speed v = fλ = 100 × 0.2 = 20 m/s

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 76.
A standing wave is represented by y = A sin (100t) cos (0.01x) where y and A are in millimetres, t in seconds and x in metres. The velocity of the wave is ………
(a) 104 m/s
(b) 1 m/s
(c) 10-4 m/s
(d) not derivable from the above information
Answer:
(a) 104 m/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 176

Question 77.
Two waves of the same frequency and intensity superimpose with each other in opposite phases. Then after superposition the ……
(a) intensity increases to four times
(b) intensity increase to two times
(c) frequency increases to four times
(d) none of the above
Answer:
(d) none of the above
Hint:
Since the waves are in opposite phases, the resultant intensity will be zero. The frequency remains the same. So, the correct choice is (d).

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 78.
Two open organ pipes of lengths 50 cm and 50.5 cm produce 3 beats/s. Then the velocity of sound is …….
(a) 300 m/s
(b) 30 m/s
(c) 303 m/s
(d) 30.3 m/s
Answer:
(c) 303 m/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 1865

Question 79.
If the ratio of the amplitudes of two waves is 4 : 3, then the ratio of maximum and minimum intensities is …….
(a) 16 : 9
(b) 49 : 16
(c) 7 : 1
(d) 49 :1
Answer:
(d) 49 : 1

Question 80.
An air column in a pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency 256 Hz, if the length of the column in centimeter is (velocity of sound in air = 340 m/s)
(a) 21.25
(b) 125
(c) 62.50
(d) 33.2
Answer:
(d) 33.2
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 187

Question 81.
Two sound waves with wavelengths 5.0 cm and 5.5 cm, respectively each propagate in a gas with velocity 330 m/s. The number of beats per second will be ……..
(a) 0
(b) 1
(c) 6
(d) 12
Answer:
(c) 6
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 188

Question 82.
Two vibrating tuning forks produce progressive waves given be y1 =4 sin 500 πt and y2 = 2 sin 506 πt where t is in seconds number of beats produced per minute is ……..
(a) 60
(b) 3
(c) 369
(d) 180
Answer:
(d) 180
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 189

Question 83.
The ratio of intensities of two waves is 16 : 9. If they produce interference, then the ratio of maximum and minium intensities will be ……..
(a) 4 : 3
(b) 49 : 1
(c) 64 : 27
(d) 81 : 49
Answer:
(b) 49 : 1
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 190

Question 84.
A closes organ pipe of length 20 cm is sounded with a tuning fork in resonance. What is the frequency of the tuning fork? (v = 332 m/s)
(a) 300 Hz
(b) 350 Hz
(c) 375 Hz
(d) 415 Hz
Answer:
(d) 415 Hz So,
Hint: In resonance, the frequency of the fork is equal to the frequency of the organ pipe,
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 191

Question 85.
In a resonance tube, the first resonance is obtained at 40 cm length, using a tuning fork of frequency 450 Hz. Ignoring end correction, the velocity of sound in air is
(a) 620 m/s
(b) 720 m/s
(c) 820 m/s
(d) 1020 m/s
Answer:
(b) 720 m/s
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 192

Question 86.
If we study the vibration of a pipe open at both ends, then which of the following statement is not true?
(a) open end will be antinode
(b) odd harmonics of the fundamental frequency will be generated
(c) all harmonics of the fundamental
(d) pressure change will be maximum at both ends.
Answer:
(d) pressure change will be maximum at both ends.
Hint:
Pressure change at open ends is zero.

Question 87.
The fundamental frequency of a closes organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of the organ pipe open at both the ends is ……
(a) 80 cm
(b) 100 cm
(c) 120 cm
(d) 140 cm
Answer:
(c) 120 cm
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 200

Samacheer Kalvi 11th Physics Waves 2 Mark Questions

Question 1.
Define the term wave motion?
Answer:
Wave motion is a kind of disturbances which travels through a medium due to repeated vibrations of the particles of the medium about their mean positions, the disturbance being handed over from one particle to the next.

Question 2.
What is a progressive wave?
Answer:
A wave that travels from one point of the medium to another is called a progressive wave.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 3.
What is a plane progressive harmonic wave?
Answer:
If during the propagation of a wave through a medium the particles of the medium vibrate simple harmonically about their mean positions, than the wave is said to be plane progressive harmonic wave.

Question 4.
What do you mean by phase of a wave?
Answer:
The phase of a harmonic is a quantity that gives complete information of the wave at any time and at any position.

Question 5.
Define wave velocity or phase velocity?
Answer:
The distance covered by a wave in the direction of its propagation per unit time is called the wave velocity.

Question 6.
What are stationary waves?
Answer:
When two identical waves of same amplitude and frequency travelling in opposite directionals with the same speed along the same path superpose each other, the resultant wave does not travel in the either direction and is called stationary or standing waves.

Question 7.
What is meant by threshold of heating?
Answer:
The lowest intensity of sound that can be perceived by the human ear is called threshold of hearing. For a sound of frequency 10 kHz, the threshold of hearing is 10-12 Wm-2

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 8.
What is meant by reverberation?
Answer:
The persistence of audible sound after the source has ceased to emit sound is called reverberation.

Question 9.
What is musical scale?
Answer:
A series of notes whose fundamental frequencies have definite ratios and which produce a pleasing effect on the ear when sounded in succession constitute a musical scale.

Question 10.
Define reverberation time?
Answer:
It is defined as the time which sound takes to fall in intensity to one millionth (10-6) part of its original intensity after it was stopped.

Samacheer Kalvi 11th Physics Waves Numerical Problems

Question 1.
The fundamental frequency in an open organ pipe is equal to the 3rd harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm. What is the length of the open organ pipe.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 210

Question 2.
Two cars moving in opposite directions approach each other with speed of 22 ms-1 and 16.5 ms-1 respectively. The driver of the first car blows a horn having a frequency 400 Hz. To find the frequency heard by the driver of the second car.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 211
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 212

Question 3.
The second overtone of an open organ pipe has the same frequency as the 1st overtone of a closed pipe L metre long. Then what will be the length of the open pipe.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 213
The Length of the open pipe is two times of the length of the closed pipe.

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 4.
A steel wire 0.72 m long has a mass of 5.0 × 10-3 kg. If the wire is under a tension of 60 N. What is the speed of transverse waves on the wire?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 214

Question 5.
Estimate the speed of sound ¡n air at standard temperature and pressure by using
(i) Newton’s formula and
(ii) Laplace formula. The mass of 1 moIe of air = 29. 0 × 10-3 kg. For air, γ = 1.4
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 215
(ii) According to Newton’s formula speedof sound in air at S.T.P is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 216

Question 6.
An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. What Is the percentage increase in the apparent frequency?
Answer:
Here observer moves towards the stationary source.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 217
The percentage increase in apparent frequency
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 218

Question 7.
Tube A has both ends open, while B has on end closed otherwise the two tubes are identical. What Is the ratio of fundamental frequency of the tubes A and B?
Answer:
The fundamental frequency for tube A with both ends open is fA = \(\frac{v}{2 \mathrm{L}}\)
The fundamental frequency for tube B with one end closed is fB = \(\frac{v}{4 \mathrm{L}}\)
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 219

Question 8.
A train moves towards a stationary observer with speed 34 mIs. The train sounds a whistle and its frequency registered by the observer is f1. If the train’s speed is reduced to 17 m/s, the frequency registered f2. If the speed of sound is 340 m/s, then find the ratio f1/f2
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 220

Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

Question 9.
A police car with a siren of frequency 8 kHz ¡s moving with uniform velocity 36 km/h towards a tall building which reflect the sound waves. The speed of sound In air is 320m/s. What is the frequency of the siren heard by the car driver?
Answer:
(a) Frequency received by the building.
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 2211
The wall (source) reflect this frequency, So frequency heard by the car driver is
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 222

Question 10.
The displacement y of a wave travelling in the x-direction ¡s given by y = 10-4 sin (600t – 2x + π/3)
Where x is expressed in metres and t is seconds. What is the speed of the wave motion (in ms-1)?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves 223

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Students can Download Physics Chapter 8 Heat and Thermodynamics Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Samacheer Kalvi 11th Physics Heat and Thermodynamics Textual Evaluation Solved

Samacheer Kalvi 11th Physics Heat and Thermodynamics Multiple Choice Questions
Question 1.
In hot summer after a bath, the body’s …….
(a) internal energy decreases
(b) internal energy increases
(c) heat decreases
(d) no change in internal energy and heat
Answer:
(a) internal energy decreases

Question 2.
The graph between volume and temperature in Charles’ law is …….
(a) an ellipse
(b) a circle
(c) a straight line
(d) a parabola
Answer:
(c) a straight line

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 3.
When a cycle tyre suddenly bursts, the air inside the tyre expands. This process is …..
(a) isothermal
(b) adiabatic
(c) isobaric
(d) isochoric
Answer:
(b) adiabatic

Question 4.
An ideal gas passes from one equilibrium state (P1, V1, T1, N) to another equilibrium state
(2P1, 3V1, T2, N). Then ………
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 1
Answer:
(b) \(\mathrm{T}_{1}=\frac{\mathrm{T}_{2}}{6}\)
Solution:
From ideal gas equation, PV = NkT
One equilibrium state (P1, V1, T1, N)
Another equilibrium state (P2, V2, T2, N)
P2 = 2P1, and V2 = 3V1
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 2

Question 5.
When a uniform rod is heated, which of the following quantity of the rod will increase
(a) mass
(b) weight
(c) center of mass
(d) moment of inertia
Answer:
(d) moment of inertia

Question 6.
When food is cooked in a vessel by keeping the lid closed, after some time the steam pushes the lid outward. By considering the steam as a thermodynamic system, then in the cooking process …….
(a) Q > 0, W > 0
(b) Q < 0, W > 0
(c) Q > 0, W < 0
(d) Q < 0, W < 0 Answer: (a) Q > 0, W > 0

Question 7.
When you exercise in the morning, by considering your body as thermodynamic system, which of the following is true?
(a) ∆U > 0, W > 0
(b) ∆U < 0, W > 0
(c) ∆U < 0, W < 0
(d) ∆U = 0, W > 0
Answer:
(b) ∆U < 0, W > 0

Question 8.
A hot cup of coffee is kept on the table. After some time it attains a thermal equilibrium with the surroundings. By considering the air molecules in the room as a thermodynamic system, which of the following is true?
(a) ∆U > 0, Q = 0
(b) ∆U > 0, W < 0 (c) ∆U > 0, Q > 0
(d) ∆U = 0, Q > 0
Answer:
(c) ∆U > 0, Q > 0

Question 9.
An ideal gas is taken from (Pi, Vi) to (Pf, Vf) in three different ways. Identify the process in which the work done on the gas the most.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 6
(a) Process A
(b) Process B
(c) Process C
(d) Equal work is done in Process A, B & C
Answer:
(b) Process B

Question 10.
The V-T diagram of an ideal gas which goes through a reversible cycle A ➝ B ➝ C ➝ D is shown below. (Processes D ➝ A and B ➝ C are adiabatic)
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 7
The corresponding PV diagram for the process is (all figures are schematic)
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 8
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 9

Question 11.
A distant star emits radiation with maximum intensity at 350 nm. The temperature of the star is
(a) 8280 K
(b) 5000 K
(c) 7260 K
(d) 9044 K
Answer:
(a) 8280 K
Solution:
According to Wien’s displacement law,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 10

Question 12.
Identify the state variables given here?
(a) Q, T, W
(b) P, T, U
(c) Q, W
(d) P, T, Q
Answer:
(b) P, T, U

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 13.
In an isochoric process, we have ……
(a) W = 0
(b) Q = 0
(c) ∆U = 0
(d) ∆T = 0
Answer:
(a) W = 0

Question 14.
The efficiency of a heat engine working between the freezing point and boiling point of water is ….. [NEET 2018]
(a) 6.25%
(b) 20%
(c) 26.8%
(d) 12.5%
Answer:
(c) 26.8%
Solution:
The freezing point of water = 0°C = 273 K
Boiling point of water = 100°C = 373 K
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 11

Question 15.
An ideal refrigerator has a freezer at temperature -12°C. The coefficient of performance of the engine is 5. The temperature of the air (to which the heat ejected) is …..
(a) 50°C
(b) 45.2°C
(c) 40.2°C
(d) 37.5°C
Answer:
(c) 40.2°C
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 12

Samacheer Kalvi 11th Physics Heat and Thermodynamics Short Answer Questions

Question 1.
‘An object contains more heat’- is it a right statement? If not why?
Answer:
When heated, an object receives heat from the agency. Now object has more internal energy than before. Heat is the energy in transit and which flows from an object at higher temperature to an object lower temperature. Heat is not a quantity. So the statement I would prefer “an object contains more thermal energy”.

Question 2.
Obtain an ideal gas law from Boyle’s and Charles’ law.
Answer:
Boyle’s law: When the gas is kept at constant temperature, the pressure of the gas is inversely proportional to the volume \(\mathrm{P} \propto \frac{1}{\mathrm{V}}\)
Charles’ law: When the gas is kept at constant pressure, the volume of the gas is directly proportional to absolute temperature V ∝ T.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 3.
Define one mole.
Answer:
One mole of any substance is the amount of that substance which contains Avogadro number (NA) of particles (such as atoms or molecules).

Question 4.
Define specific heat capacity and give its unit.
Answer:
Specific heat capacity of a substance is defined as the amount of heat energy required into raise the temperature of 1 kg of a substance by 1 Kelvin or 1°C
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 17
The SI unit for specific heat capacity is J kg-1 K-1

Question 5.
Define molar specific heat capacity.
Answer:
Molar specific heat capacity is defined as heat energy required to increase the temperature of one mole of substance by IK or 1°C. Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 18

Question 6.
What is a thermal expansion?
Answer:
Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature.

Question 7.
Give the expressions for linear, area and volume thermal expansions.
Answer:
Linear Expansion: In solids, for a small change in temperature ∆T, the fractional change in length \(\left(\frac{\Delta \mathrm{L}}{\mathrm{L}_{0}}\right)\) is directly proportional to ∆T.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 20
Area Expansion: For a small change in temperature ∆T the fractional change in area \(\left(\frac{\Delta \mathrm{A}}{\mathrm{A}_{0}}\right)\) of a substance is directly proportional to ∆T and it can be written as
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 21
Volume Expansion: For a small change in temperature ∆T the fractional change in volume \(\left(\frac{\Delta \mathrm{V}}{\mathrm{V}_{0}}\right)\) of a substance is directly proportional to ∆T.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 22

Question 8.
Define latent heat capacity. Give its unit.
Answer:
Latent heat capacity of a substance is defined as the amount of heat energy required to change the state of a unit mass of the material.

Question 9.
State Stefan-Boltzmann law.
Answer:
Stefan Boltzmann law states that, the total amount of heat radiated per second per unit area of a black body is directly proportional to the fourth power of its absolute temperature.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 23

Question 10.
What is Wien’s law?
Answer:
Wien’s law states that, the wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 24

Question 11.
Define thermal conductivity. Give its unit.
Answer:
The quantity of heat transferred through a unit length of a material in a direction normal to Unit surface area due to a unit temperature difference under steady state conditions is known as thermal conductivity of a material.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 12.
What is a black body?
Answer:
A black body is one which neither reflects nor transmits but absorbs whole of the heat radiation incident on it.
The absorptive power of a perfect black body is unity.

Question 13.
What is a thermodynamic system? Give examples.
Answer:
Thermodynamic system: A thermodynamic system is a finite part of the universe. It is a collection of large number of particles (atoms and molecules) specified by certain parameters called pressure (P), Volume (V) and Temperature (T). The remaining part of the universe is called surrounding. Both are separated by a boundary.
Examples: A thermodynamic system can be liquid, solid, gas and radiation.

Question 14.
What are the different types of thermodynamic systems?
Answer:

  1. Open system can exchange both matter and energy with the environment.
  2. Closed system exchange energy but not matter with the environment.
  3. Isolated system can exchange neither energy nor matter with the environment.

Question 15.
What is meant by ‘thermal equilibrium’?
Answer:
Two systems are said to be in thermal equilibrium with each other if they are at the same temperature, which will not change with time.

Question 16.
What is mean by state variable? Give example.
Answer:
In thermodynamics, the state of a thermodynamic system is represented by a set of variables called thermodynamic variables.
Examples: Pressure, temperature, volume and internal energy etc.

Question 17.
What are intensive and extensive variables? Give examples.
Answer:
Extensive variable depends on the size or mass of the system.
Example : Volume, total mass, entropy, internal energy, heat capacity etc.
Intensive variables do not depend on the size or mass of the system.
Example: Temperature, pressure, specific heat capacity, density etc.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 18.
What is an equation of state? Give an example.
Answer:
Equation of state: The equation which connects the state variables in a specific manner is called equation of state. An ideal gas obeys the equation PV = NkT at thermodynamic equilibrium.
For example, if we push the piston of a gas container, the volume of the gas will decrease but pressure will increase or if heat is supplied to the gas, its temperature will increase, pressure and volume of the gas may also increase.

Question 19.
State Zeroth law of thermodynamics.
Answer:
The zeroth law of thermodynamics states that if two systems, A and B, are in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.

Question 20.
Define the internal energy of the system.
Answer:
The internal energy of a thermodynamic system is the sum of kinetic and potential energies of all the molecules of the system with respect to the center of mass of the system.

Question 21.
Are internal energy and heat energy the same? Explain.
Answer:
Internal energy and thermal energy do not mean the same thing, but they are related. Internal energy is the energy stored in a body. It increases when the temperature of the body rises, or when the body changes from solid to liquid or from liquid to gas.
“Heat is the energy transferred from one body to another as a result of a temperature difference.”

Question 22.
Define one calorie.
Answer:
One calorie is defined as the amount of heat energy needed to raise the temperature of one gram of water by one degree Celsius at a pressure of one atmosphere.

Question 23.
Did joule converted mechanical energy to heat energy? Explain.
Answer:
Joule essentially converted mechanical energy to internal energy. In his experiment potential energy is converted to rotational kinetic energy of paddle wheel and this rotational kinetic energy is converted to internal energy of water.

Question 24.
State the first law of thermodynamics.
Answer:
This law states that ‘Change in internal energy (∆U) of the system is equal to heat supplied to the system (Q) minus the work done by the system (W) on the surroundings’.

Question 25.
Can we measure the temperature of the object by touching it?
Answer:
No, we can’t measure the temperature of the object touching it. Because the temperature is the degree of hotness or coolness of a body. Only we can sense the hotness or coolness of the object.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 26.
Give the sign convention for Q and W.
Answer:
System gains heat Q is positive
System loses heat Q is negative
Work done on the system W is negative
Work done by the system W is positive

Question 27.
Define the quasi-static process.
Answer:
A quasi-static process is an infinitely slow process in which the system changes its variables (P, V, T) so slowly such that it remains in thermal, mechanical and chemical equilibrium with its surroundings throughout.

Question 28.
Give the expression for work done by the gas.
Answer:
When a gas expands against pressure, it does work on the surroundings. The work done in expansion for volume V1 to V2 is given by
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 25
If the pressure remains constant during expansion,
Then W = P(V2 – V1) = P∆V
If the volume remains constant, then W = 0.
If there is no external pressure, then no work is done. For example, when a gas expands freely in vaccum, no work is done by it.

Question 29.
What is PV diagram?
Answer:
PV diagram is a graph between pressure P and volume V of the system. The P-V diagram is used to calculate the amount of work done by the gas during expansion or on the gas during compression.

Question 30.
Explain why the specific heat capacity at constant pressure is greater than the specific heat capacity at constant volume.
Answer:
It implies that to increase the temperature of the gas at constant volume requires less heat than increasing the temperature of the gas at constant pressure. In other words s is always greater than sv.

Question 31.
Give the equation of state for an isothermal process.
Answer:
It is a process in which the temperature remains constant but the pressure and volume of a thermodynamic system will change. The ideal gas equation is PV = µRT

Question 32.
Give an expression for work done in an isothermal process.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 26

Question 33.
Express the change in internal energy in terms of molar specific heat capacity.
Answer:
When the gas is heated at constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU. dU = µωdT

Question 34.
Apply first law for
(a) an isothermal
(b) adiabatic
(c) isobaric processes.
Answer:
(a) For an isothermal process since temperature is constant, the internal energy is also constant. This implies that dU or ∆U = 0.
For an isothermal process, the first law of thermodynamics can be written as follows,
Q = W
(b) This is a process in which no heat flows into or out of the system (Q = 0). But the gas can expand by spending its internal energy or gas can be compressed through some external work. So the pressure, volume and temperature of the system may change in an adiabatic process. For an adiabatic process, the first law becomes ∆U = W.
(c) The first law of thermodynamics for isobaric process is given by
∆U = Q – P∆Y
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 28

Question 35.
Give the equation of state for an adiabatic process.
Answer:
The equation of state for an adiabatic process is given by
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 29
Here γ is called adibatic exponent
(γ = Cp/Cγ) which depends on the nature of the gas

Question 36.
Give an equation state for an isochoric process.
Answer:
The equation of state for an isochoric process is given by
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 30
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 31
We can infer that the pressure is directly proportional to temperature. This implies that the P-T graph for an isochoric process is a straight line passing through origin.

Question 37.
If the piston of a container is pushed fast inward. Will the ideal gas equation be valid in the intermediate stage? If not, why?
Answer:
When the piston is compressed so quickly that there is no time to exchange heat to the surrounding, the temperature of the gas increases rapidly. In this intermediate stage the ideal gas equation be not valid. Because this equation can be relates the pressure, volume and temperature of thermodynamic system at equilibrium.

Question 38.
Draw the PV diagram for:
(a) Isothermal process
(b) Adiabatic process
(c) isobaric process
(d) Isochoric process
(a) Isothermal process:
Answer:
(a) Isothermal process
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 32
(b) Adiabatic process:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 33
(c) isobaric process:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 34
(d) Isochoric process:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 35

Question 39.
What is a cyclic process?
Answer:
This is a thermodynamic process in which the thermodynamic system returns to its initial state after undergoing a series of changes. Since the system comes back to the initial state, the change in the internal energy is zero. In cyclic process, heat can flow in to system and heat flow out of the system. From the first law of thermodynamics, the net heat transferred to the system is equal to work done by the gas.
Qnet = Qin – Qout = W (for a Cyclic Process)

Question 40.
What is meant by a reversible and irreversible processes?
Answer:
Reversible processes: A thermodynamic process can be considered reversible only if it possible to retrace the path in the opposite direction in such a way that the system and surroundings pass through the same states as in the initial, direct process.
Irreversible processes: All natural processes are irreversible. Irreversible process cannot be plotted in a PV diagram, because these processes cannot have unique values of pressure, temperature at every stage of the process.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 41.
State Clausius form of the second law of thermodynamics.
Answer:
“Heat always flows from hotter object to colder object spontaneously”. This is known as the Clausius form of second law of thermodynamics.

Question 42.
State Kelvin-Planck statement of second law of thermodynamics.
Answer:
Kelvin-Planck statement: It is impossible to construct a heat engine that operates in a cycle, whose sole effect is to convert the heat completely into work. This implies that no heat engine in the universe can have 100% efficiency.

Question 43.
Define heat engine.
Answer:
Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.

Question 44.
What are processes involves in a Carnot engine?
Answer:
There are four processes involves in a carnot engine:

  1. source
  2. sink
  3. insulating stand
  4. working substance

Question 45.
Can the given heat energy be completely converted to work in a cyclic process? If not, when can the heat can completely converted to work?
Answer:
According to first law of thermodynamics work can be completely converted into heat. Since the system comes back to the initial stage, the change in the internal energy is zero. In cyclic process, heat can flow in to system and heat flow out of the system. The net heat transferred to the system is equal to work done by the gas.
Qnet = Qin = Qout = W (for 3 Cyclic Process)

Question 46.
State the second law of thermodynamics in terms of entropy.
Answer:
“For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural process should occur.

Question 47.
Why does heat flow from a hot object to a cold object?
Answer:
Because entropy increases when heat flows from hot object to cold object. If heat were to flow from a cold to a hot object, entropy will decrease leading to violation of second law thermodynamics.

Question 48.
Define the coefficient of performance.
Answer:
It is defined as the ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.

Samacheer Kalvi 11th Physics Heat and Thermodynamics Long Answer Questions

Question 1.
Explain the meaning of heat and work with suitable examples.
Answer:
Meaning of heat: When an object at higher temperature is placed in contact with another object at lower temperature, there will be a spontaneous flow of energy from the object at higher temperature to the one at lower temperature. This energy is called heat. This process of energy transfer from higher temperature object to lower temperature object is called heating. Due to flow of heat sometimes the temperature of the body will increase or sometimes it may not increase.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 351
Meaning of work: When you rub your hands against each other the temperature of the hands increases. You have done some work on your hands by rubbing. The temperature of the hands increases due to this work. Now if you place your hands on the cheek, the temperature of the cheek increases. This is because the hands are at higher temperature than the cheek. In the above example, the temperature of hands is increased due to work and temperature of the cheek is increased due to heat transfer from the hands to the chin. It is shown in the Figure. By doing work on the system, the temperature in the system will increase and sometimes may not. Like heat, work is also not a quantity and through the work energy is transferred to the system. So we cannot use the word ‘the object contains more work’ or ‘less work’.
Either the system can transfer energy to the surrounding by doing work on surrounding or the surrounding may transfer energy to the system by doing work on the system. For the transfer of energy from one body to another body through the process of work, they need not be at different temperatures.

Question 2.
Discuss the ideal gas laws.
Answer:
Boyle’s law: For a given gas at low pressure (density) kept in a container of volume V, experiments revealed the following information.
When the gas is kept at constant temperature, the pressure of the gas is inversely proportional to the volume \(\mathrm{P} \propto \frac{1}{\mathrm{V}}\)
Charles’ law: When the gas is kept at constant pressure, the volume of the gas is directly proportional to absolute temperature V ∝ T.
By combining these two equations we have
PV = CT. Here C is a positive constant.
We can infer that C is proportional to the number of particles in the gas container by considering the following argument. If we take two containers of same type of gas with same volume V, same pressure P and same temperature T, then the gas in each container obeys the above equation. PV = CT. If the two containers of gas is considered as a single system, then the pressure and temperature of this combined system will be same but volume will be twice and number of particles will also be double.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 36
For this combined system, V becomes 2V, so C should also double to match with the ideal gas equation \(\frac{P(2 V)}{T}=2 C\). It implies that C must depend on the number of particles in the gas and also should have the dimension of \(\left[\frac{\mathrm{pv}}{\mathrm{T}}\right]=\mathrm{JK}^{-1}\). So we can write the constant C as k times the number of particles N.
Here k is the Boltzmann constant (1.381 × 10-23 JK-1) and it is found to be a universal constant. So the ideal gas law can be stated as follows
PV = NkT …(1)
The equation (1) can also be expressed in terms of mole.
Suppose if a gas contains p mole of particles then the total number of particles can be written as
N = µNA …… (2)
where NA is Avogadro number (6.023 × 1023 mol-1)
Substituting for N from equation (2), the equation (1) becomes PV = µNAkT. Here NAk = R called universal gas constant and its value is 8.314 J /mol. K
So the ideal gas law can be written for µ mole of gas as
PV = μRT …(3)
This is called the equation of state for an ideal gas. It relates the pressure, volume and temperature of thermodynamic system at equilibrium.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 3.
Explain in detail the thermal expansion.
Answer:
Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature.
All three states of matter (solid, liquid and gas) expand when heated. When a solid is heated, its atoms vibrate with higher amplitude about their fixed points. The relative change in the size of solids is small. Railway tracks are. given small gaps so that in the summer, the tracks expand and do not buckle. Railroad tracks and bridges have expansion joints to allow them to expand and contract freely with temperature changes.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 37
Liquids, have less intermolecular forces than solids and hence they expand more than solids. This is the principle behind the mercury thermometers.
In the case of gas molecules, the intermolecular forces are almost negligible and hence they expand much more than solids. For example in hot air balloons when gas particles get heated, they expand and take up more space.
The increase in dimension of a body due to the increase in its temperature is called thermal expansion.
The expansion in length is called linear expansion. Similarly the expansion in area is termed as area expansion and the expansion in volume is tenned as volume expansion.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 38
Linear Expansion: In solids, for a small change in temperature ∆T, the fractional change in length \(\left(\frac{\Delta \mathrm{L}}{\mathrm{L}_{0}}\right)\) is directly proportional to ∆T
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 39
Where, αL = coefficient of linear expansion,
∆L = Change in length,
L = Original length,
∆T = Change in temperature.
Area Expansion: For a small change in temperature ∆T the fractional change in area \(\left(\frac{\Delta \mathrm{A}}{\mathrm{A}_{0}}\right)\) of a substance is directly proportional to ∆T and it can be written as
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 40
Where, αA = coefficient of area expansion.
∆A = Change in area,
A = Original area,
∆T = Change in temperature.
Volume Expansion: For a small change in temperature AT the fractional change in volume \(\left(\frac{\Delta \mathrm{V}}{\mathrm{V}_{0}}\right)\) of a substance is directly proportional to ∆T.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 412
Where, αv = coefficient of volume expansion,
∆V = Change in volume,
V = Original volume,
∆T = Change in temperature,
Unit of coefficient of linear, area and volumetric expansion of solids is °C-1 or K-1

Question 4.
Describe the anomalous expansion of water: How is it helpful in our lives?
Answer:
Anomalous expansion of water : Liquids expand on heating and contract on cooling at moderate temperatures. But water exhibits an anomalous behavior. It contracts on heating between 0°C and 4°C. The volume of the given amount of water decreases as it is cooled from room temperature, until it reach 4°C. Below 4°C the volume increases and so the density decreases. This means that the water has a maximum density at 4°C. This behavior of water is called anomalous expansion of water.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 421
In cold countries during the winter season, the surface of the lakes will be at lower temperature than the bottom as shown in the Figure. Since the solid water (ice) has lower density than its liquid form, below 4°C, the frozen water will be on the top surface above the liquid water (ice floats). This is due to the anomalous expansion of water. As the water in lakes and ponds freeze only at the top the species living in the lakes will be safe at the bottom.

Question 5.
Explain Calorimetry and derive an expression for final temperature when two thermodynamic systems are mixed.
Answer:
Calorimetry: Calorimetry means the measurement of the amount of heat released or absorbed by thermodynamic system during the heating process. When a body at higher temperature is brought in contact with another body at lower temperature, the heat lost by the hot body is equal to the heat gained by the cold body. No heat is allowed to escape to the surroundings. It can be Anomalous expansion of water in lakes mathematically expressed as
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 431
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 4312
Heat gained or lost is measured with a calorimeter. Usually the calorimeter is an insulated container of water. A sample is heated at high temperature (T1) and immersed into water at room temperature (T2) in the calorimeter. After some time both sample and water reach a final equilibrium temperature Tf. Since the calorimeter is insulated, heat given by the water.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 125
Qgain = – Qlost
Note the sign convention. The heat lost is denoted by negative sign and heat gained is denoted as positive.
From the definition of specific heat capacity
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 462
Here S1 and s2 specific heat capacity of hot sample and water respectively.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 4512

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 6.
Discuss various modes of heat transfer.
Answer:
There are three modes of heat transfer: Conduction, Convection and Radiation.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 4612
Conduction: Conduction is the process of direct transfer of heat through matter due to temperature difference. When two objects are in direct contact with one another, heat will be transferred from the hotter object to the colder one. The objects which allow heat to travel easily through them are called conductors.
Convection: Convection is the process in which heat transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.
Boiling water in a cooking pot is an example of convection. Water at the bottom of the pot receives more heat. Due to heating, the water expands and the density of water decreases at the bottom. Due to this decrease in density, molecules rise to the top. At the same time the molecules at the top receive less heat and become denser and come to the bottom of the pot. This process goes on continuously. The back and forth movement of molecules is called convection current.
To keep the room warm, we use room heater. The air molecules near the heater will heat up and expand. As they expand, the density of air molecules will decrease and rise up while the higher density cold air will come down. This circulation of air molecules is called convection current.
Radiation: When we keep our hands near the hot stove we feel the heat even though our hands are not touching the hot stove. Here heat transferred from the hot stove to our hands is in the form of radiation. We receive energy from the sun in the form of radiations. These radiations travel through vaccum and reach the Earth. It is the peculiar character of radiation which requires no medium to transfer energy from one object to another. The conduction or convection requires medium to transfer the heat.
Radiation is a form of energy transfer from one body to another by electromagnetic waves.
Example:
1. Solar energy from the Sun.
2. Radiation from hot stove.

Question 7.
Explain in detail Newton’s law of cooling.
Answer:
Newton’s law of cooling: Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 3812
The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,
T = Temperature of the object
Ts = Temperature of the surrounding
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 3912
From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.
Let us consider an object of mass m and specific heat capacity s at temperature T. Let Ts be the temperature of the surroundings. If the temperature falls by a small amount dT in time dt, then the amount of heat lost is,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 401
Dividing both sides of equation (2) by dt
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 411
Where a is some positive constant. From equation (3) and (4)
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 423
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 432
Integrating equation (5) on both sides,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 442
Where b1, is the constant of integration. Taking exponential both sides, we get
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 455

Question 8.
Explain Wien’s law and why our eyes are sensitive only to visible rays?
Answer:
Wien’s law and Vision:
The Sun is approximately taken as a black body. Since any object above 0 K will emit radiation, Sun also emits radiation. Its surface temperature is about 5700 K. By substituting this value in the equation of Wien’s law.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 465
It is the wavelength at which maximum intensity is 508 nm. Since the Sun’s temperature is around 5700 K, the spectrum of radiations emitted by Sun lie between 400 nm to 700 nm which is the visible part, of the spectrum. It is shown in Figure.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 47
The humans evolved under the Sun by receiving its radiations. The human eye is sensitive only in the visible not in infrared or X-ray ranges in the spectrum.
Suppose if humans had evolved in a planet near the star Sirius (9940K), then they would have had the ability to see the Ultraviolet rays!

Question 9.
Discuss the:
(a) thermal equilibrium
(b) mechanical equilibrium
(c) chemical equilibrium
(d) thermodynamic equilibrium.
Answer:
(a) Thermal equilibrium: When a hot cup of coffee is kept in the room, heat flows from coffee to the surrounding air. After sometime the coffee reaches the same temperature as the surrounding air and there will be no heat flow from coffee to air or air to coffee. It implies that the coffee and surrounding air are in thermal equilibrium with each other. Two systems are said to be in thermal equilibrium with each other if they arc at the same temperature, which Mechanical equilibrium will not change with time.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 48
(b) Mechanical equilibrium: Consider a gas container with piston. When some mass is placed on the piston, it will move downward due to downward gravitational force and after certain humps and jumps the piston will come to rest at a new position. When the downward gravitational force given by the piston is balanced by the upward force exerted by the gas, the system is said to be in mechanical equilibrium. A system is said to be in mechanical equilibrium if no unbalanced force acts on the thermo dynamic system or on the surrounding by thermodynamic system.
(c) Chemical equilibrium: If there is no net chemical reaction between two thermodynamic systems in contact with each other then it is said to be in chemical equilibrium.
(d) Thermodynamic equilibrium: If two systems are set to be in thermodynamic equilibrium, then the systems are at thermal, mechanical and chemical equilibrium with each other. In a state of thermodynamic equilibrium the macroscopic variables such as pressure, volume and temperature will have fixed values and do not change with time.

Question 10.
Explain Joule’s Experiment of the mechanical equivalent of heat.
Answer:
Joule’s mechanical equivalent of heat: The temperature of an object can be increased by heating it or by doing some work on it.
In the eighteenth century, James Prescott Joule showed that mechanical energy can be converted into internal energy and vice versa.
In his experiment, two masses were attached with a rope and a paddle wheel as shown in Figure. When these masses fall through a distance h due to gravity, both the masses lose potential energy equal to 2 mgh. When the masses fall, the paddle wheel turns. Due to the turning of wheel inside water, frictional force comes in between the water and the paddle wheel.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 49
This causes a rise in temperature of the water. This implies that gravitational potential energy is converted to internal energy of water. The temperature of water increases due to the work done by the masses. In fact, Joule was able to show that the mechanical work has the same effect as giving heat. He found that to raise 1 g of an object by 1°C, 4.186 J of energy is required. In earlier days the heat was measured in calorie.
1 cal = 4.186 J
This is called Joule’s mechanical equivalent of heat.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 11.
Derive the expression for the work done in a volume change in a thermodynamic system.
Answer:
Work done in volume changes: Consider a gas contained in the cylinder fitted with a movable piston. Suppose the gas is expanded quasi-statically by pushing the piston by a small distance dx. Since the expansion occurs quasi-statically the pressure, temperature and internal energy will have unique values at every instant.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 50
The small work done by the gas on the piston
dW = Fdx …(1)
The force exerted by the gas on the piston F = PA.
Here A is area of the piston and P is pressure exerted by the gas on the piston.
Equation (1) can be rewritten as Work done by the gas
dW = PA dx …(2)
But Adx = dV= change in volume during this expansion process.
So the small work done by the gas during the expansion is given by
dW = PdV ….(3)
dV is positive since the volume is increased. Here, dW is positive.
In general the work done by the gas by increasing the volume from Vi to Vf is given by
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 52
Suppose if the work is done on the system, then Vi > Vf. Then, W is negative.
Note here the pressure P is inside the integral in equation (4). It implies that while the system is doing work, the pressure need not be constant. To evaluate the integration we need to first express the pressure as a function of volume and temperature using the equation of state.

Question 12.
Derive Mayer’s relation for an ideal gas.
Answer:
Mayer’s relation: Consider p mole of an ideal gas in a container with volume V, pressure P and temperature T.
When the gas is heated at constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU.
If Cv is the molar specific heat capacity at constant volume, from equation.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 53
Suppose the gas is heated at constant pressure so that the temperature increases by dT. If ‘Q’ is the heat supplied in this process and ‘dV’ the change in volume of the gas.
Q = µCPdT …(3)
If W is the workdone by the gas in this process, then
W = P dV …(4)
But from the first law of thermodynamics,
Q = dU + W …. (5)
Substituting equations (2), (3) and (4) in (5), we get,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 54
For mole of ideal gas, the equation of state is given by
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 55
Since the pressure is constant, dP = 0
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 56
This relation is called Mayer’s relation It implies that the molar specific heat capacity of an ideal gas at constant pressure is greater than molar specific heat capacity at constant volume. The relation shows that specific heat at constant pressure (sp) is always greater than specific heat at constant volume (sv).

Question 13.
Explain in detail the isothermal process.
Answer:
Isothermal process: It is a process in which the temperature remains constant but the pressure and volume of a thermodynamic system will change. The ideal gas equation is
PV = µRT
Here, T is constant for this process So the equation of state for isothermal process is given by
PV= Constant …(1)
This implies that if the gas goes from one equilibrium state (P1, V1) to another equilibrium state (P2, V2) the following relation holds for this process
P1V1 = P2V2 …(2)
Since PV = constant, P is inversely proportional to \(v\left(P \propto \frac{1}{V}\right)\). This implies that PV graph is a hyperbola. The pressure-volume graph for constant temperature is also called isotherm. We know that for an ideal gas the internal energy is a function of temperature only. For an isothermal process since temperature is constant, the internal energy is also constant. This implies that dU or ∆U = 0.
For an isothermal process, the first law of thermodynamics can be written as,
Q = W …(3)
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 561
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 57
From equation (3), we infer that the heat supplied to a gas is used to do only external work. It is a common misconception that when there is flow of heat to the system, the temperature will increase. For isothermal process this is not true. The isothermal compression takes place when the piston of the cylinder is pushed. This will increase the internal energy which will flow out of the system through thermal contact.

Question 14.
Derive the work done in an isothermal process.
Answer:
Work done in an isothermal process: Consider an ideal gas which is allowed to expand quasi-statically at constant temperature from initial state (Pi, Vi) to the final state (Pf, Vf). We can calculate the work done by the gas during this process. The work done by the gas,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 58
As the process occurs quasi-statically, at every stage the gas is at equilibrium with the surroundings. Since it is in equilibrium at every stage the ideal gas law is valid. Writing pressure in terms of volume and temperature,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 59
Substituting equation (2) in (1) we get
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 60
In equation (3), we take uRT out of the integral, since it is constant throughout the isothermal process.
By performing the integration in equation (3), we get
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 61
As a result the work done on the gas in an isothermal compression is negative.
In the PV diagram the work done during the isothermal expansion is equal to the area under the graph.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 62
Similarly for an isothermal compression, the area under the PV graph is equal to the work done on the gas which turns out to be the area with a negative sign.

Question 15.
Explain in detail an adiabatic process.
Answer:
Adiabatic process: This is a process in which no heat flows into or out of the system (Q = 0). But the gas can expand by spending its internal energy or gas can be compressed through some external work. So the pressure, volume and temperature of the system may change in an adiabatic process.
For an adiabatic process, the first law becomes ∆U = W.
This implies that the work is done by the gas at the expense of internal energy or work is done on the system which increases its internal energy.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 63
The adiabatic process can be achieved by the following methods:
(i) Thermally insulating the system from surroundings so that no heat flows into or out of the system; for example, when thermally insulated cylinder of gas is compressed (adiabatic compression) or expanded (adiabatic expansion) as shown in the Figure.
(ii) If the process occurs so quickly that there is no time to exchange heat with surroundings even though there is no thermal insulation. A few examples are shown in Figure.
The equation of state for an adiabatic process is given by
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 64
Here γ is called adiabatic exponent (γ = Cp/Cv) which depends on the nature of the gas. The equation (1) implies that if the gas goes from an equilibrium state (Pi, Vi) to another equilibrium state (Pf, Vf) adiabatically then it satisfies the relation
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 65
The PV diagram for an adiabatic process is also called adiabat. But actually the adiabatic curve is steeper than isothermal curve.
We can also rewrite the equation (1) in terms of T and V. From ideal gas equation, the pressure P = \(\frac{\mu \mathrm{RT}}{\mathrm{V}}\). Substituting this equation in the equation (1), we have
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 66
Note here that is another constant. So it can be written as
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 67
The equation implies that if the gas goes from an initial equilibrium state (Ti, Vi) to final equilibrium state (Tf, Vf) adiabatically then it satisfies the relation
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 68
The equation of state for adiabatic process can also be written in terms of T and P as
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 69
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 70

Question 16.
Derive the work done in an adiabatic process.
Answer:
Work done in an adiabatic process: Consider µ moles of an ideal gas enclosed in a cylinder having perfectly non conducting walls and base. A frictionless and insulating piston of cross sectional area A is fitted in the cylinder.
Let W be the work done when the system goes from the initial state (Pi, Vi, Ti) to the final state (Pf, Vf, Tf) adiabatically.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 71
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 711
By assuming that the adiabatic process occurs quasi-statically, at every stage the ideal gas law is valid. Under this condition, the adiabatic equation of state is PVγ = constant (or)
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 72
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 73
In adiabatic expansion, work is done by the gas. i.e., Wadia is positive. As Ti > Tf the gas
cools during adiabatic expansion.
In adiabatic compression, work is done on the gas. i.e., Wadia is negative. As Ti< Tf the
temperature of the gas increases during adiabatic compression.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 731
To differentiate between isothermal and adiabatic curves in PV diagram, the adiabatic curve is drawn along with isothermal curve for Tf and Ti. Note that adiabatic curve is steeper than isothermal curve. This is because γ > 1 always.

Question 17.
Explain the isobaric process and derive the work done in this process.
Isobaric process: This is a thermodynamic process that occurs at constant pressure. Even though pressure is constant in this process, temperature, volume and internal energy are not constant. From the ideal gas equation, we have
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 75
In an isobaric process the temperature is directly proportional to volume.
V ∝ T (Isobaric process) …. (2)
This implies that for a isobaric process, the V-T graph is a straight line passing through the origin.
If a gas goes from a state (Vi, Ti) to (Vf, Tf) at constant pressure, then the system satisfies the following equation
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 76
Examples for Isobaric process:
(i) When the gas is heated and pushes the piston so that it exerts a force equivalent to atmospheric pressure plus the force due to gravity then this process is isobaric.
(ii) Most of the cooking processes in our kitchen are isobaric processes. When the food is cooked in an open vessel, the pressure above the food is always at atmospheric pressure. The PV diagram for an isobaric process is a horizontal line parallel to volume axis.
Figure (a) represents isobaric process where volume decreases figure
(b) represents isobaric process where volume increases.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 77
The work done in an isobaric process: Work done by the gas
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 78
In an isobaric process, the pressure is constant, so P comes out of the integral,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 79
Where ∆V denotes change in the volume. If ∆V is negative, W is also negative. This implies that the work is done on the gas. If ∆V is positive, W is also positive, implying that work is done by the gas equation.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 80
The equation (6) can also be rewritten using the ideal gas equation.
From ideal gas equation
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 81
Substituting this in equation (6) we get
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 82.
In the PV diagram, area under the isobaric curve is equal to the work done in isobaric process. The shaded area in the above diagram is equal to the work done by the gas.
The first law of thermodynamics for isobaric process is given by
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 83

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 18.
Explain in detail the isochoric process.
Answer:
Isochoric process: This is a thermodynamic process in which the volume of the system is kept constant. But pressure, temperature and internal energy continue to be variables.
The pressure – volume graph for an isochoric process is a vertical line parallel to pressure axis as shown in Figure.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 84
The equation of state for an isochoric process is given by
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 85
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 851
We can infer that the pressure is directly proportional to temperature. This implies that the P-T graph for an isochoric process is a straight line passing through origin. If a gas goes from state (Pi, Ti) to (Pf, Tf) at constant volume, then the system satisfies the following equation
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 86

For an isochoric processes, ∆V = 0 and W = 0. Then the first law becomes
∆U = Q …(3)
Implying that the heat supplied is used to increase only the internal energy. As a result the temperature increases and pressure also increases.
Suppose a system loses heat to the surroundings through conducting walls by keeping the volume constant, then its internal energy decreases. As a result the temperature decreases; the pressure also decreases.

Question 19.
What are the limitations of the first law of thermodynamics?
Answer:
Limitations of first law of thermodynamics: The first law of thermodynamics explains well the inter convertibility of heat and work. But it does not indicate the direction of change.
For example,
(a) When a hot object is in contact with a cold object, heat always flows from the hot object to cold object but not in the reverse direction. According to first law, it is possible for the energy to flow from hot object to cold object or from cold object to hot object. But in nature the direction of heat flow is always from higher temperature to lower temperature.
(b) When brakes are applied, a car stops due to friction and the work done against friction is converted into heat. But this heat is not reconverted to the kinetic energy of the car. So the first law is not sufficient to explain many of natural phenomena.

Question 20.
Explain the heat engine and obtain its efficiency.
Answer:
Heat Engine: In the modem technological world, the role of automobile engines plays a vital role in for transportation. In motor bikes and cars there are engines which take in petrol or diesel as input, and do work by rotating wheels. Most of these automobile engines have efficiency not greater than 40%. The second law of thermodynamics puts a fundamental restriction on efficiency of engines. Therefore understanding heat engines is very important.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 700
Reservoir: It is defined as a thermodynamic system which has very large heat capacity. By taking in heat from reservoir or giving heat to reservoir, the reservoir’s temperature does not change.
Example: Pouring a tumbler of hot water in to lake will not increase the temperature of the lake. Here the lake can be treated as a reservoir.
When a hot cup of coffee attains equilibrium with the open atmosphere, the temperature of the atmosphere will not appreciably change. The atmosphere can be taken as a reservoir.
We can define heat engine as follows: Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.
A heat engine has three parts:
(a) Hot reservoir
(b) Working substance
(c) Cold reservoir A Schematic diagram for heat engine is given below:
1. Hot reservoir (or) Source: It supplies heat to the engine. It is always maintained at a high temperature TH.
2. Working substance: It is a substance like gas or water, which converts the heat supplied into Work.
3. Cold reservoir (or) Sink: The heat engine ejects some amount of heat (QL) in to cold reservoir after it doing work. It is always maintained at a low temperature TL.
The heat engine works in a cyclic process. After a cyclic process it returns to the same state. Since the heat engine returns to the same state after it ejects heat, the change in the internal energy of the heat engine is zero.
The efficiency of the heat engine is defined as the ratio of the work done (out put) to the heat absorbed (input) in one cyclic process.
Let the working substance absorb heat QH units from the source and reject QL units to the sink after doing work W units.
We can write. Input heat = Work done + ejected heat
QH = W + QL
W = QH – QL
Then the efficiency of heat engine
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 200
Note here that QH, QL and W all are taken as positive, a sign convention followed in this expression.
Since QL < QH, the efficiency (η) always less than 1. This implies that heat absorbed is not completely converted into work. The second law of thermodynamics placed fundamental restrictions on converting heat completely into work.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 21.
Explain in detail Carnot heat engine.
Answer:
In the year 1824 a young French engineer Sadi Carnot proved that a certain reversible engine operated in cycle between hot and cold reservoir can have maximum efficiency. This engine is called Carnot engine.
A reversible heat engine operating in a cycle between two temperatures in a particular way is called a Carnot Engine. The carnot engine has four parts which are given below.
(i) Source: It is the source of heat maintained at constant high temperature TH. Any amount of heat can be extracted from it, without changing its temperature.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 201
(ii) Sink: It is a cold body maintained at a constant low temperature TL. It can absorb any amount of heat.
(iii) Insulating stand: It is made of perfectly non-conducting material. Heat is not conducted through this stand.
(iv) Working substance: It is an ideal gas enclosed in a cylinder with perfectly non-conducting walls and perfectly conducting bottom. A non-conducting and frictionless piston is fitted in it.

Question 22.
Derive the expression for Carnot engine efficiency.
Answer:
Efficiency of a Carnot engine: Efficiency is defined as the ratio of work done by the working substance in one cycle to the amount of heat extracted from the source.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 202
Applying isothermal conditions, we get,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 203
Here we omit the negative sign. Since we are interested in only the amount of heat (QL) ejected into the sink, we have
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 204
Substituting equation (5) in (4), we get
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 205
Note : TL and TH should be expressed in Kelvin scale.
Important results:
1. η is always less than 1 because TL is less than TH. This implies the efficiency cannot be 100%.
2. The efficiency of the Carnot’s engine is independent of the working substance. It depends only on the temperatures of the source and the sink. The greater the difference between the two temperatures, higher the efficiency.
3. When TH = TL the efficiency η = 0. No engine can work having source and sink at the same temperature.

Question 23.
Explain the second law of thermodynamics in terms of entropy.
Answer:
The quantity \(\frac{Q}{T}\) is called entropy. It is a very important thermodynamic property of a system.

It is also a state variable. \(\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}\) is the entropy received by the Carnot engine from hot reservoir is entropy given out by the Carnot engine to the cold reservoir. For reversible engines (Carnot Engine) both entropies should be same, so that the change in entropy of the Carnot engine in one cycle is zero. But for all practical engines 1 ike diesel and petrol engines which are not reversible engines, they satisfy the relation \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}>\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}\)
In fact we can reformulate the second law of thermodynamics as follows “For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural process should occur.
Entropy increases when heat flows from hot object to cold object. If heat were to flow from a cold to a hot object, entropy will decrease leading to violation of second law thermodynamics. Entropy is also called ‘measure of disorder’. All natural process occur such that the disorder should always increases.
Consider a bottle with a gas inside. When the gas molecules are inside the bottle it has less disorder. Once it spreads into the entire room it leads to more disorder. In other words when the gas is inside the bottle the entropy is less and once the gas spreads into entire room, the entropy increases. From the second law of thermodynamics, entropy always increases. If the air molecules go back in to the bottle, the entropy should decrease, which is not allowed by the second law of thermodynamics. The same explanation applies to a drop of ink diffusing into water. Once the drop of ink spreads, its entropy is increased. The diffused ink can never become a drop again. So the natural processes occur in such a way that entropy should increase for all irreversible process.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 24.
Explain in detail the working of a refrigerator.
Answer:
Refrigerator: A refrigerator is a Carnot’s engine working in the reverse order.
Working Principle: The working substance (gas) absorbs a quantity of heat QL from the cold body (sink) at a lower temperature TL. A certain amount of work W is done on the working substance by the compressor and a quantity of heat QH is rejected to the hot body (source) ie, the atmosphere at TH. When you stand beneath of refrigerator, you can feel warmth air. From the first law of thermodynamics, we have
QL + W = QH ….. (1)
As a result the cold reservoir (refrigerator) further cools down and the surroundings (kitchen or atmosphere) gets hotter.
Coefficient of performance (COP) (β): COP is a measure of the efficiency of a refrigerator. It is defined as the ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 220
Inferences.
1. The greater the COP, the better is the condition of the refrigerator. A typical refrigerator has COP around 5 to 6.
2. Lesser the difference in the temperatures of the cooling chamber and the atmosphere, higher is the COP of a refrigerator.
3. In the refrigerator the heat is taken from cold object to hot object by doing external work. Without external work heat cannot flow from cold object to hot object. It is not a violation of second law of thermodynamics, because the heat is ejected to surrounding air and total entropy of (refrigerator + surrounding) is always increased.

Samacheer Kalvi 11th Physics Heat and Thermodynamics Numerical Problems

Question 1.
Calculate the number of moles of air is in the inflated balloon at room temperature as shown in the figure.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 221
The radius of the balloon is 10 cm, and pressure inside the balloon is 180 kPa
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 2211

Question 2.
In the planet Mars, the average temperature is around – 53°C and atmospheric pressure is 0.9 kPa. Calculate the number of moles of the molecules in unit volume in the planet Mars? Is this greater than that in earth?
Answer:
T = – 53°C = 220 K
P = 0.9 × 103 Pa
V = 1 m3
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 222

Question 3.
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V2 and contains ideal gas at pressure P2 and temperature T2. If the partition is removed without doing any work on the gases, calculate the final equilibrium temperature of the container.
Answer:
Let T be the equilibrium temperature and let n1 and n2 be the number of moles in vessels 1 and 2 respectively. As there is no loss of energy,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 223
Substituting n1 and n2 values and solving, we get
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 225

Question 4.
The temperature of a uniform rod of length L having a coefficient of linear expansion αL is changed by ∆T. Calculate the new moment of inertia of the uniform rod about axis passing through its center and perpendicular to an axis of the rod.
Answer:
Moment of inertia of a uniform rod of mass and length l about its perpendicular bisector. Moment of inertia of the rod
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 2251
Increase in length of the rod when temperature is increased by ∆T, is given by
L’ = L(1 + αL∆T)
New moment of inertia of the rod
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 226

Question 5.
Draw the TP diagram (P – x axis, T – y axis), VT(T – x axis, V – y axis) diagram for
(a) Isochoric process
(b) Isothermal process
(c) Isobaric process
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 227
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 228
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 2289

Question 6.
A man starts bicycling in the morning at a temperature around 25°C, he checked the pressure of tire which is equal to be 500 kPa. Afternoon he found that the absolute pressure in the tyre is increased to 520 kPa. By assuming the expansion of tyre is negligible, what is the temperature of tyre at afternoon?
Answer:
For ideal gas equation of state
PV = nRT
P1 = 500 kPa, T1 = 25°C = 25 + 273 = 298K, P2 = 520 kPa, T2 = ?
Expansion of tyre is negligible (Vconstant)
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 229

Question 7.
Normal human body of the temperature is 98.6°F. During high fever if the temperature increases to 104°F, what is the change in peak wavelength that emitted by our body? (Assume human body is a black body).
Answer:
Normal human body temperature (T) = 98.6°F
Convert Fahrenheit into Kelvin,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 230
So, T = 98.6° F = 310 K
From Wien’s displacement law
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 231
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 232

Question 8.
In an adiabatic expansion of the air, the volume is increased by 4%, what is percentage change in pressure? (For air γ = 1.4)
Answer:
From equation for adiabatic process,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 233

Question 9.
In a petrol engine, (internal combustion engine) air at atmospheric pressure and temperature of 20°C is compressed in the cylinder by the piston to 1/8 of its original volume. Calculate the temperature of the compressed air. (For air γ = 1.4)
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 234

Question 10.
Consider the following cyclic process consist of isotherm, isochoric and isobar which is given in the figure.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 235
Draw the same cyclic process qualitatively in the V-T diagram where T is taken along x – direction and V is taken along y – direction. Analyze the nature of heat exchange in each process.
Answer:
Process 1 to 2 = increase in volume. So heat must be added.
Process 2 to 3 = Volume remains constant. Increase in temperature.
The given heat is used to increase the internal energy.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 236
Process 3 to 1 : Pressure remains constant. Volume and Temperature are reduced. Heat flows out of the system.
It is an isobaric compression where the work is done on the system.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 11.
An ideal gas is taken in a cyclic process as shown in the figure.
Calculate
(a) work done by the gas.
(b) work done on the gas
(c) Net work done in the process
Answer:
(a) Work done by the gas (along AB)
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 237
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 2371
(c) Net work done in the process = Area under the curve AB
= Rectangle area + triangle area
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 238

Question 12.
For a given ideal gas 6 × 105J heat energy is supplied and the volume of gas is increased from 4 m3 to 6 m3 at atmospheric pressure. Calculate
(a) the work done by the gas
(b) change ¡n internal energy of the gas
(c) graph this process ¡n PV and TV diagram.
Answer:
Heat energy supplied to gas Q = 6 × 105J
Change in volume ∆V = (6 – 4) = 2 m3
1 atm = 1.0 13 × 105 Nm-2
(a) Work done by the gas W = P × ∆V = 1.013 × 105 × 2 = 2.026 × 105
W = 202.6kJ1
(b) Change in internal energy of the gas
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 239
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 240

Question 13.
Suppose a person wants to increase the efficiency of the reversible heat engine that is operating between 100°C and 300°C. He had two ways to increase the efficiency,
(a) By decreasing the cold reservoir temperature from 100°C to 50°C and keeping the hot reservoir temperature constant
(b) by increasing the temperature of the hot reservoir from 300°C to 350°C by keeping the cold reservoir temperature constant. Which is the suitable method?
Answer:
Heat engine operates at initial temperature = 100°C + 273 = 373 K
Final temperature = 300°C + 273 = 573 K
At melting point = 273 K
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 241
(a) By decreasing the cold reservoir, efficiency
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 242
(b) By increasing the temperature of hot reservoir, efficiency,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 243
Method (a) More efficiency than method (b).

Question 14.
A Carnot engine whose efficiency is 45% takes heat from a source maintained at a temperature of 327°C. To have an engine of efficiency 60% what must be the intake temperature for the same exhaust (sink) temperature?
Answer:
Efficiency of Carnot engine (η1) = 45% = 0.45
Initial intake temperature (T1) = 327°C = 600 K
New efficiency (η2) = 60% = 0.6
Efficiency of Carnot engine is given by
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 244
T1 is temperature of source ; T2 is temperature of sink
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 4613

Question 15.
An ideal refrigerator keeps its content at 0°C while the room temperature is 27°C. Calculate its coefficient of performance.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 261

Samacheer Kalvi 11th Physics Heat and Thermodynamics Textual Evaluation Solved Additional Questions Solved

I. Choose the correct answer from the following:

Question 1.
The coefficient of volume expansion of a solid is x times the coefficient of superficial expansion. Then x is
(a) 1.5
(b) 2
(c) 2.5
(d) 3
Answer:
(a) 1.5
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 27

Question 2.
A solid metal ball has a spherical cavity. If the ball is heated, the volume of the cavity will
(a) increase
(b) decrease
(c) remain unaffected
(d) remain unaffected but the shape of the cavity will change.
Answer:
(a) increase

Question 3.
A metal sheet with a circular hole is heated. The hole will
(a) contract
(b) expand
(c) remain unaffected
(d) contract or expand depending on the value of the linear expansion coefficient.
Answer:
(b) expand

Question 4.
The length of a metal rod at 0°C is 0.5m. When it is heated, its length increases by 2.7 mm. The final temperature of the rod is (coefficient of linear expansion of the metal = 90 × 106/°C) ……
(a) 20°C
(b) 30°C
(c) 40°C
(d) 60°C
Answer:
(d) 60°C
Solution:
lt = l0(1 + αt)
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 281
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 282

Question 5.
A bimetal made of copper and iron strips welded together is straight at room temperature. It is held vertically with iron strip towards left and copper strip towards right. If this bimetal is heated, it will
(a) remain straight
(b) bend towards right
(c) bend towards left
(d) bend forward
Answer:
(c) bend towards left
Solution:
Since αcopper > αiron, the bimetal will bend towards iron, i.e., towards left.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 6.
When water is heated from 0°C to 10°C, its volume …..
(a) decreases
(b) increases
(c) first increase and then decrease
(d) first decreases and then increases.
Answer:
(d) first decreases and then increases.

Question 7.
A block of wood is floating on water at 0°C with a certain volume V above water level. The temperature of water is slowly raised to 20°C. How does the volume V change with the rise of temperature?
(a) remain unchanged
(b) decrease continuously
(c) decrease till 4°C and then increase
(d) increase till 4°C and then decrease.
Answer:
(a) increase till 4°C and then decrease.
Solution:
The density of water increases from 0° to 4°C and then decreases. Therefore, the volume V of the block above water level will increase till 4°C and then decrease.

Question 8.
An iron tyre is to be fitted on a wooden wheel 0.1m in diameter. The diameter of the tyre is 6 mm smaller than that of the wheel. The tyre should be heated by a temperature of (coefficient of volume expansion of iron is 3.6 × 10-5/°C)
(a) 167°C
(b) 334°C
(c) 500°C
(d) 1000°C
Answer:
(a) 167°C
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 2301

Question 9.
A steel rod of length 25 cm has a cross-sectional area of 0.8 cm2. The force required to stretch this rod by the same amount as the expansion produced by heating it through 10°C is (coefficient of linear expansion of steel is 10 51°C and Young’s modulus of steel is 2 × 1010 N/m2) …….
(a) 40 N
(b) 80 N
(c) 120 N
(d) 160 N
Answer:
(d) 160N
Solution:
The required force is given by
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 2312

Question 10.
Which of the following will make the volume of an ideal gas four times?
(a) double the absolute temperature and double the pressure.
(b) Halve the absolute temperature and double the pressure.
(c) Quarter the absolute temperature at constant pressure.
(d) Quarter the pressure at constant temperature.
Answer:
(d) Quarter the pressure at constant temperature.

Question 11.
A perfect gas at 27°C is heated at constant pressure so as to double its volume. The temperature of the gas becomes.
(a) 54° C
(b) 150 K
(c) 327° C
(d) 327 K
Answer:
(c) 327°C
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 2321

Question 12.
An air bubble doubles in radius on rising from the bottom of a lake to its surface. If the atmospheric pressure is equal to that of a column of water of height H, the depth of lake is
(a) H
(b) 2H
(c) 7H
(d) 8H
Answer:
(c) 7H
Solution:
Since the radius becomes double, the volume becomes eight times. Therefore, according to Boyle’s law, the pressure becomes one-eighth. Now, the pressure at the surface Hρg. Therefore pressure at the bottom must be 8 Hρg. Hence the depth of the lake is 7H.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 13.
The mass of 1 litre of helium under a pressure of 2 atm and at a temperature of 27°C is
(a) 0.16 g
(b) 0.32 g
(c) 0.48 g
(d) 0.64 g
Answer:
(b) 0.32 g
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 2331

Question 14.
Pressure exerted by a perfect gas is equal to …….
(a) mean kinetic energy per unit volume
(b) half of mean kinetic energy per unit volume
(c) one-third of mean kinetic energy per unit volume
(d) two-thirds of mean kinetic energy per unit volume
Answer:
(d) two-thirds of mean kinetic energy per unit volume
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 2341

Question 15.
Two vessels A and B contain the same ideal gas. The volume of B is twice that of A, the pressure in B is twice that in A and the temperature of B is twice that of A. The ratio of the number of gas molecules in A and B is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(a) 1 : 2
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 812

Question 16.
According to Boyle’s law, PV = C when the temperature of the gas remains constant. The value of C depends on
(a) temperature of the gas
(b) nature of the gas
(c) quantity of the gas
(d) both temperature and quantity of the gas.
Answer:
(d) both temperature and quantity of the gas.

Question 17.
The pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 1°C. The initial temperature was
(a) 250 K
(b) 250°C
(c) 25 K
(d) 25°C
Answer:
(c) 25 K
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 2361

Question 18.
A temperature difference of 25°C is equivalent to a temperature difference of
(a) 25°F
(b) 45°F
(c) 67°F
(d) 77°F
Answer:
(b) 45°F
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 2372

Question 19.
A temperature at which both the Fahrenheit and the centigrade scales have the same value is
(a) 40°
(b) -40°
(c) 20
(d) – 20°
Answer:
(b) – 40°
Solution: Let the required temperature be t. then, \(\frac{t}{5}=\frac{t-32}{9} \Rightarrow t=-40^{\circ}\)

Question 20.
If the temperature of patient is 40°C, his temperature on the Fahrenheit scale will be
(a) 72°F
(b) 96°F
(c) 100°F
(d) 104°F
Answer:
(d) 104°F
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 250

Question 21.
The correct value of 0°C on the Kelvin scale is ……..
(a) 273.15 K
(b) 272.85 K
(c) 273 K
(d) 273.2 K
Answer:
(a) 273.15 K

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 22.
When a gas in a closed vessel was heated so as to increase its temperature by 5°C, there occurred an increase of 1% in its pressure, the original temperature of the gas was …….
(a) 50°C
(b) 227°C
(c) 273°C
(d) 500°C
Answer:
(b) 227°C
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 251

Question 23.
A perfect gas at 27°C is heated at constant pressure so as to double its volume. The temperature of the gas will be …….
(a) 600°C
(b) 54°C
(c) 327°C
(d) 300°C
Answer:
(c) 327°C
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 252

Question 24.
Temperature can be expressed as a derived quantity in terms of ……
(a) length and mass
(b) mass and time
(c) length, mass and time
(d) none of these
Answer:
(d) none of these

Question 25.
The equation of state corresponding to 8 g of O2 is
(a) PV = 8RT
(b) PV = RT/4
(c) PV = RT
(d) PV = RT/2
Answer:
(b) PV = RT/4
Solution:
8g of O2 is 1/4 of a mole of O2, which is 32g. Thus, the required equation of state is PV = \(\frac{1}{4} \mathbf{R} \mathbf{T}\).

Question 26.
At a given volume and temperature, the pressure of a gas ….
(a) varies inversely as its mass
(b) varies inversely as the square of its mass
(c) varies linearly as its mass
(d) is independent of its mass
Answer:
(c) varies linearly as its mass

Question 27.
Oxygen boils at -183°C. This temperature in Fahrenheit scale is
(a) -215.7°
(b) – 297.4°
(c) -310.6°
(d) – 373.2°
Answer:
(b) – 297.4°
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 255

Question 28.
A centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers 140°. What is the decrease in temperature as registered by the centigrade thermometer?
(a) 80°
(b) 60°
(c) 40°
(d) 30°
Answer:
(c) 40°
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 256

Question 29.
The change in temperature of a body is 50°C. The change on the kelvin scale is …….
(a) 50K
(b) 323K
(c) 70K
(d) 30K
Answer:
(a) 50K

Question 30.
Mercury thermometers can be used to measure temperature up to ……..
(a) 260°C
(b) 100°C
(c) 360°C
(d) 500°C
Answer:
(c) 360°C

Question 31.
For an ideal gas the inter particle interaction is ……..
(a) attractive
(b) repulsive
(c) very large
(d) zero
Answer:
(d) zero

Question 32.
Device used to measure very high temperature is …….
(a) Pyrometer
(b) Thermometer
(c) Bolometer
(d) calorimeter
Answer:
(a) Pyrometer

Question 33.
Two metal rods A and B are having their initial lengths in the ratio 2 : 3, and coefficients of linear expansion in the ratio 3 : 4. When they are heated through same temperature difference,the ratio of their linear expansions is ……
(a) 1 : 2
(b) 2 : 3
(c) 3 : 4
(d) 4 : 3
Answer:
(a) 1 : 2
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 257

Question 34.
Boyles’ law is applicable in ……
(a) isochoric process
(b) isothermal process
(c) isobaric process
(d) both (a) and (b)
Answer:
(b) isothermal process

Question 35.
A rod, when heated from 0°C to 50°C, expands by 1.0 mm. Another rod, twice as long as the first at 0°C and of the same material, is heated from 0°C to 25°C. The second rod will expand by ……
(a) 0.5 mm
(b) 1.0 mm
(c) 2.0 mm
(d) 4.0 mm
Answer:
(b) 1.0 mm
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 258

Question 36.
A container contains hydrogen gas at pressure P and temperature T. Another identical container contains helium gas at pressure 2P and temperature T/2. The ratio of the number of molecules of the two gases is
(a) 1 : 4
(b) 4 : 1
(c) 1 : 2
(d) 2 : 1
Answer:
(a) 1 : 4
Solution:
The ratio of the number of molecules is same as the ratio of the number of moles.
Now n = \(\frac{P V}{R T}\)
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 259

Question 37.
Density of water is maximum at the temperature of
(a) 32°F
(b) 39.2°F
(c) 42°F
(d) 40°F
Answer:
(b) 39.2°F
Solution:
The density of water is maximum at 4°C. Let F be the corresponding temperature on the Fahrenheit scale. Then using the equation
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 260

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 38.
The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, is (R is the gas constant) …….
(a) PV = (5/32) RT
(b) PV = 5RT
(c) PV = (5/2)RT
(d) PV = (5/16)RT
Answer:
(a) PV = (5/32) RT
Solution:
Number of moles, n = \(\frac{5}{32}\)

Question 39.
A bimetallic strip consists of metals X and Y. It is mounted rigidly at the base as shown. The metal X has a higher coefficient of expansion compared to that for metal Y. When the bimetallic strip is placed in a cold bath?
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 2612
(a) it will bend towards the right
(b) it will bend towards the left
(b) it will not bend but shrink
(d) it will neither bend nor shrink
Answer:
(b) it will bend towards the left.
Solution:
In cold bath, the metal X will contract more than the metal Y. Therefore, the strip will bend towards the left.

Question 40.
An ideal gas is expanding such that PT2 = constant the coefficient of volume expansion of the gas is
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 262
Answer:
\(\frac{3}{\mathrm{T}}\)
Solution:
Coefficient of volume expansion \(\alpha_{\mathrm{v}}=\frac{d \mathrm{V}}{\mathrm{v} d \mathrm{T}}\)
PT2 = K and PV = nRT
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 263

Question 41.
A metallic solid sphere is rotating about its diameter as axis of rotation. If the temperature is increased by 200°C, the percentage increase in its moment of inertia is : (coefficient of linear expansion of the metal = 10-5/°C)
(a) 0.1
(b) 0.2
(c) 0.3
(d) 0.4
Answer:
(d) 0.4
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 264

Question 42.
The difference between volume and pressure coefficients of an ideal gas is ……
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 265
Answer:
(d) Zero

Question 43.
Which of the following instruments is used in the measurement of temperatures above 2000°C?
(a) Gas thermometer
(b) Pyrometer
(c) Bolometer
(d) Thermo-electric Pile
Answer:
(b) Pyrometer

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 44.
At 0°C, Pressure measured by barometer is 760 mm. What will be the pressure at 100°C?
(a) 760
(b) 730
(c) 780
(d) none of these
Answer:
(d) none of these
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 266

Question 45.
The temperature on the new scale, corresponding to a temperature of 39°C on the Celsius scale?
(a) 73°W
(b) 117°W
(c) 200°W
(d) 139°W
Answer:
(b) 117°W
Solution:
Let t be the required temperature. Then,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 267

Question 46.
Two balloons are filled, one with pure He gas and the other with air. If the pressure and temperature in both the balloons are same the number of molecules per unit volume is
(a) more in the He filled balloon
(b) more in the air filled balloon.
(c) same in both the balloon.
(d) in the ratio 1 : 4.
Answer:
(c) same in both the balloons.
Solution: Assuming ideal gas behavior, the number of moles per unit volume is \(\frac{n}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{RT}}\) Since P and T are same in both the balloon, \(\frac{n}{V}\) is also same in both.

Question 47.
Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on the kinetic energy of molecules?
(a) increase
(b) no change
(c) decrease
(d) can’t be determined
Answer:
(b) no change
Solution:
K.E of an ideal gas depends only on the temperature. Hence, it remains the same.

Question 48.
One mole of gas occupies a volume of 200 ml. at 100 mm pressure. What is the volume occupied by two moles of this gas at 400 mm pressure and at same temperature?
(a) 50 ml
(b) 100 ml
(c) 200 ml
(d) 400 ml
Answer:
(b) 100 ml
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 268
⇒ Volume of 2 moles of the gas at 400 mm pressure = 2 × 50 = 100 ml

Question 49.
There is a change in length when a 33000 N tensile force is applied on a steel rod of area of cress-section 10-3 m2. The change of temperature required to produce the same elongation, if the steel rod is heated, is (modulus of elasticity of steel = 3 × 1011 N/m2, coefficient of linear expansion of steel = 1.1 × 10-5/°C) …….
(a) 20°C
(b) 15°C
(c) 10°C
(d) 0°C
Answer:
(c) 10°C
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 269

Question 50.
In the given (V – T) diagram, what is the relation between pressures P1 and P2?
(a) P2 = P1
(b) P2 > P1
(c) P2 < P1
(d) cannot be predicted
Answer:
(c) P2 < P1
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 270
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 271

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 51.
Boiling water is changing into steam. Under this condition the specific heat of water is
(a) zero
(b) one
(c) infinite
(d) less than one
Answer:
(c) infinite
Solution:
In order to change boiling water into steam, heat has to be given but there is no increase of temperature. Therefore, under this condition the specific heat of water is infinite.

Question 52.
The first law of thermodynamics is concerned with the conservation of ……
(a) number of molecules
(b) energy
(c) number of moles
(d) temperature
Answer:
(b) energy

Question 53.
The gas law \(\frac{P V}{T}\) = constant is true for
(a) isothermal changes only
(b) adiabatic changes only
(c) both isothermal and adiabatic changes
(d) neither isothermal nor adiabatic changes
Answer:
(c) both isothermal and adiabatic changes

Question 54.
The pressure-temperature relationship for an ideal gas undergoing adiabatic change is …..
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 4614
Answer:
(a) \(\mathrm{p}^{1-\gamma} \mathrm{T}^{\gamma}\) = constant

Question 55.
For a certain gas the ratio of specific heats is given to be γ = 1.5. For this gas ……..
(a) Cv = 3R
(b) CP = 3R
(c) CV = 5R
(d) CP = 5R
Answer:
(b) CP = 3R
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 4618

Question 56.
Cooking takes longest time ………
(a) at the sea level
(b) at Shimla
(c) at mount Everest (if tried)
(d) in a submarine 100 m below the surface of water.
Answer:
(c) at mount Everest (if tried)

Question 57.
A closed bottle containing water at room temperature is taken to the moon and then the lid is opened. The water will ……
(a) freeze
(b) boil
(c) decompose into hydrogen and oxygen
(d) not change at all.
Answer:
(b) boil
Solution:
There is no atmosphere on the moon and so there is no pressure

Quarter 58.
A gas receives an amount of heat equal to 110 joules and performs 40 joules of work. The change in the internal energy of the gas is …….
(a) 70 J
(b) 150 J
(c) 110 J
(d) 40 J
Answer:
(a) 70 J

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 59.
For a mono-atomic gas, the molar specific heat at constant pressure divided by the molar gas constant R, is equal to ……
(a) 2.5
(b) 1.5
(c) 5.0
(d) 3.5
Answer:
(a) 2.5

Question 60.
Heat capacity of a substance is infinite. It means …….
(a) infinite heat is given out
(b) infinite heat is taken in
(c) no change in temperature whether heat is taken in or given out
(d) all of these
Answer:
(c) no change in temperature whether heat is taken in or given out
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 285

Question 61.
We consider a thermodynamic system. If ∆U represent the increase in its energy and W the work done by the system, which of the following statements is true?
(a) ∆U = – W in an isothermal process
(b) ∆U = – W in an adiabatic process
(c) ∆U = W in an isothermal process
(d) ∆U = W in an adiabatic process
Answer:
(b) ∆U = – W in an adiabatic process
Solution:
According to the first law of thermodynamics ∆Q = AU + W
In an adiabatic process, ∆Q = 0. Therefore, ∆U = – W

Question 62.
The first operation involved in a carnot cycle is
(a) isothermal expansion
(b) adiabatic expansion
(c) isothermal compression
(d) adiabatic compression
Answer:
(a) isothermal expansion

Question 63.
During an adiabatic process, if the pressure of an ideal gas is proportional to the cube of its temperature, the ratio \(\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}\) is ….
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 286
Answer:
(d) \(\frac{3}{2}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 287

Question 64.
In a given process on an ideal gas, dW = 0 and dQ < 0. Then for the gas ……..
(a) The temperature will decrease.
(b) the volume will decrease.
(c) the pressure will remain constant.
(d) the temperature will increase.
Answer:
(a) The temperature will decrease.
Solution:
According to the first law of thermodynamics, the internal energy decreases. Hence the temperature will decrease.

Question 65.
In a carnot heat engine 8000J of heat is absorbed from a source at 400 K and 6500 J of heat is rejected to the sink. The temperature of the sink is …..
(a) 325 K
(b) 100 K
(c) 200 K
(d) 273 K
Answer:
(a) 325 K
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 288

Question 66.
2 Kg of water of 60°C is mixed with 1 kg of water at 30°C kept in a vessel of heat capacity 220 J K-1. The specific heat of water is 4200 J Kg-1K ‘. Then the final temperature is nearly.
(a) 35°C
(b) 45°C
(c) 55°C
(d) 50°C
Answer:
(d) 50°C
Solution:
According to the principle of calorimetry,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 289

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 67.
A carnot engine absorbs heat at 127°C and rejects heat at 87°C. The efficiency of engine is
(a) 10%
(b) 30%
(c) 50%
(d) 70%
Answer:
(a) 10%
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 290

Question 68.
The first law of thermodynamics confirms the law of ……
(a) conservation of momentum
(b) conservation of energy
(c) flow of heat in a particular direction
(d) conservation of heat energy and mechanical energy
Answer:
(b) conservation of energy

Question 69.
The internal energy of an ideal gas depends on
(a) only pressure
(b) only volume
(c) only temperature
(d) none of these
Answer:
(c) only temperature

Question 70.
An ideal gas heat engine operators in a carnot’s cycle between 227°C and 127°C. It absorbs 6 × 104J at high temperature. The amount of heat converted into work is
(a) 2.4 × 104 J
(b) 4.8 × 104J
(c) 1.2 × 104J
(d) 6 × 104J
Answer:
(c) 1.2 × 104J
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 291

Question 71.
In an isochoric process
(a) ∆U = ∆Q
(b) ∆Q = ∆W
(c) ∆U = ∆W
(d) None of these
Answer:
(a) ∆U = ∆Q
Solution:
In an isochoric process, volume remains constant. Therefore, no work is done by or on the system. So, ∆W = 0 Hence ∆U = ∆Q.

Question 72.
The molar specific heat at constant pressure of an ideal gas is (7/2) R. The ratio specific heats at constant pressure to that at constant volume is ………
(a) 8/7
(b) 5/7
(c) 9/7
(d) 7/5
Answer:
(d) 7/5
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 292

Question 73.
If energy dQ is supplied to a gas isochorically, increase in internal energy is dU. Then ….
(a) dQ > dU
(b) dQ < dU
(c) dQ = dU
(d) dQ = -dU
Answer:
(c) dQ = dU

Question 74.
A diatomic ideal gas is used in a camot engine as the working substance. If during the adiabatic expansion part of the cycle, the volume of the gas increases from V to 32 V, the efficiency of the engine is ……
(a) 0.25
(b) 0.5
(c) 0.75
(d) 0.99
Answer:
(c) 0.75
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 293

Question 75.
The mechanical equivalent of heat J is:
(a) a constant
(b) a physical quantity
(c) a conversion factor
(d) none of the above
Answer:
(c) a conversion factor

Question 76.
Which of the following process is reversible?
(a) transfer of heat by radiation
(b) Transfer of heat by conduction
(c) Electrical heating of nichrome wire
(d) Isothermal compression
Answer:
(d) Isothermal compression

Question 77.
An ideal gas heat engine operates in camot cycle between 227°C and 127°C. It absorbs 6 × 104 cal of heat converted to work is ……
(a) 1.2 × 104 cal
(b) 4.8 × 104 cal
(c) 6 × 104 cal
(d) 2.4 × 104 cal
Answer:
(a) 1.2 × 104 cal
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 294

Question 78.
Ten moles of an ideal gas at constant temperature 600 K is compressed from 100 l to 10 L. The work done in the process is …..
(a) 4.11 × 104 J
(b) -4.11 × 104 J
(c) 11.4 × 104 J
(d) – 11.4 × 104 J
Answer:
(d) -11.4 × 104 J
Solution:
The process is isothermal. The work done is,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 295

Question 79.
A gas is compressed at a constant pressure of 50 N/m2 from a volume 4 m3. Energy of 100 J is then added to the gas by heating. Its internal energy is …..
(a) increased by 400 J
(b) increased by 200 J
(c) increased by 100 J
(d) decreased by 200 J
Answer:
(a) increased by 400 J
Solution:
∆U = ∆Q – ∆W = ∆Q – P∆V = 100 – 50 (4 – 10) = 400 J

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 80.
If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cyclic process, then ……
(a) W = 0
(b) Q = W = O
(c) E = 0
(d) Q = 0
Answer:
(c) E = 0
Solution:
In a cyclic process, a system starts in one state and comes back to the same state. Therefore, the change in internal energy is zero.

Question 81.
A carnot engine takes heat from a reservoir at 627°C and rejects heat to a sink at 27°C. Its efficiency is
(a) 3/5
(b) 1/3
(c) 2/3
(d) 200/209
Answer:
(c) 2/3

Question 82.
A carnot engine operates with source at 127°C and sink at 27°C. If the source supplies 40 KJ of heat energy, the work done by the engine is
(a) 1 KJ
(b) 4 KJ
(c) 10 KJ
(d) 30 KJ
Answer:
(c) 10 KJ
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 296

Question 83.
The ratio of the specific heat \(\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma\) in term of degrees of freedom (n) is given by:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 4615
Answer:
(b) \(\left(1+\frac{2}{n}\right)\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 4616

Question 84.
The heat required to increase the temperature of 4 moles of a mono-atomic ideal gas from 273 K to 473 K constant volume is ….
(a) 200 R
(b) 400 R
(c) 800 R
(d) 1200 R
Answer:
(d) 1200 R
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 4617

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 85.
The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is – 20°C, the temperature of the surroundings to which it rejects heat is ……
(a) 21°C
(b) 31°C
(c) 41°C
(d) 11°C
Answer:
(b) 31°C
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 299

II. Write brief answer to the following questions:

Question 1.
What is meant by ‘heat’?
Answer:
When an object at higher temperature is placed in contact with another object at lower temperature, there will be a spontaneous flow of energy from the object at higher temperature to the one at lower temperature. This energy is called heat.

Question 2.
What is meant by ‘temperature’? Give its unit.
Answer:
Temperature is the degree of hotness or coolness of a body. Hotter the body higher is its temperature. The temperature will determine the direction of heat flow when two bodies are in thermal contact. The SI unit of temperature is kelvin (K).

Question 3.
Define Avogadro’s number NA?
Answer:
The Avogadro’s number NA is defined as the number of carbon atoms contained in exactly 12g of 12C

Question 4.
Define heat capacity (or) thermal capacity?
Answer:
The heat capacity of a body is defined as the amount of heat required to raise its temperature through one degree.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 5.
What is meant by ‘Triple point of a substance’?
Answer:
The triple point of a substance is the temperature and pressure at which the three phases (gas, liquid and solid) of that substance coexist in thermodynamic equilibrium.

Question 6.
What do you meant by change of state of a substance?
Answer:
The transition of a substance from one state to another by heating or cooling it is called change of state.

Question 7.
Define coefficient of linear expansion. Give its unit.
Answer:
The coefficient of linear expansion of the material of a solid rod is defined as the increase in length per unit original length per degree rise in its temperature.
The unit of αL is °C-1 (or KT-1)

Question 8.
Define coefficient of area expansion? Give its unit.
Answer:
The coefficient of area expansion of a metal sheet is defined as the increase in its surface area per unit original surface area per degree rise in its temperature.
The unit of αA is °Cα-1 (or) KT-1

Question 9.
Define coefficient of Volume expansion? Give its unit.
Answer:
The coefficient of Volume expansion of a substance is defined as the increase in volume per unit original volume per degree rise in its temperature.
The unit of αv is °C-1 (or) KT-1

Question 10.
What is meant by conduction?
Answer:
Conduction is the process of direct transfer of heat through matter due to temperature difference. When two objects are in direct contact with one another, heat will be transferred from the hotter object to the colder one. The objects which allow heat to travel easily through them are called conductors.

Question 11.
What is meant by Convection?
Answer:
Convection is the process in which heat transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 12.
What is meant by Radiation? Give example.
Answer:
Radiation is a form of energy transfer from one body to another by electromagnetic waves.
Example:
1. Solar energy from the Sun.
2. Radiation from room heater.

Question 13.
State Newton’s Law of cooling.
Answer:
Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.

Question 14.
What do you meant by absolute zero of temperature?
Answer:
The lowest temperature of 0 K at which a gas is supposed to have zero volume (and zero pressure) and at which entire molecular motion stops is called absolute zero of temperature.

Question 15.
State Prevost theory of heat exchange.
Answer:
Prevost theory states that all bodies emit thermal radiation at all temperatures above absolute zero irrespective of the nature of the surroundings.

Question 16.
What is meant by ‘Mechanical equilibrium’?
Answer:
A system is said to be in mechanical equilibrium if no unbalanced force acts on the thermo dynamic system or on the surrounding by thermodynamic system.

Question 17.
What is meant by ‘chemical equilibrium’?
Answer:
Chemical equilibrium: If there is no net chemical reaction between two thermodynamic systems in contact with each other then it is said to be in chemical equilibrium.

Question 18.
What is meant by ‘thermodynamic equilibrium’.
Answer:
In a state of thermodynamic equilibrium the macroscopic variables such as pressure, volume and temperature will have fixed values and do not change with time.

Question 19.
Briefly explain how such a quasi-static process can be carried out.
Answer:
Quasi-static process: Consider a system of an ideal gas kept in a cylinder of volume V at pressure P and temperature T. When the piston attached to the cylinder moves outward the volume of the gas will change. As a result the temperature and pressure will also change because all three variables P,T and V are related by the equation of state PV = NkT. If a block of some mass is kept on the piston, it will suddenly push the piston downward. The pressure near the piston will be larger than other parts of the system. It implies that the gas is in non-equilibrium state. We cannot determine pressure, temperature or internal energy of the system until it reaches another equilibrium state. But if the piston is pushed very slowly such that at every stage it is still in equilibrium with surroundings, we can use the equation of state to calculate the internal energy, pressure or temperature. This kind of process is called quasi-static process.
A quasi-static process is an infinitely slow process in which the system changes its variables (P,V,T) so slowly such that it remains in thermal, mechanical and chemical equilibrium with its surroundings throughout. By this infinite, slow variation, the system is always almost close to equilibrium state.

Question 20.
Define specific heat capacity at constant pressure.
Answer:
Specific heat capacity at constant pressure (sP): The amount of heat energy required to raise the temperature of one kg of a substance by 1 K or 1°C by keeping the pressure constant is called specific heat capacity of at constant pressure.

Question 21.
Define specific heat capacity at constant volume.
Answer:
Specific heat capacity at constant volume (sV): The amount of heat energy required to raise the temperature of one kg of a substance by 1 K or 1°C by keeping the volume constant is called specific heat capacity at constant volume.

Question 22.
Define molar specific heat capacity at constant volume.
Answer:
The amount of heat required to rise the temperature of one mole of a substance by 1K or 1 °C at constant volume is called molar specific heat capacity at constant volume.

Question 23.
Define molar specific heat capacity at constant pressure.
Answer:
The amount of heat required to rise the temperature of one mole of a substance by 1K or 1°C at constant pressure is called molar specific heat capacity at constant pressure.

Question 24.
What is a isobaric process?
Answer:
It is a thermodynamic process which occurs at a constant pressure.

Question 25.
What is a isochoric process?
Answer:
It is a thermodynamic process which occurs at a constant volume.

Samacheer Kalvi 11th Physics Heat and Thermodynamics Numerical Problems

Question 1.
Copper wire of length l increase in length by 1% when heated from temperature T1 to T2. Find the percentage change in area when a copper plate of dimensions 2l × l is heated from T1 to T2
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 300

Question 2.
A steel rod of length 25 cm has a cross-sectional area of 0.8 cm2. Find the force required to stretch this rod by the same amount as the expansion produced by heating it through 10°C. (coefficient of linear expansion of steel is 10-5/°C and Young’s modulus of steel is 2 × 1010Nm-2)
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 301

Question 3.
The temperature at the bottom of a 40 m deap lake is 12°C and that at the surface is 35°C. An air bubble of volume 1.0 cm3 rises from the bottom to the surface. Find its volume, (atmospheric pressure = 10 m of water)
Answer:
Let P1, V1 and T1 be the pressure, bubble volume and absolute temperature at the bottom of the lake and let P2, V2 and T2 be the corresponding quantities at the surface. Then
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 302

Question 4.
Ajar ‘A’ is filled with an ideal gas characterised by parameters P, V and T and another jar B is filled with an ideal gas with parameters 2P, \(\frac{\mathrm{V}}{4}\) and 2T. Find the ratio of the number of molecules in jar A and B.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 303

Question 5.
By what percentage should the pressure of a given mass of a gas be increased so as to decrease its volume by 10% at a constant temperature?
Answer:
According to Boyle’s law,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 304

Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

Question 6.
A flask is filled with 13g of an ideal gas at 27°C and its temperature is raised to 52°C. Find the mass of the gas that hias to be released to maintain the temperature of the gas in the flask at 52°C and the pressure is same as the initial pressure.
Answer:
Let n be the initial number of moles and ri be the final number of moles. Since pressure and volume remain the same, we have
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 305

Question 7.
A rod of metal-1 of length 50.0 cm elongates by 0.10 cm when it is heated from 0°C to 100°C. Another rod of metal-2 of length 80.0 cm elongates by 0.08 cm for the same rise in temperature. A third rod of length 50.0 cm, made by welding pieces of rod 1 rad and 2 rad placed end to end, elongates by 0.03 cm when it is heated from 0°C to 50°C. Then what is the length of metal-1 in the third rod at 0°C?
Answer:
For metal-1, For metal-2,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 306
Let the lengths of metal-1 and metal- 2 in the third rod at 0°C be l1 and l2, respectively.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 307

Question 8.
A balloon is filled at 27°C and 1 atm pressure by 500 m3 He. Then find the volume of He at -3°C and 0.5 mm Hg pressure.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 308

Question 9.
During an experiment an ideal gas is found to obey an additional law VP2 = constant. The gas is initially at temperature T and Volume V. Find the resulting temperature when it expands to volume 2V.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 309

Question 10.
Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weights of A and B is:
Answer:
According to the ideal gas equation for one mole,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 311
Where ρ is the density and M is the molecular weight of the gas.
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 3110

Question 11.
For a gas the difference between the two specific heats is 4150 J/Kg K. What is the specific heat of the gas at constant volume if the ratio of specific heat is 1.4?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 312

Question 12.
A mass of ideal gas at pressure P is expanded isothermally to four times the original volume and then slowly compressed adiabatically to its original volume. Assuming γ(= CP/CV) to be 1.5, find the new pressure of the gas.
Answer:
Let the initial volume be V. After isothermal expansion, pressure = \(\frac{P}{4}\), volume = 4V
Let P’ is the pressure after adiabatic compression. The volume then is V. Therefore,
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 313

Question 13.
A carnot engine operating between temperature T1 and T2 has efficiency \(\frac{1}{6}\). When T2 is lowered by 62 K, its efficiency increases to \(\frac{1}{3}\). Then find the values of T1 and T2.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 314
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 315

Question 14.
A perfect gas goes from state A to state B by absorbing 8 × 105 J of heat and doing 6.5 × 105 J of external work. It is now transferred between the same two states in another process in which it absorbs 10s J of heat. In second process. Find the work done in the second process.
Answer:
According to the first law of thermodynamics
∆U = ∆Q – ∆W
In the first process ∆U = 8 × 105 – 6.5 × 105 = 1.5 × 105 J
Now ∆U, being a state function, remains the same in the second process,
∆W = ∆Q – ∆U = 1 × 105 – 1.5 × 105
∆W = – 0.5 × 105J
The negative sign shows that work is done on the gas.

Question 15.
A carnot engine, having efficiency of \(\eta=\frac{1}{10}\) as heat engine, is used as a refrigerator. If the work done on the system is 10 J, then find the amount of energy absorbed from the reservoir at lower temperature.
Answer:
As heat engine, let Q1 be the heat energy absorbed from the source and Q2 be the energy rejected to the sink. Then
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 316
From equation (1) and (2),
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 317

Question 16.
An ideal gas compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas?
Answer:
Work done on the gas = Area under the curve
Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics 318
Wadiabatic > Wisothermal > Wisobaric
Wisochoric is obviously zero because in an isochoric process there is no change in Volume.

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Students can Download Physics Chapter 9 Kinetic Theory of Gases Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Samacheer Kalvi 11th Physics Kinetic theory of Gases Textual Evaluation Solved

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Multiple Choice Questions

Question 1.
A particle of mass m is moving with speed u in a direction which makes 60° with respect to x axis. It undergoes elastic collision with the wall. What is the change in momentum in x and y direction?
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 1
Answer:
\(\Delta p_{x}=-2 m u, \Delta p_{y}=0\)
Solution:
The change in momentum of the molecule in x direction
\(\Delta p_{x}\) = Final momentum – Initial momentum ,
= After collision – Before collision
= – mu – mu = – 2mu
The change in momentum of the molecule in Y direction \(\Delta p_{y}\) = 0

Question 2.
A sample of ideal gas is at equilibrium. Which of the following quantity is zero?
(a) rms speed
(b) average speed
(c) average velocity
(d) most probable speed
Answer:
(c) average velocity

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 3.
An ideal gas is maintained at constant pressure. If the temperature of an ideal gas increases
from 100K to 1000K then the rms speed of the gas molecules
(a) increases by 5 times
(b) increases by \(\sqrt{10}\) times
(c) remains same
(d) increases by 7 times
Answer:
(b) increases by \(\sqrt{10}\) times
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 2

Question 4.
Two identically sized rooms A and B are.connected by an open door. If the room A is air conditioned such that its temperature is 4° lesser than room B, which room has more air in it?
(a) Room A
(b) RoomB
(c) Both room has same air
(d) Cannot be determined
Answer:
(a) Room A

Question 5.
The average translational kinetic energy of gas molecules depends on ……
(a) number of moles and T
(b) only on T
(c) P and T
(d) P only
Answer:
(a) number of moles and T

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 6.
If the internal energy of an ideal gas U and volume V are doubled then the pressure ……
(a) doubles
(b) remains same
(c) halves
(d) quadruples
Answer:
(b) remains same
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 3

Question 7.
The ratio \(\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}\) for a gas mixture consisting of 8 g of helium and 16 g of oxygen is …… [Physics Olympiad – 2005]
(a) 23/15
(b) 15/23
(c) 27/17
(d) 17/27
Answer:
(c) 27/17
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 7
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 5

Question 8.
A container has one mole of monoatomic ideal gas. Each molecule has /degrees of freedom.
What is the ratio of \(\gamma=\frac{\mathbf{C}_{\mathbf{P}}}{\mathbf{C}_{\mathbf{V}}}\) = ?
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 6
Answer:
(d) \(\frac{f+2}{f}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 78

Question 9.
If the temperature and pressure of a gas is doubled the mean free path of the gas molecules
(a) remains same
(b) doubled
(c) tripled
(d) quadruples
Answer:
(a) remains same
Solution:
Mean free path of the molecule
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 8

Question 10.
Which of the following shows the correct relationship between the pressure and density of an ideal gas at constant temperature?
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 9
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 10

Question 11.
A sample of gas consists of µ1 moles of monoatomic molecules, µ2 moles of diatomic molecules and µ3 moles of linear triatomic molecules. The gas is kept at high temperature. What is the total number of degrees of freedom?
(a) [3µ1 + 7(µ2 + µ3)] NA
(b) [3µ1 + 7µ2 + 6µ3] NA
(c) [7µ1 + 3(µ2 + µ3)] NA
(d) [3µ1 + 6(µ2 + µ3)] NA
Answer:
(a) [3µ1 + 7(µ2 + µ3)] NA

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 12.
If SP and SV denote the specific heats of nitrogen gas per unit mass at constant pressure and constant volume respectively, then …….. [JEE 2007]
(a) SP – SV = 28 R
(b) SP – SV = R/28
(c) SP – SV = R/14
(d) SP – SV = R
Answer:
(b) SP – SV = R/28

Question 13.
Which of the following gases will have least rms speed at a given temperature?
(a) Hydrogen
(b) Nitrogen
(c) Oxygen
(d) Carbon dioxide
Answer:
(d) Carbon dioxide

Question 14.
For a given gas molecule at a fixed temperature, the area under the Maxwell-Boltzmann distribution curve is equal to
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 1012
Answer:
(a) \(\frac{\mathrm{PV}}{k T}\)

Question 15.
The following graph represents the pressure versus number density for ideal gas at two different temperatures T1 and T2. The graph implies …….
(a) T1 = T2
(b) T1 > T2
(c) T1 < T2
(d) Cannot be determined
Answer:
(b) T1 > T2

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Short Answer Questions

Question 1.
What is the microscopic origin of pressure?
Answer:
According to the kinetic theory of a gases, the pressure exerted by the molecules depends on
(i) Number density n = \(\frac{N}{V}\)
(ii) Mass of the molecule
(iii) Mean square speed
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 11
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 12

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 2.
What is the microscopic origin of temperature?
Answer:
The average K.E per molecule \(\overline{\mathrm{KE}}=\epsilon=\frac{3}{2} k \mathrm{T}\)
The equation implies that the temperature of a gas is a measure of the average translational K.E. per molecule of the gas.

Question 3.
Why moon has no atmosphere?
Answer:
Moon has no atmosphere. The escape speed of gases on the surface of Moon is much less than the root mean square speeds of gases due to low gravity. Due to this all the gases escape from the surface of the Moon.

Question 4.
Write the expression for rms speed, average speed and most probable speed of a gas
molecule.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 122

Question 5.
What is the relation between the average kinetic energy and pressure?
Answer:
The internal energy of the gas U = \(\frac{3}{2} \mathrm{N} k \mathrm{T}\)
The above equation can also be written as U = \(\frac{3}{2} \mathrm{PV}\)
since PV = NkT
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 13
From the equation (1), we can state that the pressure of the gas is equal to two thirds of internal energy per unit volume or internal energy density \(\left(u=\frac{U}{V}\right)\)
Writing pressure in terms of mean kinetic energy density
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 14
where ρ = nm = mass density (Note n is number density)
Multiply and divide R.H. S of equation (2) by 2, we get
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 15
From the equation (3), pressure is equal to \(\frac{2}{3}\) of mean kinetic energy per unit volume.

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 6.
Define the term degrees of freedom.
Answer:
The minimum number of independent coordinates needed to specify the position and configuration of a thermo-dynamical system in space is called the degree of freedom of the system.

Question 7.
State the law of equipartition of energy.
Answer:
According to kinetic theory, the average kinetic energy of system of molecules in thermal equilibrium at temperature T is uniformly distributed to all degrees of freedom (x or y or z
directions of motion) so that each degree of freedom will get \(\frac{1}{2}\) kT of energy. This is called law of equipartition of energy.

Question 8.
Define mean free path and write down its expression.
Answer:
The average distance travelled by the molecule bfetween collisions is called mean free path (λ).
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 16

Question 9.
Deduce Chailes’ law based on kinetic theory.
Answer:
Charles’ law: From PV = \(\frac{2}{3} U\). For a fixed pressure, the volume of the gas is proportional to internal energy of the gas or average kinetic energy of the gas and the average kinetic energy is directly proportional to absolute temperature. It implies that
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 17
This is Charles’ law.

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 10.
Deduce Boyle’s law based on kinetic theory.
Answer:
Boyle’s law: From PV = \(\frac{2}{3} U\)
But the internal energy of an ideal gas is equal to N times the average kinetic energy (ε) of each molecule. U = Nε. For a fixed temperature, the average translational kinetic energy ε will remain constant. It implies that
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 18
Thus PV = constant Therefore, pressure of a given gas is inversely proportional to its volume provided the temperature remains constant. This is Boyle’s law.

Question 11.
Deduce Avogadro’s law based on kinetic theory.
Answer:
Avogadro’s law: This law states that at constant temperature and pressure, equal volumes of all gases contain the same number of molecules. For two different gases at the same temperature and pressure, according to kinetic theory of gases,
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 19
At the same temperature, average kinetic energy per molecule is the same for two gases.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 199
Dividing the equation (1) by (2) we get N1 = N2
This is Avogadro’s law. It is sometimes referred to as Avogadro’s hypothesis or Avogadro’s Principle.

Question 12.
List the factors affecting the mean free path.
Answer:

  1. Mean free path increases with increasing temperature. As the temperature increases, the average speed of each molecule will increase. It is the reason why the smell of hot sizzling food reaches several metre away than smell of cold food.
  2. Mean free path increases with decreasing pressure of the gas and diameter of the gas molecules.

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 13.
What is the reason for Brownian motion?
Answer:
According to kinetic theory, any particle suspended in a liquid or gas is continuously bombarded from all the directions so that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner.

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Long Answer Questions

Question 1.
Write down the postulates of kinetic theory of gases.
Answer:

  1. All the molecules of a gas are identical, elastic spheres.
  2. The molecules of different gases are different.
  3. The number of molecules in a gas is very large and the average separation between them is larger than size of the gas molecules.
  4. The molecules of a gas are in a state of continuous random motion.
  5. The molecules collide with one another and also with the walls of the container.
  6. These collisions are perfectly elastic so that there is no loss of kinetic energy during collisions.
  7. Between two successive collisions, a molecule moves with uniform velocity.
  8. The molecules do not exert any force of attraction or repulsion on each other except during collision. The molecules do not possess any potential energy and the energy is wholly kinetic.
  9. The collisions are instantaneous. The time spent by a molecule in each collision is very small compared to the time elapsed between two consecutive collisions.
  10. These molecules obey Newton’s laws of motion even though they move randomly.

Question 2.
Derive the expression of pressure exerted by the gas on the walls of the container.
Answer:
Expression for pressure exerted by a gas: Consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side l.
The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy, but a change in momentum occurs.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 25
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 26
The molecules of the gas exert pressure on the walls of the container due to collision on it. During each collision, the molecules impart certain momentum to the wall. Due to transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time.
A molecule of mass m moving with a velocity \(\vec{v}\) having components (vx, vy, vz) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with same speed and its x-component is reversed. This is shown in the figure. The components of velocity of the molecule after collision are (- vx, vy, vz).
The x-component of momentum of the molecule before collision = mvx
The x-component of momentum of the molecule after collision = – mvx
The change in momentum of the molecule in x direction = Final momentum – initial momentum = – mvx – mvx = – 2mvx
According to law of conservation of linear momentum, the change in momentum of the wall = 2mvx
The number of molecules hitting the right side wall in a small interval of time At.
The molecules within the distance of vx∆t from the right side wall and moving towards’ the right will hit the wall in the time interval ∆t. The number of molecules that will hit the right side wall in a time interval At is equal to the product of volume (Avx∆t) and number density of the molecules (n). Here A is area of the wall and n is number of molecules per unit volume \(\frac{\mathrm{N}}{\mathrm{V}}\).
We have assumed that the number density is the same throughout the cube.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 30
Not all the n molecules will move to the right, therefore on an average only half of the n molecules move to the right and the other half moves towards left side.
The number of molecules that hit the right side wall in a time interval ∆t
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 31
In the same interval of time At, the total momentum transferred by the molecules
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 32
From Newton’s second law, the change in momentum in a small interval of time gives rise to force.
The force exerted by the molecules on the wall (in magnitude)
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 33
Pressure, P = force divided by the area of the wall
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 34
Since all the molecules are moving completely in random manner, they do not have same speed. So we can replace the term \(v_{x}^{2}\) by the average \(\overline{v_{x}^{2}}\) in equation (4).
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 35
Since the gas is assumed to move in random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So, \(\overline{v_{x}^{2}}=\overline{v_{y}^{2}}=\overline{v_{z}^{2}}\) . The mean square speed is written as
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 36

Question 3.
Explain in detail the kinetic interpretation of temperature.
Answer:
To understand the microscopic origin Of temperature in the same way, from pressure exerted by a gas,
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 37
Comparing the equation (1) with ideal gas equation PV = NkT,
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 38
Multiply the above equation by \(\frac{3}{2}\) on both sides,
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 39
R.H.S. of the equation (3) is called average kinetic energy of a single molecule (\(\overline{\mathrm{KE}}\)).
The average kinetic energy per molecule
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 40
Equation (3) implies that the temperature, of a gas is a measure of the average translatiohal kinetic energy per molecule of the gas.
Equation (4) is a very important result from kinetic theory of gas. We can infer the following from this equation.
(i) The average kinetic energy of the molecule is directly proportional to absolute temperature of the gas. The equation (3) gives the connection between the macroscopic world (temperature) to microscopic world (motion of molecules).
(ii) The average kinetic energy of each molecule depends only on temperature of the gas hot on mass of the molecule. In other words, if the temperature of an ideal gas is measured using thermometer, the average kinetic energy of each molecule can be calculated without seeing the molecule through naked eye.
By multiplying the total number of gas molecules with average kinetic energy of each molecule, the internal energy of the gas is obtained.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 355
Here, we understand that the internal energy of an ideal gas depends only on absolute temperature and is independent of pressure and volume.

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 4.
Describe the total degrees of freedom for monoatomie molecule, diatomic molecule and triatomic molecule,
Monoatomie molecule: A monoatomie molecule by virtue of its nature has only three translational degrees of freedom.
Therefore f = 3
Example: Helium, Neon, Argon
Diatomic molecule: There are two cases.
(i) At Normal temperature: A molecule of a diatomic gas consists of two atoms bound to each other by a force of attraction. Physically the molecule can be regarded as a system of two point masses fixed at the ends of a massless elastic spring. The center of mass lies in the center of the diatomic molecule. So, the motion of the center of mass requires three translational degrees of freedom (figure a). In addition, the diatomic molecule can rotate about three mutually perpendicular axes (figure b). But the moment of inertia about its own axis of rotation is negligible. Therefore, it has only two rotational degrees of freedom (one rotation is about Z axis and another rotation is about Y axis). Therefore totally there are five degrees of freedom.
f = 5

2. At High Temperature: At a very high temperature such as 5000 K, the diatomic molecules possess additional two degrees of freedom due to vibrational motion [one due to kinetic energy of vibration and the other is due to potential energy] (figure c ). So totally there are seven degrees of freedom.
f = 7
Examples: Hydrogen, Nitrogen, Oxygen.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 1013
Triatomic molecules: There are two cases.
Linear triatomic molecule:
In this type, two atoms lie on either side of the central atom. Linear triatomic molecule has three translational degrees of freedom. It has two rotational degrees of freedom because it is similar to diatomic molecule except there is an additional atom at the center. At normal temperature, linear triatomic molecule will have five degrees of freedom. At high temperature it has two additional vibrational degrees of freedom. So a linear triatomic molecule has seven degrees of freedom.
Example: Carbon dioxide.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 41
Non-linear triatomic molecule: In this case, the three atoms lie at the vertices of a triangle.
It has three translational degrees of freedom and three rotational degrees of freedom about three mutually orthogonal axes. The total degrees of freedom.
f = 6
Example: Water, Sulphurdioxide.

Question 5.
Derive the ratio of two specific heat capacities of monoatomic, diatomic and triatomic molecules.
Answer:
Application of law of equipartition energy in specific heat of a gas, Meyer’s relation CP – CV = R connects the two specific heats for one mole of an ideal gas.
Equipartition law of energy is used to calculate the value of CP – CV and the ratio between them \(\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}\). Here γ is called adiabatic exponent.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 42
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 43
Note that the CV and CP are higher for diatomic molecules than the mono atomic molecules. It implies that to increase the temperature of diatomic gas molecules by 1°C it require more heat energy than monoatomic molecules.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 44

(iii) Triatomic molecule
(a) Linear molecule:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 444
(b) Non-linear molecule:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 445
Note that according to kinetic theory model of gases the specific heat capacity at constant volume and constant pressure are independent of temperature. But in reality it is not sure.
The specific heat capacity varies with the temperature.

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 6.
Explain in detail the Maxwell Boltzmann distribution function.
Answer:
Maxwell-Boltzmann: In speed distribution function
Consider an atmosphere, the air molecules are moving in random directions. The speed of each molecule is not the same even though macroscopic parameters like temperature and pressure are fixed. Each molecule collides with every other molecule and they exchange their speed. In the previously we calculated the rms speed of each molecule and not the speed of each molecule which is rather difficult. In this scenario we can find the number of gas molecules that move with the speed of 5 ms-1 to 10 ms-1 or 10 ms-1 to 15 ms-1 etc. In general our interest is to find how many gas molecules have the range of speed from v to v + dv. This is given by Maxwell’s speed distribution function.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 401
The above expression is graphically shown as follows:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 412
From the above figure, it is clear that, for a given temperature the number of molecules having lower speed increases parabolically but decreases exponentially after reaching most probable speed. The ms speed, average speed and most probable speed are indicated in the figure. It can be seen that the rms speed is greatest among the three.
(i) The area under the graph will give the total number of gas molecules in the system
(ii) Figure 2 shows the speed distribution graph for two different temperatures. As temperature increases, the peak of the curve is shifted to the right. It implies that the average speed of each molecule will increase. But the area under each graph is same since it represents the total number of gas molecules.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 420

Question 7.
Derive the expression for mean free path of the gas.
Answer:
Expression for mean free path
We know from postulates of kinetic theory that the molecules of a gas are in random motion and they collide with each other. Between two successive collisions, a molecule moves along a straight path with uniform velocity. This path is called mean free path. Consider a system of molecules each with diameter d. Let n be the number of molecules per unit volume. Assume that only one molecule is in motion and all others are at rest.
If a molecule moves with average speed v in a time t, the distance travelled is vt. In this time t, consider the molecule to move in an imaginary cylinder of volume πd2vt. It collides with any molecule whose center is within this cylinder. Therefore, the number of collisions is equal to the number of molecules in the volume of the imaginary cylinder. It is equal to πd2vtn. The total path length divided by the number of collisions in time t is the mean free path.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 431
Though we have assumed that only one molecule is moving at a time and other molecules are at rest, in actual practice all the molecules are in random motion. So the average relative speed of one molecule with respect to other molecules has to be taken into account. After some detailed calculations the correct expression for mean free path
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 1014
The equation (2) implies that the mean free path is inversely proportional to number density. When the number density increases the molecular collisions increases so it decreases the distance travelled by the molecule before collisions.
Case 1: Rearranging the equation (2) using ‘m’ (mass of the molecule)
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 45
But mn = mass per unit volume = ρ (density of the gas)
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 46
The equation (4) implies the following:
(i) Mean free path increases with increasing temperature. As the temperature increases, the average speed of each molecule will increase. It is the reason why the smell of hot sizzling food reaches several meter away than smell of cold food.
(ii) Mean free path increases with decreasing pressure of the gas and diameter of the gas molecules.

Question 8.
Describe the Brownian motion.
Answer:
Brownian motion is due to the bombardment of Brownian motion suspended particles by molecules of the surrounding fluid. But during 19th century people did not accept that every matter is made up of small atoms or molecules. In the year 1905, Einstein gave systematic theory of Brownian motion based on kinetic theory and he deduced the average size of molecules.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 461
According to kinetic theory, any particle suspended – in a liquid or gas is continuously bombarded from all the directions so that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner. But when we put our hand in water it causes no random motion because the mass of our hand is so large that the momentum transferred by the molecular collision is not enough to move our hand.

Factors affecting Brownian Motion:

  1. Brownian motion increases with increasing temperature.
  2. Brownian motion decreases with bigger particle size, high viscosity and density of the liquid (or) gas.

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Numerical Problems

Question 1.
A fresh air is composed of nitrogen N2 (78%) and oxygen O2 (21%). Find the rms speed of N2 and O2 at 20°C.
Answer:
Absolute temperature T = 20°C + 273 = 293K
Gas constant R = 8.32 J mol-1 K-1
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 50

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 2.
If the rms speed of methane gas in the Jupiter’s atmosphere is 471.8 m s-1, show that the surface temperature of Jupiter is sub-zero.
Answer:
RMS speed of methane gas (vrms) = 471.8 ms-1
Molar mass of methane gas (M) = 16.04 g per mol
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 51

Question 3.
Calculate the temperature at which the rms velocity of a gas triples its value at S.T.P. (Standard temperature T1 = 273 K)
Answer:
At STP temperature T1 = 273 K
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 52

Question 4.
A gas is at temperature 80°C and pressure \(5 \times 10^{-10} \mathrm{N} \mathrm{m}^{-2}\). What is the number of molecules per m3 if Boltzmann’s constant is \(1.38 \times 10^{-23} \mathrm{J} \mathrm{K}^{-1}\).
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 53
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 54

Question 5.
From kinetic theory of gases, show that Moon cannot have an atmosphere (Assume k = 1.38 × 10-23 J K-1 Temperature T = 0°C = 273 K).
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 55

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 6.
If 1020 oxygen molecules per second strike 4 cm2 of wall at an angle of 30° with the normal when moving at a speed of 2 × 103 ms-1, find the pressure exerted on the wall.
(mass of 1 atom = 2.67 × 10-26 kg).
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 56
Momentum normal to the wall at angle 30°
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 57

Question 7.
During an adiabatic process, the pressure of a mixture of monoatomic and diatomic gases is found to be proportional to the cube of the temperature. Find the value of Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 81
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 59
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 591

Question 8.
Calculate the mean free path of air molecules at STP. The diameter of N2 and O2 is about 3 × 10-10 m.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 60

Question 9.
A gas made of a mixture of 2 moles of oxygen and 4 moles of argon at temperature T. Calculate the energy of the gas in terms of RT. Neglect the vibrational modes.
Answer:
For two moles of diatomic nitrogen with no vibrational mode,
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 61
For four mole of monatomic argon,
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 62
Total energy of the gas, U = U1 + U2 = 5 RT + 6 RT
U = 11 RT

Question 10.
Estimate the total number of air molecules in a room of capacity 25 m3 at a temperature of 27°C.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 63
The number of molecules in the room,
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 64

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Additional Multiple Choice Questions

I. Choose the correct answer from the following:

Question 1.
Oxygen and hydrogen gases are at the same temperature the ratio of the average K.E of an oxygen molecule and that of a hydrogen molecule is
(a) 16
(b) 4
(c) 1
(d) \(\frac{1}{4}\)
Answer:
(c) 1
Solution:
P1V1 = P2V2, if T is constant.

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 2.
According to the kinetic theory of gases,
(a) the pressure of a gas is proportional to the rms speed of the molecules.
(b) the rms speed of the molecules of a gas is proportional to the absolute temperature.
(c) the rms speed of the molecules of a gas is proportional to the square root of the absolute temperature.
(d) the pressure of a gas is proportional to the square root of the rms speed of the molecules.
Answer:
(c) the rms speed of the molecules of a gas is proportional to the square root of the absolute temperature.
Solution:
The rms speed of the molecules of a gas is proportional to the square root of the absolute temperature. .

Question 3.
Pressure exerted by a perfect gas is equal to
(a) mean K.E. per unit volume.
(b) half of mean K.E. per unit volume.
(c) one-third of mean K.E. per unit volume.
(d) two-third of mean K.E. per unit volume.
Answer:
(d) two-third of mean K.E. per unit volume.
Solution: P = \(\frac{2}{3}\) K.E. (average K.E. of the gas per unit volume).

Question 4.
The temperature of an ideal gas is increased from 27°C to 927°C. The root mean square speed of its molecules becomes.
(a) 3 times
(b) double
(c) 4 times
(d) 6 times
Answer:
(b) double
1200
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 551
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 552

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 5.
Two gases are enclosed in a container at constant temperature. One of the gases, which is diatomic, has relative molecular mass eight times the other, which is monoatomic. The ratio pf the rms speed of the molecules of the monoatomic gas to that of the molecules of the diatomic gas is
(a) 8
(b) 4
(c) 2\(\sqrt{2}\)
(d) 2
Answer:
(c) 2\(\sqrt{2}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 571

Question 6.
If the absolute temperature of a gas is increased 3 times the rms velocity of the molecules will be ………
(a) 3 times
(b) 9 times
(c) 73 times
(d) 76 times
Answer:
(c) 73 times

Question 7.
At a given temperature which of the following gases possesses maximum rms velocity of molecules?
(a) H2
(b) O2
(c) N2
(d) CO2
Answer:
(a) H2
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 601

Question 8.
Two vessels have equal volumes. One of them contains hydrogen at one atmosphere and the other helium at two atmospheres. If both the samples are at the same temperature, the rms velocity of the hydrogen molecules is …..
(a) equal to that of the helium molecules
(b) twice that of the helium molecules
(c) half that of the helium molecules
(d) \(\sqrt{2}\) times that of the helium molecules
Answer:
(d) \(\sqrt{2}\) times that of the helium molecules
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 612

Question 9.
A gas is enclosed in a container which is then placed on a fast moving train. The temperature of the gas …..
(a) rises
(b) remains unchanged
(c) falls
(d) becomes unsteady
Answer:
(c) falls

Question 10.
The mean translational K.E. of a perfect gas molecule at absolute temperature T is (K is Boltzmann constant)
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 621
Answer:
(c) \(\frac{3}{2} k \mathrm{T}\)

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 11.
Ajar has mixture of hydrogen and Oxygen gases in the ratio 1 : 5. The ratio of mean kinetic energies of hydrogen and Oxygen molecules is ……..
(a) 1 : 5
(b) 5 : 1
(c) 1 : 1
(d) 1 : 25
Answer:
(c) 1 : 1
Solution:
The mean kinetic energy depends only on the temperature.

Question 12.
The pressure exerted on the walls of the container by a gas is due to the fact that the gas molecules ……..
(a) lose their K.E
(b) Stick to the walls
(c) are accelerated towards the walls
(d) change their momenta due to collision with the walls.
Answer:
(d) change their momenta due to collision with the walls.

Question 13.
Pressure exerted by a gas is …….
(a) independent of the density of the gas
(b) inversely proportional to the density of the gas
(c) directly proportional to the density of the gas
(d) directly proportional to the square of the density of the gas.
Answer:
(c) directly proportional to the density of the gas

Question 14.
Four molecules have speed 2 km/s, 3 km/s, 4 km/s and 5 km/s. The rms speed of these molecules in km/s is …….
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 1015
Answer:
(a) \(\sqrt{\frac{27}{2}}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 71

Question 15.
A real gas behaves as an ideal gas at …….
(a) low pressure and high temperature
(b) high pressure and low temperature
(c) low pressure and low temperature
(d) high pressure and high temperature.
Answer:
(a) low pressure and high temperature

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 16.
The kinetic theory of gases breaks down most at ………
(a) low pressure and high temperature
(b) high pressure and low temperature
(c) low pressure and low temperature
(d) high pressure and high temperature.
Answer:
(b) high pressure and low temperature

Question 17.
Two different ideal gases are enclosed in two different vessels at the same pressure. If ρ1 and ρ2 are their densities and v1 and v2 their rms speeds, respectively then ____ is equal to ……
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 754
Answer:
(a) \(\sqrt{\frac{\rho_{2}}{\rho_{1}}}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 752

Question 18.
A cylinder of capacity 20 litres is filled with hydrogen gas. The total average K.E. of translatory motion of its molecules is 1.5 × 105 J. The pressure of hydrogen in the cylinder is …….
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 76
(d) 5 × 106 Nm-2
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 77

Question 19.
The molecular weights of oxygen and hydrogen are 32 and 2, respectively. The rms velocities of their molecules at a given temperatures, will be in the ratio
(a) 4 : 1
(b) 1 : 4
(e) 1 : 16
(6) 16 : 1
Answer:
(b) 1 : 4
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 781

Question 20.
The average energy of a molecules of a monoatomic gas at temperature T is (K Boltzmann
constant)
(a) \(\frac{1}{2}\)kT
(b) kT
(c) \(\frac{3}{2}\)kT
(d) \(\frac{5}{2}\)kT
Answer:
(c) \(\frac{3}{2}\)kT

Question 21.
The temperature at which the molecules of nitrogen will have the same rms velocity as the
molecules of oxygen at 127° C is
(a) 77°C
(b) 350°C
(c) 273°C
(d) 457°C
Answer:
(a) 77°C
Solution:
Let the required temperature be T. Then
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 79>

Question 22.
The temperature of a gas is raised from 27°C to 927°C. The root mean square speed of its molecules …….
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 80
(b) gets halved
(c) remains the same
(d) gets doubled
Answer:
(d) gets doubled
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 811

Question 23.
The temperature at which the K.E of a gas molecules is double its value at 27°C is
(a) 54°C
(b) 300 K
(c) 327°C
(d) 108°C
Answer:
(c) 327°C
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 82

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 24.
The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K the root mean square velocity of the gas molecules is v, at 480 K it becomes
(a) 4v
(b) 2v
(c) \(\frac{v}{2}\)
(d) \(\frac{v}{4}\)
Answer:
(b) 2v
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 83

Question 25.
The average translational K.E. of O2(molar mass 32) molecules at a particular temperature is 0.048 eV. The translational K.E. of N2 (molar mass 28) molecules in eV at the same temperature is ……….
(a) 0.0015
(b) 0.003
(c) 0.048
(d) 0.768
Answer:
(c) 0.048
Solution:
Average translational K.E. of a gas molecule = \(\frac{3}{2}\)kT.
It is independent of molecular mass.

Question 26.
The K.E. of one mole of a gas at normal temperature and pressure is Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 842 ……..
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 841
Solution:
(d) 3.4 × 103J
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 85

Question 27.
The average K.E. of a hydrogen gas molecule at STP will be (Boltzmann constant KB= 1.38 × 10-23 JK-1) ………
(a) 0.186 × 10-28 J
(b) 0.372 × 10-20 J
(c) 0.56 × 10-20 J
(d) 5.6 × 10-20 J
Answer:
(c) 0.56 × 10-20 J
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 86

Question 28.
The rms speed of the particles of fume of mass 5 × 10-17 kg executing Brownian motion in air at STP is ……
(a) 1.5 ms-1
(b) 3.0 ms-1
(c) 1.5 cm s-1
(d) 3.0 cm s-1
Answer:
(c) 1.5 cm s-1
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 87

Question 29.
To what temperature should the hydrogen at room temperature (27°C) be heated at constant pressure so that the RMS velocity of its molecules becomes double its previous value?
(a) 1200°C
(b) 927°C
(c) 600°C
(d) 108°C
Answer:
(b) 927°C
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 88

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 30.
A vessel contains oxygen at 400 K. Another similar vessel contains an equal mass of hydrogen at 300K. The ratio of the rms speeds of molecules of hydrogen and oxygen is
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 89
Answer:
(d) \(2 \sqrt{3}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 891

Question 31.
A chamber contains a mixture of helium gas (He) and hydrogen gas (H2). The ratio of the root- mean-square speeds of the molecules of He and H2 is ……..
(a) 2
(b) \(\sqrt{2}\)
(c) \(\frac{1}{\sqrt{2}}\)
(d) \(\frac{1}{2}\)
Answer:
(c) \(\frac{1}{\sqrt{2}}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 91

Question 32.
On colliding with the walls in a closed container, the ideal gas molecules.
(a) transfer momentum to the walls
(b) lose momentum completely
(c) move with smaller speeds
(d) perform Brownian motion.
Answer:
(a) transfer momentum to the walls

Question 33.
The speeds of 5 molecules of a gas (in arbitrary units) are as follows: 2, 3, 4, 5, 6 The root mean square speed for these molecule is
(a) 2.91
(b) 3.52
(c) 4.00
(d) 4.24
Answer:
(d) 4.24
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 92

Question 34.
At absolute zero temperature, the K.E. of the molecules becomes ………
(a) zero
(b) maximum
(c) minimum
(d) none of these
Answer:
(a) zero

Question 35.
If the rms speed of the molecules of a gas is 1000 ms-1 the average speed of the molecule is
(a) 1000 ms-1
(b) 922 ms-1
(c) 780 ms-1
(d) 849 ms-1
Answer:
(b) 922 ms-1
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 93

Question 36.
The gas having average molecular speed four times that of SO2 (molecular mass 64) is
(a) He (molecular mass 4)
(b) O2 (molecular mass 32)
(c) H2 (molecular mass 2)
(d) CH4 (molecular mass 16)
Answer:
(a) He (molecular mass 4)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 94

Samacheer Kalvi 11th Physics Kinetic Theory of Gases 2 Marks Questions

Question 1.
State Avogadro’s law.
Answer:
It states that equal volumes of all gases under similar conditions of temperature and pressure, contain equal number of molecules.

Question 2.
Define root mean square speed (vrms). Write down its equations.
Answer:
Root mean square speed is defined as the square root of the mean of the square of speeds of all molecules. It is denoted by Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 95
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 96

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 3.
Define Avogadro’s number.
Answer:
It is defined as the number of particles present in one mole of the substance. It is denoted by NA
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 97

Question 4.
Define Average speed. Write it equation.
Answer:
Average speed is defined as the mean (or) average of all the speeds of molecules.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 98

Question 5.
Define most probable speed of the gas. Write its expressions.
Answer:
It is defined as the speed acquired by most of the molecules of the gas.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 99

Question 6.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 100
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 101

Question 7.
What is the reason “No hydrogen in Earth’s atmosphere” ?
Answer:
As the root mean square speed of hydrogen is much less than that of nitrogen, it easily escapes from the earth’s atmosphere.
In fact, the presence of nonreactive nitrogen instead of highly combustible hydrogen deters
many disastrous consequences.

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 8.
What is Brownian motion?
Answer:
The motion of the particles in a random and zig-zag mannar in a fluid is called Brownian motion.

Question 9.
What does the universal gas constant R signify? Give its value.
Answer:
The universal gas constant R signifies the workdone by (or on) a gas per mole per kelvin. Its value is R = 8.314 J mol-1 K-1

Question 10.
What is Boltzmann’s constant? Give its value.
Answer:
Boltzmann’s constant is defined as the gas constant per molecule.
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 102

Question 11.
When do the real gases obey more correctly the gas equation: PV = nRT?
Answer:
An ideal gas is one whose molecules have zero volume and no mutual force between them. At low pressure, the volume of a gas is large and so the volume occupied by the molecules is negligible in comparison to the volume of the gas. At high temperature, the molecules have large velocities and so the intermolecular force has no influence on their motion. Hence at low pressure and high temperature, the behaviour of real gases approach the ideal gas behaviour.

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Additional Numerical Problems

Question 1.
If the rms speed of hydrogen molecules at 300 K is 1930 ms-1. Then what is the rms speed of oxygen molecules
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 103

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 2.
The rms velocity of the molecules in a sample of helium is 5/7 times that of molecules in a sample of hydrogen. If the temperature of hydrogen is 0°C. Then, what is the temperature of helium?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 104
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 105

Question 3.
A cylinder of fixed capacity 44.8 litres contain helium gas at standard pressure and temperature. What is the amount of heat needed to raise the temperature of the gas by 15°C? (R = 8.31 J mol-1 K-1)
Answer:
Volume of 1 mole of He at STP = 22.4 litres
Total volume of He at STP = 44.8 litres
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 106
Molar specific heat of He (monoatomic gas) at constant volume,
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 107

Question 4.
An insulated container containing monoatomic gas of molar mass m is moving with a velocity v0. If the container is suddenly stopped. Find the change in temperature.
Answer:
Suppose the container has n moles of the monoatomic gas. Then the loss in K.E. of the gas
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 108
If the temperature of the gas changes by ∆T, then heat gained by the gas,
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 109

Question 5.
Estimate the total number of molecules inclusive of oxygen, nitrogen, water vapour and other constituents in a room of capacity 25.0 m3 at a temperature of 27°C and 1 atmospheric pressure. (kB = 1.38 × 10-23 JK-1)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 110
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 111

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 6.
Estimate the average energy of a helium atom at
(i) room temperature (27°C)
(ii) the temperature on the surface of the sun (6000 K) and
(iii) the temperature of 107 K. (kB = 1.38 × 10-23JK-1)
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 112

Question 7.
The molecules of a given mass of a gas have rms velocity of 200 ms 1 at 27°C and 1.0 × 10s Nm-2 pressure. When the temperature and pressure of the gas are respectively. 127 °C and 0.05 × 10s Nm-2.
Find the rms velocity of its molecules in ms-1
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 113

Question 8.
At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere? (Mass of oxygen molecule (m) = 2.76 × 10-26 kg
Boltzmann’s constant (kB) = 1.38 × 10-23 JK-1)
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 114
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 115

Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases

Question 9.
The temperature of a gas is raised from 27°C to 927°C. What is the rms molecular speed?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 116

Question 10.
A gaseous mixture consists of 16 g of helium and 16 g of oxygen. Find the ratio \(\frac{\mathbf{C}_{\mathbf{P}}}{\mathbf{C}_{\mathbf{V}}}\) of the
mixture.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 9 Kinetic Theory of Gases 117

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5

Evaluate the following limits, if necessary use l’Hopital Rule.

Question 1.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 1
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 2

Question 2.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 3
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 4

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5

Question 3.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 5
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 6

Question 4.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 7
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 8

Question 5.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 9
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 10

Question 6.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 11
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 12

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5

Question 7.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 13
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 14

Question 8.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 15
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 16

Question 9.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 17
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 18

Question 10.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 99
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 20
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 200

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5

Question 11.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 21
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 22

Question 12.
If an initial amount A0 of money is invested at an interest rate r compounded n times a year, the value of the investment after t years is Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 367. If the interest is compounded continuously, (that is as n ➝ ∞), show that the amount after t years is A = A0ert.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 368
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 24

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 Additional Problems

Question 1.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 25
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 26
Note that here l’Hopital’s rule, applied yields the result

Question 2.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 27
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 28
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 29

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5

Question 3.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 30
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 31

Question 4.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 32
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 33

Question 5.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 34
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 36

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5

Question 6.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 366
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 37
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.5 38

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Students can Download Physics Chapter 10 Oscillations, Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Samacheer Kalvi 11th Physics Oscillations Textual Evaluation Solved

Samacheer Kalvi 11th Physics Oscillations Multiple Choice Questions

Question 1.
In a simple harmonic oscillation, the acceleration against displacement for one complete oscillation will be [model NSEP 2000-01 ]
(a) an ellipse
(b) a circle
(c) a parabola
(d) a straight line
Answer:
(d) a straight line

Question 2.
A particle executing SHM crosses points A and B with the same velocity. Having taken 3 s in passing from A to B, it returns to B after another 3s. The time period is
(a) 15s
(b) 6s
(c) 12s
(d) 9s
Answer:
(c) 12s
Hint:
Time period of Oscillation = 2 × (time taken to go from A to B + the next time taken to return at B)
= 2 × (3 + 3)
= 2 × 6
Time period = 12s

Question 3.
The length of a second’s pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on surface of planet X such that the acceleration of the planet X is n times greater than the Earth
(a) 0.9 n
(b) \(\frac{0.9}{n} \mathrm{m}\)
(c) 0.9 n2m
(d) \(\frac{0.9}{n^{2}}\)
Answer:
(a) 0.9 n
Hint:
Second’s pendulum on the surface of planet,
Time period, T = 2 sec
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 4.
A simple pendulum is suspended from the roof of a school bus which moves in a horizontal direction with an acceleration a, then the time period is …..
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 2
Answer:
\(\mathrm{T} \propto \frac{1}{g^{2}+a^{2}}\)
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 4

Question 5.
Two bodies A and B whose masses are in the ratio 1:2 are suspended from two separate massless springs of force constants kA and kB respectively. If the two bodies oscillate vertically such that their maximum velocities are in the ratio 1 : 2, the ratio of the amplitude A to that of B is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 3
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 5
Hint:
The maximum velocity in the Oscillation is given as vmax = Aω
The ratio of maximum velocity, vA: vB = 1 : 2
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 6

Question 6.
A spring is connected to a mass m suspended from it and its time period for vertical oscillation is T. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 7
Answer:
(b) \(\mathrm{T}^{\prime}=\frac{\mathrm{T}}{\sqrt{2}}\)
Hint:
Spring constant of spring depends upon number of coil in the spring! When you cut the spring then the number of coil remain half of the original, so
k’ = 2k
Hint:
According to law of accelaration, the time period of simple pendulum
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 8

Question 5.
Two bodies A and B whose masses are in the ratio 1 : 2 are suspended from two separate massless springs of force constants kA and kB respectively. If the two bodies oscillate vertically such that their maximum velocities are in the ratio 1 : 2, the ratio of the amplitude A to that of B is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 9
Answer:
(b) \(\sqrt{\frac{k_{\mathrm{B}}}{8 k_{\mathrm{A}}}}\)
Hint:
The maximum velocity in the Oscillation is given as vmax = Aω
The ratio of maximum velocity, vA : vB = 1 : 2
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 999

Question 6.
A spring is connected to a mass m suspended from it and its time period for vertical oscillation is T. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 10
Answer:
(b) \(\mathrm{T}^{\prime}=\frac{\mathrm{T}}{\sqrt{2}}\)
Hint:
Spring constant of spring depends upon number of coil in the spring! When you cut the spring then the number of coil remain half of the original, so
k’ = 2k
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 11

Question 7.
The time period for small vertical oscillations of block of mass m when the masses of the pulleys are negligible and spring constant k1 and k2 is ….
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 12
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 14

Question 8.
A simple pendulum has a time period T1. When its point of suspension is moved vertically upwards according as y = kt2, where y is vertical distance covered and k = 1 ms-2, its time period becomes T2. Then Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1513 [IIT 2005]
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1514
Answer:
(c) \(\frac{6}{5}\)
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 16

Question 9.
An ideal spring of spring constant k, is suspended from the ceiling of a room and a block of mass M is fastened to its lower end. If the block is released when the spring is un-stretched, then the maximum extension in the spring is : [IIT 2002]
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 17
Answer:
(c) \(2 \frac{\mathrm{Mg}}{k}\)
Hint:
Work by gravity + Work by spring = Change in kinetic energy
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 18

Question 10.
A pendulum is hung in a very high building oscillates to and fro motion freely like a simple harmonic oscillator. If the acceleration of the bob is 16 ms-2 at a distance of 4 m from the mean position, then the time period is [NEET 2018 model]
(a) 2s
(b) 1s
(c) 2πs
(d) πs
Answer:
(d) πs
Hint:
Acceleration, a = 16 m/s2
displacement, y = 4m
According to SHM,
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 40

Question 11.
A hollow sphere is filled with water. It is hung by a long thread. As the water flows out of a hole at the bottom, the period of oscillation will
(a) first increase and then decrease
(b) first decrease and then increase
(c) increase continuously
(d) decrease continuously
Answer:
(a) first increase and then decrease

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 12.
The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are ………
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 41
Answer:
(c) \(\mathrm{kg} \mathrm{s}^{-1}\)
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 42

Question 13.
When a damped harmonic oscillator completes 100 oscillations, its amplitude is reduced to \(\frac{1}{3}\) of its initial value. What will be its amplitude when it completes 200 oscillations?
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 43
Answer:
(d) \(\frac{1}{9}\)
Hint:
In damped vibration, amplitude at any instant t is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 44

Question 14.
Which of the following differential equations represents a damped harmonic oscillator?
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 45
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 46

Question 15.
If the inertial mass and gravitational mass of the simple pendulum of length / are not equal, then the time period of the simple pendulum is …..
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 47
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 48

Samacheer Kalvi 11th Physics Oscillations Short Answer Questions

Question 1.
What is meant by periodic and non-periodic motion? Give any two examples, for each motion.
Answer:
Periodic motion: Any motion which repeats itself in a fixed time interval is known as periodic motion. Examples: Hands in pendulum clock, swing of a cradle.
Non-Periodic motion: Any motion which does not repeat itself after a regular interval of time is known as non-periodic motion. Example: Occurrence of Earth quake, eruption of volcano.

Question 2.
What is meant by force constant of a spring?
Answer:
The force constant (or) spring factor is defined as the restoring force produced per unit displacement.

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 3.
Define time period of simple harmonic motion.
Answer:
Time period: The time period is defined as the time taken by a particle to complete one oscillation. It is usually denoted by T. For one complete revolution, the time taken is t = T, therefore,
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 49

Question 4.
Define frequency of simple harmonic motion.
Answer:
The number of oscillations produced by the particle per second is called frequency. It is denoted by f. SI unit for frequency is s-1 or hertz (Hz).
Mathematically, frequency is related to time period by f = \(\frac{1}{\mathrm{T}}\)

Question 5.
What is an epoch?
Answer:
The phase of a vibrating particle corresponding to time t = 0 is called initial phase or epoch. At, t = φ, φ = φ0.
The constant φ0 is called initial phase or epoch. It tells about the initial state of motion of the vibrating particle.

Question 6.
Write short notes on two springs connected in series.
Answer:
Phase: The phase of a vibrating particle at any instant completely specifies the state of the particle. It expresses the position and direction of motion of the particle at that instant with respect to its mean position the phase of the vibrating particle. At time t = 0 s (initial time), the phase φ = φ0 is called epoch (initial phase) where φ0 is called the angle of epoch.

Question 7.
Write short notes on two springs connected in parallel.
Answer:
When two or more springs are connected in parallel, we can replace (by removing) all these springs with an equivalent spring (effective spring) whose net effect is same as if all the springs are in parallel connection.

Question 8.
Write down the time period of simple pendulum.
Answer:
The angular frequency of this oscillator (natural frequency of this system) is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 50
The frequency of oscillations is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 51
and time period of oscillations is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 52

Question 9.
State the laws of simple pendulum?
Answer:
Law of length: For a given value of acceleration due to gravity, the time period of a simple pendulum is directly proportional to the square root of length of the pendulum.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 53
Law of acceleration: For a fixed length, the time period of a simple pendulum is inversely proportional to square root of acceleration due to gravity.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 54

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 10.
Write down the equation of time period for linear harmonic oscillator.
Answer:
From Newton’s second law, we can write the equation for the particle executing simple harmonic motion
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 55
Comparing the equation with simple harmonic motion equation, we get
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 56
which means the angular frequency or natural frequency of the oscillator is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 57
The frequency of the oscillation is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 58
and the time period of the oscillation is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 59

Question 11.
What is meant by free oscillation?
Answer:
When the oscillator is allowed to oscillate by displacing its position from equilibrium position, it oscillates with a frequency which is equal to the natural frequency of the oscillator. Such an oscillation or vibration is known as free oscillation or free vibration.

Question 12.
Explain damped oscillation. Give an example.
Answer:
The oscillations in which the amplitude decreases gradually with the passage of time are called damped Oscillations.
Example:

  1. The oscillations of a pendulum or pendulum oscillating inside an oil filled container.
  2. Electromagnetic oscillations in a tank circuit.
  3. Oscillations in a dead beat and ballistic galvanometers.

Question 13.
Define forced oscillation. Give an example.
Answer:
The body executing vibration initially vibrates with its natural frequency and due to the presence of external periodic force, the body later vibrates with the frequency of the applied periodic force. Such vibrations are known as forced vibrations.
Example: Sound boards of stringed instruments.

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 14.
What is meant by maintained oscillation? Give an example.
Answer:
While playing in swing, the oscillations will stop after a few cycles, this is due to damping. To avoid damping we have to supply a push to sustain oscillations. By supplying energy from an external source, the amplitude of the oscillation can be made constant. Such vibrations, are known as maintained vibrations.
Example: The vibration of a tuning fork getting energy from a battery or from external power supply.

Question 15.
Explain resonance. Give an example.
Answer:
The frequency of external periodic force (or driving force) matches with the natural frequency of the vibrating body (driven). As a result the oscillating body begins to vibrate such that its amplitude increases at each step and ultimately it has a large amplitude. Such a phenomenon is known as resonance and the corresponding vibrations are known as resonance vibrations. Example: The breaking of glass due to sound

Samacheer Kalvi 11th Physics Oscillations Long Answer Questions

Question 1.
What is meant by simple harmonic oscillation? Give examples and explain why every simple harmonic motion is a periodic motion whereas the converse need not be true.
Answer:
Simple harmonic motion is a special type of oscillatory motion in which the acceleration or force on the particle is directly proportional to its displacement from a fixed point and is always directed towards that fixed point. In one dimensional case, let x be the .displacement of the particle and ax be the acceleration of the particle, then
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 60
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 61
where b is a constant which measures acceleration per unit displacement and dimensionally it is equal to T-2.
By multiplying by mass of the particle on both sides of equation (1) and from Newton’s second law, the force is
Fx = -kx …(3)
where k is a force constant which is defined as force per unit length. The negative sign indicates that displacement
and force (or acceleration) are in opposite directions. This means that when the displacement of the particle is taken towards right of equilibrium position (x takes positive value), the force (or acceleration) will point towards equilibrium (towards left) and similarly, when the displacement of the particle is taken towards left of equilibrium position (x takes negative value), the force (or acceleration) will point towards equilibrium (towards right). This type of force is known as restoring force because it always directs the particle executing simple harmonic motion to restore to its original (equilibrium or mean) position. This force (restoring force) is central and attractive whose center of attraction is the equilibrium position.
In order to represent in two or three dimensions, we can write using vector notation
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 62
where \(\vec{r}\) is the displacement of the particle from the chosen origin. Note that the force and displacement have a linear relationship. This means that the exponent of force \(\overrightarrow{\mathrm{F}}\) and the exponent of displacement \(\vec{r}\) are unity. The sketch between cause (magnitude of force | \(\overrightarrow{\mathbf{F}}\) |) and effect (magnitude of displacement | \(\vec{r}\) |) is a straight line passing through second and fourth quadrant.
By measuring slope \(\frac{1}{k}\), one can find the numerical value of force constant k.

Question 2.
Describe Simple Harmonic Motion as a projection of uniform circular motion.
Answer:
The projection of uniform circular motion on a diameter of SHM
Consider a particle of mass m moving with uniform speed v along the circumference of a circle whose radius is r in anti-clockwise direction (as shown in figure). Let us assume that the origin of the coordinate system coincides with the center O of the circle. If ω is the angular velocity of the particle and θ the angular displacement of the particle at any instant of time t, then θ = ωt. By projecting the uniform circular motion on its diameter gives a simple harmonic motion. This means that we can associate a map (or a relationship) between uniform circular (or revolution) motion to vibratory motion. Conversely, any vibratory motion or revolution can be mapped to uniform circular motion. In other words, these two motions are similar in nature.

Let us first project the position of a particle moving on a circle, on to its vertical diameter or on to a line parallel to vertical diameter as shown in figure. Similarly, we can do it for horizontal axis or a line parallel to horizontal axis.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 63
As a specific example, consider a spring mass system (or oscillation of pendulum). When the spring moves up and down (or pendulum moves to and fro), the motion of the mass or bob is mapped to points on the circular motion.
Thus, if a particle undergoes uniform circular motion then the projection of the particle on the diameter of the circle (or on a line parallel to the diameter) traces straight line motion which is simple harmonic in nature. The circle is known as reference circle of the simple harmonic motion. The simple harmonic motion can also be defined as the motion of the projection of a particle on any diameter of a circle of reference.

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 3.
What is meant by angular harmonic oscillation? Compute the time period of angular harmonic oscillation.
Answer:
Time period and frequency of angular SHM:
When a body is allowed to rotate freely about a given axis then the oscillation is known as the angular oscillation. The point at which the resultant torque acting on the body is taken to be zero is called mean position. If the body is displaced from the mean position, then the resultant torque acts such that it is proportional to the angular displacement and this torque has a tendency to bring the body towards the mean position.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 64
Let \(\vec{\theta}\) be the angular displacement of the body and the resultant torque \(\vec{\tau}\) acting on the body is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 25
\(\kappa\) is the restoring torsion constant, which is torque per unit angular displacement. If I is the
moment of inertia of the body and a is the angular acceleration then
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 65

Question 4.
Write down the difference between simple harmonic motion and angular simple harmonic motion.
Answer:
Comparision of simple harmonic motion and angular harmonic motion
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 66

Question 5.
Discuss the simple pendulum in detail.
Answer:
Simple pendulum: A pendulum is a mechanical system which exhibits periodic motion. It has a bob with mass m suspended by a long string (assumed to be massless and inextensible string) and the other end is fixed on a stand. At equilibrium, the pendulum does not oscillate and hangs vertically downward. Such a position is known as mean position or equilibrium position. When a pendulum is displaced through a small displacement from its equilibrium position and released, the bob of the pendulum executes to and fro motion. Let l be the length of the pendulum which is taken as the distance between the point of suspension and the centre of gravity of the bob. Two forces act on the bob of the pendulum at any displaced position.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 67
(i) The gravitational force acting on the body (\(\overrightarrow{\mathrm{F}}=m \vec{g}\)) which acts vertically downwards.
(ii) The tension in the string \(\overrightarrow{\mathrm{T}}\) which acts along the string to the point of suspension. Resolving the gravitational force into its components:
(a) Normal component: The component along the string but in opposition to the direction of tension, Fas = mg cos θ
(b) Tangential component: The component perpendicular to the string i.e., along tangential direction of arc of swing, Fps = mg sin θ
Therefore, The normal component of the force is, along the string,
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 68
From the figure, we can observe that the tangential component Wps of the gravitational force always points towards the equilibrium position i.e., the direction in which it always points opposite to the direction of displacement of the bob from the mean position. Hence, in this case, the tangential force is nothing but the restoring force. Applying Newton’s second law along tangential direction, we have
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 642
where, s is the position of bob which is measured along the arc. Expressing arc length in terms of angular displacement i.e.,
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 652
Because of the presence of sin 0 in the above differential equation, it is a non-linear differential equation (Here, homogeneous second order). Assume “the small oscillation approximation”, sin θ ≈ θ, the above differential equation becomes linear differential equation
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1515
This is the well known oscillatory differential equation. Therefore, the angular frequency of this oscillator (natural frequency of this system) is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 672

Question 6.
Explain the horizontal oscillations of a spring.
Answer:
Horizontal oscillations of a spring-mass system: Consider a system containing a block of mass m attached to a massless spring with stiffness constant or force constant or spring constant k placed on a smooth horizontal surface (frictionless surface) as shown in figure. Let x0 be the equilibrium position or mean position of mass m when it is left undisturbed. Suppose the mass is displaced through a small displacement x towards right from its equilibrium position and then released, it will oscillate back and forth about its mean position Let F be the restoring force (due to stretching of the spring) which is proportional to the amount of displacement of block. For one dimensional motion, mathematically, we have
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 682
where negative sign implies that the restoring force will always act opposite to the direction of the displacement. This equation is called Hooke’s law. Notice that, the restoring force is linear with the displacement (i.e., the exponent of force and displacement are unity). This is not always true; in case if we apply a very large stretching force, then the amplitude
of oscillations becomes very large (which means, force is proportional to displacement containing higher powers of x) and
therefore, the oscillation of the system ìs not linear and hence, it is called non-linear oscillation. We restrict ourselves only to linear oscillations throughout our discussions, which means Hooke’s law is valid (force and displacement have a linear relationship).
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 69
From Newton’s second law, we can write the equation for the particle executing simple harmonic motion
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 70
Comparing the equation with simple harmonic motion equation, we get
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 701
The frequency of the oscillation is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 71>
and the time period of the oscillation is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 712

Question 7.
Describe the vertical oscillations of a spring.
Answer:
Vertical oscillations of a spring: Let us consider a massless spring with stiff ness constant or force constant k attached to a ceiling as shown in figure. Let the length of the spring before loading mass m be L. If the block of mass m is attached to the other end of spring, then the spring elongates by a length l. Let F1 be the restoring force due to stretching of spring. Due to mass m, the gravitational force acts vertically downward. We can draw free-body diagram for this system as shown in figure. When the system is under equilibrium,
F1 + mg = 0 …(1)
But the spring elongates by small displacement /, therefore,
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 72
Substituting equation (2) in equation (1), we get
-kl + mg = 0
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 73
Suppose we apply a very small external force on the mass such that the mass further displaces downward by a displacement y, then it will oscillate up and down. Now, the restoring force due to this stretching of spring (total extension of spring is y +1) is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 74
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 75
The above equation is in the form of simple harmonic differential equation. Therefore, we get the time period as
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 77

Question 8.
Write short notes on the oscillations of liquid column in U-tube.
Answer:
Oscillation of liquid in a U-tube:
Consider a U-shaped glass tube which consists of two open arms with uniform cross sectional area A. Let us pour a non- viscous uniform incompressible liquid of density ρ in the U-shaped tube to a height h as shown in the figure. If the liquid and tube are not disturbed then the liquid surface will be in equilibrium position O.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 76
It means the pressure as measured at any point on the liquid is the same and also at the surface on the arm (edge of the tube on either side), which balances with the atmospheric pressure. Due to this the level of liquid in each arm will be the same. By blowing air one can provide sufficient force in one arm, and the liquid gets disturbed from equilibrium position O, which means, the pressure at blown arm is higher than the other arm. This creates difference in pressure which will cause the liquid to oscillate for a very short duration of time about the mean or equilibrium position and finally comes to rest. Time period of the oscillation is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 78

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 9.
Discuss in detail the energy in simple harmonic motion.
Answer:
Energy in simple harmonic motion:
(a) Expression for Potential Energy
For the simple harmonic motion, the force and the displacement are related by Hooke’s law
\(\overrightarrow{\mathrm{F}}=-k \vec{r}\)
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 79
Since force is a vector quantity, in three dimensions it has three components.
Further, the force in the above equation is a conservative force field; such a force can be derived from a scalar function which has only one component. In one dimensional case
F = — kx …… (1)
The work done by the conservative force field is independent of path. The potential energy U can be calculated from the following expression.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 80
Comparing (1) and (2), we get
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 81
This work done by the force F during a small displacement dx stores as potential energy
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 82
where ω is the natural frequency of the oscillating system. For the particle executing simple harmonic motion from equation y = A sift ωt, we get
x = A sin ωt
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 83
This variation of U is shown in figure.

(b) Expression for Kinetic Energy
Kinetic energy
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 84
Since the particle is executing simple harmonic motion, from equation
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 85
This variation with time is shown in figure.

(c) Expression for Total Energy
Total energy is the sum of kinetic energy and potential energy
E = KE + U …(11)
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 86
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 87
Alternatively, from equation (5) and equation (10), we get the total energy as
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 88
which gives the law of conservation of total energy.
Thus the amplitude of simple harmonic oscillator, can be expressed in terms of total energy.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 89

Question 10.
Explain in detail the four different types of oscillations.
Answer:
Free oscillations: When the oscillator is* allowed to oscillate by displacing its position from equilibrium position, it oscillates with a frequency which is equal to the natural frequency of the oscillator. Such an oscillation or vibration is known as free oscillation or free vibration. In this case, the amplitude, frequency and the energy of the vibrating object remains constant.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 90
Examples:

  1. Vibration of a tuning fork.
  2. Vibration in a stretched string.
  3. Oscillation of a simple pendulum.
  4. Oscillations of a spring-mass system.

Damped oscillations: During the oscillation of a simple pendulum (in previous case), we have assumed that the amplitude of the oscillation is constant and also the total energy of the oscillator is constant. But in reality, in a medium, due to the presence of friction and air drag, the amplitude of oscillation decreases as time progresses. It implies that the oscillation is not sustained and the energy of the SHM decreases gradually indicating the loss of energy.

The energy lost is absorbed by the surrounding medium. This type of oscillatory motion is known as damped oscillation. In other words, if an oscillator moves in a resistive medium, its amplitude goes on decreasing and the energy of the oscillator is used to do work against the resistive medium. The motion of the oscillator is said to be damped and in this case, the resistive force (or damping force) is proportional to the velocity of the oscillator.

Examples:

  1. The oscillations of a pendulum (including air friction) or pendulum oscillating inside an oil filled container.
  2. Electromagnetic oscillations in a tank circuit.
  3. Oscillations in a dead beat and ballistic galvanometers.

Maintained oscillations: While playing in swing, the oscillations will stop aft er a few cycles, this is due to damping. To avoid damping we have to supply a push to sustain oscillations. By supplying energy from an external source, the amplitude of the oscillation can be made constant. Such vibrations are known as maintained vibrations.
Example: The vibration of a tuning fork getting energy from a battery or from external power supply.
Forced oscillations: Any oscillator driven by an external periodic agency to overcome the damping is known as forced oscillator or driven oscillator. In this type of vibration, the body executing vibration initially vibrates with its natural frequency and due to the presence of external periodic force, the body later vibrates with the frequency of the applied periodic force. Such vibrations are known as forced vibrations.
Example: Sound boards of stringed instruments.

Samacheer Kalvi 11th Physics Oscillations Numerical Problems

Question 1.
Consider the Earth as a homogeneous sphere of radius R and a straight hole is bored in it through its centre. Show that a particle dropped into the hole will execute a simple harmonic motion such that its time period is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 92
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 91
Answer:
Oscillations of a particle dropped in a tunnel along the diameter of the earth.
Consider earth to be a sphere of radius R and centre O. A straight tunnel is dug along the diameter of the earth. Let ‘g’ be the value of acceleration due to gravity at the surface of the earth.
Suppose a body of mass ‘m’ is dropped into the tunnel and it is at point R i.e., at a depth d below the surface of the earth at any instant.
If g’ is acceleration due to gravity at P.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 93
If y is distance of the body from the centre of the earth, then
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 955
Negative sign indicates that the force acts in the opposite direction of displacement.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 956

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 2.
Calculate the time period of the oscillation of a particle of mass m moving in the potential defined as Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 96
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 961

Question 3.
Consider a simple pendulum of length l = 0.9 m which is properly placed on a trolley rolling down on a inclined plane which is at θ = 45° with the horizontal. Assuming that the inclined plane is frictionless, calculate the time period of oscillation of the simple pendulum.
Answer:
Length of the pendulum l = 0.9 m
Inclined angle θ = 45°
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 97

Question 4.
A piece of wood of mass m is floating erect in a liquid whose density is p. If it is slightly pressed down and released, then executes simple harmonic motion. Show that its time period of oscillation is Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 98
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 99

Question 5.
Consider two simple harmonic motion along x and y-axis having same frequencies but different amplitudes as x = A sin (ωt + φ) (along x axis) and y = B sin ωt (along y axis).
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1516
Note: when a particle is subjected to two simple harmonic motion at right angle to each other the particle may move along different paths. Such paths are called Lissajous figures.
Answer:
(a) \(y=\frac{\mathrm{B}}{\mathrm{A}} x\), equation is a straight line passing through origin with positive slope.
(b) \(y=-\frac{B}{A} x\), equation is a straight line passing through origin with negative slope.
(c) \(\frac{x^{2}}{\mathrm{A}^{2}}+\frac{y^{2}}{\mathrm{B}^{2}}=1\), equation is an ellipse whose center is origin. A2 B2
(d) \(x^{2}+y^{2}=\mathrm{A}^{2}\), equation is a circle whose center is origin.
(e) \(\frac{x^{2}}{\mathrm{A}^{2}}+\frac{y^{2}}{\mathrm{B}^{2}}-\frac{2 x y}{\mathrm{AB}} \frac{1}{\sqrt{2}}=\frac{1}{2}\), equation is an ellipse which (oblique ellipse which means tilted ellipse)

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 6.
Show that for a particle executing simple harmonic motion
(a) the average value of kinetic energy is equal to the average value of potential energy.
(b) average potential energy = average kinetic energy = \(\frac{1}{2}\) (total energy)
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 102
Suppose a particle of mass m executes SHM of period T. The displacement of the particles at any instant t is given by y = A sin ωt
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 103
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 104

Question 7.
Compute the time period for the following system if the block of mass m is slightly displaced vertically down from Its equilibrium position and then released. Assume that the pulley is light and smooth, strings and springs are light.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 105
Case (a): Pulley is fixed rigidly here. When the mass displace by y and the spring will also stretch by y. Therefore, F = T = ky
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1051
Case (b): Mass displace by y, pulley also displaces by y. T = 4ky
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 106

Samacheer Kalvi 11th Physics Oscillations Additional Questions 

Question 1.
The total energy of a particle vibrating in SHM is proportional to the square of its ……..
(a) velocity
(b) acceleration
(c) amplitude
(d) none of these
Answer:
(a) veloctiy

Question 2.
In order to double the period of a simple pendulum ……….
(a) its length should doubled
(b) its length should be quadrupled.
(c) the mass of its bob should be doubled.
(d) the mass of its bob should be quadrupled.
Answer:
(b) its length should be quadrupled.

Question 3.
A simple harmonic oscillator has amplitude A and time period T. Its maximum speed is …….
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 107
Answer:
(d) \(\frac{2 \pi \mathrm{A}}{\mathrm{T}}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 108

Question 4.
A simple harmonic oscillator has a period of 0.01 s and an amplitude of 0.2 m. The magnitude of the velocity in m/s at the centre of oscillation is ……..
(a) 20π
(b) 40π
(c) 60π
(d) 80π
Answer:
(b) 40π
Solution:
Velocity is maximum at the centre of oscillation and is given by
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 109

Question 5.
A particle is executing SHM. Then the graph of acceleration as a function of displacement is ……..
(a) straight line
(b) circle
(c) ellipse
(d) hyperbola
Answer:
(a) straight line
Solution:
In SHM, F ∝ y ⇒ a ∝ y; Thus the graph is a straight line.

Question 6.
A particle is executing SHM. Then the graph of velocity as a function of displacement is …….
(a) straight line
(b) circle
(c) ellipse
(d) hyperbola
Answer:
(c) ellipse
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 110

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 7.
The amplitude of a vibrating body situated in a resisting medium ………
(a) decreases linearly with time
(b) decrease exponentially with time
(c) decreases with time in some other manner
(d) remains constant with time
Answer:
(b) decreases exponentially with time

Question 8.
The frequency of a vibrating body situated in air …….
(a) is the same as its natural frequency
(b) is higher than its natural frequency
(c) is lower than its natural frequency
(d) can have any value
Answer:
(c) is lower than its natural frequency

Question 9.
The equation Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1101 represents the equation of motion for a ……. vibration
(a) free
(b) damped
(c) forced
(d) resonant
Answer:
(b) damped

Question 10.
The displacement equation of an oscillator is y = 5 sin (0.2 7 πt + 0.5 π) in SI units. The time period of oscillation is
(a) 10 s
(b) 1 s
(c) 0.2 s
(d) 0.5 s
Answer:
(a) 10 s
Solution:
Comparing with the standard equation y = A sin (ωt + φ)
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 112

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 11.
A loaded spring vibrates with a period T. The spring is divided into four equal parts and the same load is suspended from one as these parts. The new time period is ………
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 121
Answer:
(b) \(\frac{\mathrm{T}}{2}\)
Solution:
Let the force constant of the spring be k. Then T = \(2 \pi \sqrt{\frac{m}{k}}\)
If the spring is divided into four equal parts, then the force constant of each part will be 4k.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 13

Question 12.
The vertical extension in a light spring by a weight of 1 kg, in equilibrium is 9.8 cm. The period of oscillation of the spring, in seconds, will be ……
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 141
Answer:
(a) \(\frac{2 \pi}{10}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 151

Question 13.
A particle executing SHM has an acceleration of 64 cm/s2 with its displacement is 4 cm. Its time period, in seconds is ……
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 161
Answer:
(a) \(\frac{\pi}{2}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 171

Question 14.
A body executes SHM with an amplitude A. Its energy is half kinetic and half potential when the displacement is ……..
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 181
Answer:
\(\frac{\mathrm{A}}{\sqrt{2}}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 191

Question 15.
The maximum displacement of a particle executing SHM is 1 cm and the maximum acceleration is (1.57)2 cm/s2. Its time period is …..
(a) 0.25s
(b) 4.0s
(c) 1.57s
(d) 3.14s
Answer:
(b) 4.0s
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 20

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 16.
The velocity of a particle, undergoing SHM is v at the mean position. If its amplitude is doubled, the velocity at the mean position will be …….
(a) 2 v
(b) 3 v
(c) \(2 \sqrt{2} v\)
(d) 4v
Answer:
(a) 2 v
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 21

Question 17.
A girl is swinging on a swing in the sitting position. How will the period of swing be affected if she stands up?
(a) The period will now be shorter
(b) The period will now be longer
(c) The period will remain unchanged
(d) The period may become longer or shorter depending upon the height of the girl
Answer:
(a) The period will now be shorter
Solution:
The effective value of I will decrease. Therefore, the time period will be shorter.

Question 18.
The equation of SHM of a particle is Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 22> where k is a positive constant. The time period of motion is given by …….
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 221
Answer:
(a) \(\frac{2 \pi}{\sqrt{k}}\)
Solution:
Here k is same as ω2

Question 19.
The amplitude of a damped oscillator becomes half in one minute. The amplitude after 3 minutes will be \(\frac{1}{x}\) time the original, where x is ………
(a) 2 × 3
(b) 23
(c) 32
(d) 3 × 22
Answer:
(b) 23
Solution:
The amplitude decreases exponentially with time and becomes half in 1 minute.
Amplitude after 3 minutes = \(\left(\frac{1}{2}\right)^{3}\) of the original value. Thus, x = 23

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 20.
When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a is ……..
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 23
Answer:
(c) \(\frac{a}{2}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 24

Question 21.
A massless spring, having force constant k, oscillates with a frequency n when a mass m is suspended from it. The spring is cut into two equal halves and a mass 2m is suspended from one of the parts. The frequency of oscillation will now be
For a simple pendulum the graph between length and time period will be …….
(a) n
(b) \(n \sqrt{2}\)
(c) \(\frac{n}{\sqrt{2}}\)
(d) 2n
Answer:
(a) n
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 251

Question 22.
For a simple pendulum the graph between the length and time period will be a ……….
(a) hyperbola
(b) Parabola
(c) Straight line
(d) none of these
Answer:
(b) Parabola

Question 23.
A particle is executing simple harmonic motion given by Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 26 The velocity of the particle when its displacement is 3 units is …….
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 261
Answer:
(a) 16 units
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 27

Question 24.
When a particle oscillates simple harmonically, its potential energy varies periodically. If the frequency of oscillation of the particle is n, the frequency of potential energy variation is ……
(a) \(\frac{n}{2}\)
(b) n
(c) 2n
(d) 4n
Answer:
(c) 2n

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 25.
A particle, moving along the x-axis, executes simple harmonic motion when the force acting on it is given by (A and k are positive constants.) …….
(a) – Akx
(b) A cos (kx)
(c) A exp (- kx)
(d) Akx
Answer:
(a) – Akx

Question 26.
The motion of a particle is expressed by the equation a = -bx, where x is the displacement from the mean position, a is the acceleration and b is a constant. The periodic time is ………
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 30
Answer:
(b) \(\frac{2 \pi}{\sqrt{b}}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 31
Question 27.
The angular velocity and the amplitude of a simple pendulum are ω and a, respectively. The ratio of its kinetic and potential energies at a displacement x from the mean position is ……….
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 32
Solution:
(d) \(\frac{a^{2}-x^{2}}{x^{2}}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 33

Question 28.
A particle is oscillating according to the equation x = 5 cos (0.5 π t) where t is in seconds. The particle moves from the position of equilibrium to the position of maximum displacement in time ………
(a) 1 s
(b) 2 s
(c) 0.5 s
(d) 4 s
Answer:
(a) 1 s
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 34
Time taken to move from the position of equilibrium to the position of maximum displacement is t = \(\frac{T}{4}\) = 1s

Question 29.
A seconds pendulum is placed in a space laboratory orbiting around the Earth at a height 3R from the Earth’s surface where R is the radius of the Earth. The time period of the pendulum will be ……..
(a) zero
(b) 2/3s
(c) 4 s
(d) infinite
Answer:
(d) infinite
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 35

Question 30.
A mass m is vertically suspended from a spring of negligible mass; the system oscillates with a frequency n. What will be the frequency of the system, if a mass 4m is suspended from the same spring?
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 36
Answer:
(a) \(\frac{n}{2}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 37

Question 31.
Two simple pendulums of lengths 0.5m and 2.0m respectively are given small linear displacement in one direction at the same time. They will again be in phase when the pendulum of shorter length has completed …….. oscillations.
(a) 5
(b) 3
(c) 1
(d) 2
Answer:
(d) 2
Solution:
The time period of the shorter pendulum is half that of the longer pendulum. Therefore, the pendulums will again be in phase ( at the mean position). When the shorter pendulum has completed 2 oscillations.

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 32.
A body is executing simple harmonic motion with an angular frequency 2 rod/sec. The velocity of the body at 20mm displacement, when the amplitude of motion is 60mm, is
(a) 90mm/s
(b) 113mm/s
(c) 118 mm/s
(d) 131 mm/s
Answer:
(b) 113 mm/s
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 38

Question 33.
If the displacement of a particle executing SHM, is given by y = 0.30 sin (220t + 0.64) in metre, then the frequency and the maximum velocity of the particle are (t is in seconds)
(a) 35 Hz, 66m/s
(b) 45 Hz, 66 m/s
(c) 58 Hz, 113 m/s
(d) 35 Hz, 132 m/s
Answer:
(a) 35 Hz, 66 m/s
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 39

Question 34.
The kinetic energy of a particle, executing SHM, is 16 J when it is at its mean position. If the amplitude of oscillations is 25 cm, and the mass of the particle is 5.12 kg, the time period of its oscillation is …….
(a) π/5 s
(b) 2π s
(c) 20π s
(d) 5π s
Answer:
(a) π/5 s
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 401

Question 35.
A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is V(x) = kx2 .Where k is a positive constant. If the amplitude of oscillation is a, then its time period T is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 412
Answer:
(b) independent of a
Solution:
Since V(x) = Kx2, the motion is simple harmonic. In SHM, the time period is independent of the amplitude of oscillation.

Question 36.
The amplitude of a damped oscillation reduces to one third of its original value a0 in 20s. The amplitude of such oscillation after a period of 40s will be ……..
(a) a0/9
(b) a0/6
(c) a0/2
(d) a0/27
Answer:
(a) a0/9
Solution:
In the first 20s, the amplitude reduces to one-third of the original value, i.e., to a0/3, In the next 20s, it will reduce to one-third of the reduced value, i.e., to a0/9.

Question 37.
Masses mA and mB hanging from the ends of strings of lengths lA and lB are executing. Simple harmonic motions. If their frequencies are related as fA= 2fB, then ……
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 421
Answer:
(c) lA = lB/4 regardless of masses.
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 431

Question 38.
Two simple harmonic motions act on a particle. These harmonic motions are x = A cos (ωt + δ); y = A cos (ωt + α) When Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1517 the resulting motion is …….
(a) A circle and the actual motion is clockwise
(b) an ellipse and the actual motion is counter clockwise
(c) a ellipse and the actual motion is clockwise
(d) a circle and the actual motion is counter clockwise
Answer:
(d) a circle and the actual motion is counter clockwise
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 441
Solution:
Squaring and adding the two equations, We get x2 + y2 = A2. This is an equation of a circle. Hence the resultant motion is circular. The motion is counter clockwise.

Question 39.
A metal bob is suspended from a coiled spring. When set into vertical vibrations on the earth. It oscillates up and down with frequency f If the same experiment is carried out in a satellite circling the Earth the frequency of vibration will be …….
(a) f
(b) zero
(c) infinite
(d) depend on the distance of the satellite from the earth.
Answer:
(a) f
Solution:
The frequency of oscillation of a mass spring system depends only on the mass and the spring constant.

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 40.
In forced oscillations of a particle, the amplitude is maximum for a frequency ω1 of the force, while the energy is maximum for a frequency ω2 of the force. Then ……
(a) ω1 < ω2
(b) ω1 > ω2 when damping is small and ω1 > ω2 when damping is large.
(c) ω1 > ω2
(d) ω1 = ω2
Answer:
(d) ω1 = ω2

Question 41.
Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion?
(a) When v is maximum, a is maximum
(b) Value of a is zero, whatever may be the value of v
(c) When v is zero, a is zero
(d) When v is maximum, a is zero
Answer:
(d) When v is maximum, a is zero
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 120

Question 42.
The function sin2 (ωt) represents
(a) a simple harmonic motion with a period π/ω
(b) a simple harmonic motion with a period 2π/ω
(c) a periodic, but not simple harmonic motion with a period π/ω
(d) a periodic, but not simple harmonic motion with a period 2π/ω
Answer:
(a) a simple harmonic motion with a period π/ω
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1211

Question 43.
A particle executing simple harmonic motion has a kinetic energy K0 cos2 ωt. The maximum
values of the potential energy and the total energy are, respectively ….
(a) k0/2 and k0
(b) k0 and 2k0
(c) k0 and k0
(d) 0 and 2k0
Answer:
(c) k0 and k0

Question 44.
A particle executing simple harmonic motion of amplitude 31.4 cm/s. The frequency of its oscillation is ………
(a) 3 Hz
(b) 4 Hz
(c) 2 Hz
(d) 1 Hz
Answer:
(d) 1 Hz
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 122

Question 45.
The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is ……..
(a) 0.5π
(b) π
(c) 0.707π
(d) zero
Answer:
(a) 0.5π

Question 46.
Which one of the following equations of motion represents simple harmonic motion?
(a) Acceleration = – k0x + k1x2
(b) Acceleration = – k(x + a)
(c) Acceleration = k(x + a)
(d) Acceleration = kx
Answer:
(b) Acceleration = k(x + a)

Question 47.
Which of the following functions represent SHM?
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 123
(a) I and III
(b) I and II
(c) only I
(d) I, II and III
Answer:
(a) I and III

Question 48.
Two simple harmonic motions of angular frequencies 100 and 1000 rad/s have the same displacement amplitude. The ratio of their maximum accelerations is ……….
(a) 1 : 10
(b) 1 : 102
(c) 1 : 103
(d) 1 : 104
Answer:
(b) 1 : 10
Solution:
The magnitude of the maximum acceleration is given by
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 124

Question 49.
The period of oscillation of a simple pendulum is T in a stationary lift. If the lift moves upwards with an acceleration of 8g, the period will ……..
(a) remain the same
(b) decrease by T/2
(c) increase by T/3
(d) none of these
Answer:
(c) increase by T/3
Solution:
Thus, the new time period is T/3. Hence the correct option is (d).

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 50.
A simple harmonic oscillator consist of a particle of mass m and an ideal spring with spring constant k The particle oscillates with a time period T. The spring is cut into two equal parts. If one part oscillates with the same particle, the time period will be …….
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 125
Answer:
(b) \(\frac{\mathrm{T}}{\sqrt{2}}\)
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1251
If the spring is cut into two equal parts, the force constant of each part becomes 2k. Therefore,
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1256

Question 51.
A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is …….
(a) 3 Hz
(b) 4 Hz
(c) 2 Hz
(d) 1 Hz
Answer:
(d) 1 Hz
Solution:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 126

Samacheer Kalvi 11th Physics Oscillations 2 Marks Questions

Question 1.
What is meant by Oscillatory motion?
Answer:
When an object or a particle moves back and forth repeatedly for some duration of time, its – motion is said to be oscillatory (or vibratory).

Question 2.
What is meant by simple harmonic motion (SHM)?
Answer:
A particle is said to execute simple harmonic motion if it moves to and fro about a mean position under the action of a restoring force which is directly proportional to its displacement from the mean position and is always directed towards the mean position.

Question 3.
What is meant by displacement in SHM?
Answer:
The distance travelled by the vibrating particle at any instant of time t from its mean position is known as displacement.
y = A sin ωt
The maximum displacement from the mean position is known as amplitude (A) of the vibrating particle.

Question 4.
Define velocity in SHM.
Answer:
Velocity: The rate of change of displacement is velocity.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 127

Question 5.
Define acceleration in SHM.
Answer:
Acceleration: The rate of change of velocity is acceleration.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 128

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 6.
What is meant by phase in SHM?
Answer:
Phase: The phase of a vibrating particle at any instant completely specifies the state of the particle. It expresses the position and direction of motion of the particle at that instant with respect to its mean position.

Question 7.
What is meant by angular oscillation?
Answer:
When a body is allowed to rotate freely about a given axis then the oscillation is known as the angular oscillation.

Question 8.
Define Amplitude.
Answer:
The maximum displacement of the oscillating particle on either side of its mean position is called its amplitude.

Samacheer Kalvi 11th Physics Oscillations 3 Marks Questions

Question 1.
Derive the expression for resultant spring constant when two springs having constant k1 and k2 are connected in series.
Answer:
Let x1 and x2 be the elongation of springs from their equilibrium position (un-stretched position) due to applied force F. Then, the net displacement of the mass point is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 135
Therefore, substituting equation (3) in equation (2), the effective spring constant can be calculated as
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 136
Suppose we have n springs connected in series, the effective spring constant in series is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 137
If all spring constants are identical i.e., k1 = k2 = … = kn = k then
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 138
This means that the effective spring constant reduces by the factor n. Hence, for springs in series connection, the effective spring constant is lesser than the individual spring constants. From equation (3), we have,
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 139>
Then the ratio of compressed distance or elongated distance x1 and x2 is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 140
The elasticity potential energy stored in first and second springs are Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1412 respectively. Then their ratio is
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1413

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 2.
Derive the expression for resultant spring constant when two springs having constant k1 and k2 are connected in parallel.
Answer:
k1 and k2 attached to a mass m as shown in figure. The results can be generalized to any number of springs in parallel.
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 142
Let the force F be applied towards right as shown in figure. In this case, both the springs elongate or compress by the same amount of displacement. Therefore, net force for the displacement of mass m is
F = – kpX …(1)
where kp is called effective spring constant.
Let the first spring be elongated by a displacement x due to force F1 and second spring be elongated by the same displacement x due to force F2, then the net force
F = – k1x – k2x …..(2)
Equating equations (2) and (1), we get
kp = k1 + k2…..(2)
Generalizing, for n springs connected in parallel,
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 143
If all spring constants are identical i.e., k1 = kk2 = … = kn = k then
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 144
This implies that the effective spring constant increases by a factor n. Hence, for the springs in parallel connection, the effective spring constant is greater than individual spring constant.

Samacheer Kalvi 11th Physics Oscillations Numerical Problems

Question 1.
A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum oscillator. The acceleration of the bob of the pendulum is 20 ms-2 at a distance of 5 m from the mean position. To find the time period of oscillation.
Answer:
Given; a = 20ms-2 ; y = 5 m a = ω2y
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 145

Question 2.
The acceleration due to gravity on the surface of the moon is 1.7 ms-2. What is the time period of simple pendulum on the moon if its time period on the earth is 3.5 s? Give g on Earth = 9.8 ms-2
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 146

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 3.
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then find its time period in seconds.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 147

Question 4.
A body of mass m is attached to lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5s. Find the value of m is kg.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 148

Question 5.
Two simple harmonic motions are represented by the equations:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 149
What is the ratio of their amplitudes ?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 150
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 1512

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 6.
A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 Nm-1. The block is pulled to a distance x = 10 cm from its equilibrium position at t = 0 on a frictionless surface from rest at t = 0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 152

Question 7.
A 5 kg collar is attached to a spring of force constant 500 Nm-1. It slides without friction on a horizontal rod as shown in figure. The collar is displaced from its equilibrium position by 10.0 cm and released.
Calculate:
(i) the period of oscillation
(ii) the maximum speed, and
(iii) the maximum acceleration of the collar.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 153

Question 8.
A 0.2 kg of mass hangs at the end of a spring. When 0.02 kg more mass is added to the end of the spring, it stretches 7 cm more. If the 0.02 kg mass is removed, what will be the period of vibration of the system?
Answer:
When 0.02 kg mass is added, the spring streches by 7 cm
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 154
When 0.02 kg mass ia removed, the period of vibration will be
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 155

Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

Question 9.
A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. What is the ratio m/M?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 156

Question 10.
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2s. Find the acceleration and velocity of the body when the displacement is 5 cm.
Answer:
Here A = 5 cm, T = 0.2s
Velocity and acceleration at any displacement x are given by
Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations 157

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4

Question 1.
Write the Maclaurin series expansion of the following functions:
(i) ex
(ii) sin x
(iii) cos x
(iv) log (1 – x); -1 ≤ x < 1
(v) tan-1 (x) ; -1 ≤ x ≤ 1
(vi) cos2 x
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 1
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 2
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 3
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 4

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4

(vi) f(x) = cos2 x
f(0) = 1
f'(x) = 2 cos x (- sin x) = – sin 2x
f'(0) = 0
f”(x) = (-cos 2x)(2)
f”(0) = -2
f”'(x) = -2[- sin 2x](2) = 4 sin 2x
f”'(0) = 0
f4 (x) = 4(cos 2x)(2) = 8 cos 2x
f4 (0) = 8
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 5

Question 2.
Write down the Taylor series expansion, of the function log x about x = 1 upto three non-zero terms for x > 0.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 6

Question 3.
Expand sin x in ascending powers x – \(\frac{\pi}{4}\) upto three non-zero terms.
Solution:
f (x) = sin x
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 7
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 8

Question 4.
Expand the polynomial f(x) = x2 – 3x + 2 in powers of x – 1
Solution:
f(x) = x2 – 3x + 2 = (x – 1) (x – 2)
f(1) = 0
f'(x) = 2x – 3 ; f'(1) = -1
f”(x) = 2 ; f”(1) = 2
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 9

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 Additional Problems

Question 1.
The Taylor’s series expansion of f(x) = sin x about x = \(\frac{\pi}{2}\) is obtained by the following way.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 10

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4

Question 2.
Obtain the Maclaurin’s series expansion for the following functions.
(i) ex
(ii) sin2 x
(iii) \(\frac{1}{1+x}\)
Solution:
(i)
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 12
(ii)
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 13
(iii)
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 14

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Students can Download Physics Chapter 6 Gravitation Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Samacheer Kalvi 11th Physics Gravitation Textual Evaluation Solved

Samacheer Kalvi 11th Physics Gravitation Multiple Choice Questions

Question 1.
The linear momentum and position vector of the planet is perpendicular to each other at
(a) perihelion and aphelion
(b) at all points
(c) only at parihelion
(d) no point
Answer:
(a) perihelion and aphelion

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 2.
If the masses of the Earth and Sun suddenly double, the gravitational force between them will
(a) remain the same
(b) increase 2 times
(c) increase 4 times
(d) decrease 2 times
Answer:
(c) increase 4 times
Solution:
The gravitation force of attraction is given by F = Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 1391
If the masses are doubled then the force will be
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 1392
The gravitational force between them will be increase 4 times.

Question 3.
A planet moving along an elliptical orbit is closest to the Sun at distance r1 and farthest away at a distance of r2 If V1 and v2 are linear speeds at these points respectively. Then the ratio \(\frac{v_{1}}{v_{2}}\) is ……. [NEET 2016]
Answer:
(a) \(\frac{r_{2}}{r_{1}}\)
Solution:
According to the Law of conservation of angular momentum
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 13933

Question 4.
The time period of a satellite orbiting Earth in a circular orbit is independent of ……..
(a) Radius of the orbit
(b) The mass of the satellite
(c) Both the mass and radius of the orbit
(d) Neither the mass nor the radius of its orbit
Answer:
(b) The mass of the satellite

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 5.
If the distance between the Earth and Sun were to be doubled from its present value, the number of days in a year would be
(a) 64.5
(b) 1032
(c) 182.5
(d) 730
Answer:
(b) 1032
Solution:
By Kepler’s law
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 1394
If the distance between the Earth and Sun were to be doubled from its present Value
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 400

Question 6.
According to Kepler’s second law, the radial vector to a planet from the Sun sweeps out equal areas in equal intervals of time. This law is a consequence of
(a) conservation of linear momentum
(b) conservation of angular momentum
(c) conservation of energy
(d) conservation of kinetic energy
Answer:
(b) conservation of angular momentum

Question 7.
The gravitational potential energy of the Moon with respect to Earth is
(a) always positive
(b) always negative
(c) can be positive or negative
(d) always zero.
Answer:
(b) always negative

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 8.
The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then ….. [NEET 2018]
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 13920
(a) KA > KB > KC
(b) KB < KA < KC
(c) KA < KB < KC
(d) KB > KA > KC
Answer:
(a) KA > KB > KC
Solution:
The kinetic energy of a planet becomes
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 21
So, kinetic energy is inversly proportional to the orbital distance. Where the distance will be closer EK is larger and the distance will be larger EK is less.

Question 9.
The work done by the Sun’s gravitational force on the Earth is ….
(a) always zero
(b) always positive
(c) can be positive or negative
(d) always negative
Answer:
(c) can be positive or negative

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 10.
If the mass and radius of the Earth are both doubled, then the acceleration due to gravity g1
(a) remains same
(b) \(\frac{g}{2}\)
(c) 2g
(d) 4g
Answer:
(b) \(\frac{g}{2}\)
Solution:
Acceleration due to gravity g’ = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
If the mass and radius of the Earth are both doubled
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 22

Question 11.
The magnitude of the Sun’s gravitational field as experienced by Earth is …..
(a) same over the year
(b) decreases in the month of January and increases in the month of July
(c) decreases in the month of July and increases in the month of January
(d) increases during day time and decreases during night time.
Answer:
(c) decreases in the month of July and increases in the month of January

Question 12.
If a person moves from Chennai to Trichy, his weight …..
(a) increases
(b) decreases
(c) remains same
(d) increases and then decreases
Answer:
(b) decreases

Question 13.
An object of mass 10 kg is hanging on a spring scale which is attached to the roof of a lift. If
the lift is in free fall, the reading in the spring scale is ……..
(a) 98 N
(b) zero
(c) 49 N
(d) 9.8 N
Answer:
(b) zero
Solution:
The lift is in freefall. It and its contents will experience apparent weightlessness just like astronauts.
The spring balance reading will change from 100 N to zero.

Question 14.
If the acceleration due to gravity becomes 4 times its original value, then escape speed
(a) remains same
(b) 2 times of original value
(c) becomes halved
(d) 4 times of original value
Answer:
(b) 2 times of original value
Solution:
Escape speed 4 times of ‘g’
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 30

Question 15.
The kinetic energy of the satellite orbiting around the Earth is
(a) equal to potential energy
(b) less than potential energy
(c) greater than kinetic energy
(d) zero
Answer:
(b) less than potential energy

Samacheer Kalvi 11th Physics Gravitation Short Answer Questions

Question 1.
State Kepler’s three laws.
Answer:
1. Law of Orbits: Each planet revolves moves around the Sun in an elliptical orbit with the Sun at one of the foci of the ellipse.
2. Law of area: The radial vector line joining the Sun to a planet sweeps equal areas in equal intervals of time.
3. Law of period: The square of the time period of revolution of a planet around the Sun in its elliptical orbit is directly proportional to the cube of the semi-major axis of the ellipse.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 25

Question 2.
State Newton’s Universal law of gravitation.
Answer:
Newton’s law of gravitation: States that the gravitational force between two masses is directly proportional to product of masses and inversely proportional to square of the distance between the masses.
In the mathematical form, it can be written as,
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 26

Question 3.
Will the angular momentum of a planet be conserved? Justify your answer.
Answer:
The triumph of the law of gravitation is that it concludes that the mango that is falling down and the Moon orbiting the Earth are due to the same gravitational force.

Question 4.
Define the gravitational field. Give its unit.
The gravitational field intensity \(\overrightarrow{\mathrm{E}}_{1}\) at a point is defined as the gravitational force experienced by unit mass at that point. It’s unit N kg-1.

Question 5.
What is meant by superposition of gravitational field?
Answer:
The total gravitational field at a point due to all the masses is given by the vector sum of the gravitational field due to the individual masses. This principle is known as superposition of gravitational fields.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 266

Question 6.
Define gravitational potential energy.
Answer:
Potential energy of a body at a point in a gravitational field is the work done by an external agent in moving the body from infinity to that point.

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 7.
Is potential energy the property of a single object? Justify.
Answer:
There is no potential energy for a single object. The gravitational potential energy depends upon the two masses and the distance between them.

Question 8.
Define gravitational potential.
Answer:
The gravitational potential is defined as the amount of work required to bring unit mass from infinity to that point.

Question 9.
What is the difference between gravitational potential and gravitational potential energy?
Answer:

Gravitational Potential Gravitational potential Energy
The amount of work required to bring unit mass from infinity to that point in point an gravitational field. The work done by an external agent in moving the body from infinity to that in an gravitational field.
The unit of V(r) is J kg-1 The unit of U(r) is J (or) joule.

Question 10.
What is meant by escape speed in the case of the Earth?
Answer:
The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity force. This can be written as, ve = \(\sqrt{2 g R_{E}}\).

Question 11.
Why is the energy of a satellite (or any other planet) negative?
Answer:
The energy of satellite is negative. Because the energy implies that the satellite is bound to the Earth by means of the attractive gravitational force.

Question 12.
What are geostationary and polar satellites?
Answer:
Geostationary Satellite: It is the satellite which appears at a fixed position and at a definite height to an observer on earth.
Polar Satellite: It is the satellite which revolves in polar orbit around the earth.

Question 13.
Define weight.
Answer:
The weight of an object \(\overrightarrow{\mathrm{W}}\) is defined as the downward force whose magnitude W is equal to that of upward force that must be applied to the object to hold it at rest or at constant velocity relative to the earth. The direction of weight is in the direction of gravitational force. So the magnitude of weight of an object is denoted as, W = N = mg.

Question 14.
Why is there no lunar eclipse and solar eclipse every month?
Answer:
If the orbits of the Moon and Earth lie on the same plane, during full Moon of every month, we can observe lunar eclipse. If this is so dining new Moon we can observe solar eclipse. But Moon’s orbit is tilted 5° with respect to Earth’s orbit. Due to this 5° tilt, only during certain periods of the year, the Sun, Earth and Moon align in straight line leading to either lunar eclipse or solar eclipse depending on the alignment.

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 15.
How will you prove that Earth itself is spinning?
Answer:
The Earth’s spinning motion can be proved by observing star’s position over a night. Due to Earth’s spinning motion, the stars in sky appear to move m circular motion about the pole star.

Samacheer Kalvi 11th Physics Gravitation Long Answer Questions

Question 1.
Discuss the important features of the law of gravitation.
Answer:
1. As the distance between two masses increases, the strength of the force tends to decrease because of inverse dependence on r2. Physically it implies that the planet Uranus experiences less |F| gravitational force from the Sun than the Earth since Uranus is at larger distance from the Sun compared to the Earth.
2. The gravitational forces between two particles always constitute an action-reaction pair. It implies that the gravitational force exerted by the Sim on the Earth is always towards the Sun. The reaction-force is exerted by the Earth on the Sun. The direction of this reaction force is towards Earth.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 401
3. The torque experienced by the Earth due to the gravitational force of the Sun is given by
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 40
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 402. It implies that angular momentum \(\overrightarrow{\mathrm{L}}\) is a constant vector.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 41
4. The expression Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 411 has one inherent assumption that both M1, and M2 are treated as point masses. When it is said that Earth orbits around the Sun due to Sun’s gravitational force, we assumed Earth and Sun to be point masses.
5. Point masses holds even for small distance.
6. There is also another interesting result. Consider a hollow sphere of mass M. If we place another object of mass ‘m’ inside this hollow sphere the force experienced by this mass ‘m’ will be zero.

Question 2.
Explain how Newton arrived at his law of gravitation from Kepler’s third law.
Answer:
Newton law of gravitation states that a particle of mass M1 attracts any other particle of mass M2 in the universe with an attractive force. The strength of this force of attraction was found to be directly proportional to the product of their masses and is inversely proportional to the square to the distance between them. In mathematical form, it can be written as:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 42
where \(\hat{r}\) is the unit vector from M1 towards M2 as shown in given figure, and G is the Gravitational constant that has the value of 6.67 × 10-11 N m2 kg-2, and r is the distance between the two masses M1 and M2.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 43
In given figure the vector \(\overrightarrow{\mathrm{F}}\) denotes the gravitational force experienced by M2 due to M1. Here the negative sign indicates that the gravitational force is always attractive in nature and the direction of the force is along the line joining the two masses. In Cartesian coordinates, the square of the distance is expressed as r2 = (x2 + y2 + z2)

Question 3.
Explain how Newton verified his law of gravitation.
Answer:
Newton inverse square law: Newton considered the orbits of the planets as circular. For circular orbit of radius r, the centripetal acceleration towards the centre is
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 44
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 1389
The velocity in terms of known quantities r and T, is
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 46
Here T is the time period of revolution of the planet. Substituting this value of v in equation (1) we get,
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 47
Substituting the value ‘a’ from (3) in Newton’s second law, F = ma, where ‘m’ is the mass of the planet.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 48
By substituting equation (6) in the force expression, we can arrive at the law of gravitation.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 49
Here negative sign implies that the force is attractive and it acts towards the center. But Newton strongly felt that according to his third law, if Earth is attracted by the Sun, then the Sim must also be attracted by the Earth with the same magnitude of force. So he felt that the Sun’s mass (M) should also occur explicitly in the expression for force (7). From this insight, he equated the constant 4π2k to GM which turned out to be the law of gravitation.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 50
Again the negative sign in the above equation implies that the gravitational force is attractive.

Question 4.
Derive the expression for gravitational potential energy.
Answer:
The gravitational force is a conservative force and hence we can define a gravitational potential energy associated with this conservative force field.
Two masses m1 and m2 are initially separated by a distance r’. Assuming m1 to be fixed in its position, work must be done on m2 to move the distance from r’ to r.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 51
To move the mass m2 through an infinitesimal displacement \(d \vec{r}\) from \(\vec{r}\) to \(\vec{r}\) + \(d \vec{r}\), work has to be done externally. This infinitesimal work is given by
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 52
The work is done against the gravitational force, therefore,
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 53
Substituting equation (2) in (1), we get
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 54
Thus the total work for displacing the particle from r’ to r is
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 55
This work done W gives the gravitational potential energy difference of the system of masses m1 and m2 when the seperation between them are r and r’ respectively.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 556
Case 1: If r < r’ : Since gravitational force is attractive, m2 is attracted by m1. Then m2 can move from r’ to r without any external Work. Here work is done by the system spending its internal energy and hence the work done is said to be negative.
Case 2: If r > r’: Work has to be done against gravity to move the object from r’ to r. Therefore work is done on the body by external force and hence work done is positive.

Question 5.
Prove that at points near the surface of the Earth, the gravitational potential energy of the object is U = mgh.
Answer:
When an object of mass m is raised to a height h, the potential energy stored in the object is mgh. This can be derived using the general expression for gravitational potential energy.
Consider the Earth and mass system, with r, the distance between the mass m and the Earth’s centre. Then the gravitational potential energy.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 60
Here r = Re + h, where Re is the radius of the Earth, h is the height above the Earth’s surface
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 61
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 62
Substituting equation (4) and (5) we get,
U = -mge + mgh …….. (6)
It is clear that the first term in the above expression is independent of the height h. For example, if the object is taken from height h1 to h2, then the potential energy at h1 is
U(h1) = – mgRe + mgh1 …(7)
and the potential energy at h2 is
U(h2) = – mgRe + mgh2 …(8)
The potential energy difference between h1 and h2 is
U(h2) – U(h1) = mg(h2 – h1) …(9)
The term mgRe in equation (7) and (8) plays no role in the result. Hence in the equation (6) the first term can be omitted or taken to zero. Thus it can be stated that the gravitational potential energy stored in the particle of mass m at a height h from the surface of the Earth is U = mgh. On the surface of the Earth, U = 0, since h is zero.

Question 6.
Explain in detail the idea of weightlessness using lift as an example.
Answer:
When a man is standing in the elevator, there are two forces acting on him.
1. Gravitational force which acts downward. If we take the vertical direction as positive y direction, the gravitational force acting on the man is \(\overrightarrow{\mathrm{F}}_{\mathrm{G}}=-m \hat{g} \hat{j}\)
2. The normal force exerted by floor on the man which acts vertically upward, \(\overrightarrow{\mathrm{N}}=\mathrm{N} \hat{j}\)
Weightlessness of freely falling bodies: Freely falling objects experience only gravitational force. As they fall freely, they are not in contact with any surface (by neglecting air friction). The normal force acting on the object is zero. The downward acceleration is equal to the acceleration due to the gravity of the Earth, i.e., (a = g)
Newton’s 2nd law acting on the man N = m(g – a)
a = g ∴ N = m(g – g) = 0.

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 7.
Derive an expression for escape speed.
Answer:
Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed vi, the initial total energy of the object is
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 75
where, ME is the mass of the Earth and RE the radius of the Earth. The term Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 78 is the potential energy of the mass M.
When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∞) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise kinetic energy can be nonzero.
Ef = 0
According to the law of energy conservation,
Ei = Ef …. (2)
Substituting (1) in (2) we get,
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 76
Consider the escape speed, the minimum speed required by an object to escape Earth’s gravitational field, hence replace vi with ve, i.e.,
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 77
From equation (4) the escape speed depends on two factors acceleration due to gravity and radius of the Earth. It is completely independent of the mass of the object. By substituting the values of g (9.8 ms-2) and Re = 6400 km, the escape speed of the Earth is ve = 11.2 kms-1. The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the sarfte initial speed to escape Earth’s gravity.

Question 8.
Explain the variation of g with latitude.
Answer:
When an object is on the surface for the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been mg. However, the object experiences an additional centrifugal force due to spinning of the Earth.
This centrifugal force is given by mω2R’.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 788
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 89
where λ is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
aPQ = ω2R’ cos λ = ω2R cos2 λ
Since R’ = R cos λ
Therefore, g’ = g – ω2 R cos2 λ
From the above expression, we can infer that at equator, λ = 0, g’ = g – ω2R. The acceleration due to gravity is minimum. At poles λ = 90; g’ = g, it is maximum. At the equator, g’ is minimum.

Question 9.
Explain the variation of g with altitude from the Earth’s surface.
Answer:
Consider an object of mass m at a height h from the surface of the Earth. Acceleration experienced by the object due to Earth is
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 90
We find that g’ < g. This means that as altitude h increases the acceleration due to gravity g decreases.

Question 10.
Explain the variation of g with depth from the Earth’s surface.
Answer:
Consider a particle of mass m which is in a deep mine on the Earth. (Example: coal mines in Neyveli). Assume the depth of the mine as d. To calculate g’ at a depth d, consider the following points.
The part of the Earth which is above the radius (Re – d) do not contribute to the acceleration. The result is proved earlier and is given as
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 91
Here M’ is the mass of the Earth of radius (Re – d)
Assuming the density of Earth ρ to be constant, \(\rho=\frac{M}{V}\)
where M is the mass of the Earth and V its volume, thus,
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 92
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 922
Here also g’ < g. As depth increase, g’ decreases. It is very interesting to know that acceleration due to gravity is maximum on the surface of the Earth but decreases when we go either upward or downward.

Question 11.
Derive the time period of satellite orbiting the Earth.
Answer:
Time period of the satellite: The distance covered by the satellite during one rotation in its orbit is equal to 2π(RE + h) and time taken for it is the time period, T. Then,
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 93
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 94
Squaring both sides of the equation (2) we get
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 95
Equation (3) implies that a satellite orbiting the Earth has the same relation between time and distance as that of Kepler’s law of planetary motion. For a satellite orbiting near the surface of the Earth, h is negligible compared to the radius of the Earth RE. Then,
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 96
By substituting the values of RE = 6.4 × 106 m and g = 9.8ms-2, the orbital time period is obtained as T ≅ 85 minutes.

Question 12.
Derive an expression for energy of satellite.
Answer:
The total energy of the satellite is the sum of its kinetic energy and the gravitational potential energy. The potential energy of the satellite is,
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 97
Here Ms – mass of the satellite, ME -mass of the Earth, RE – radius of the Earth.
The Kinetic energy of the satellite is
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 98
Here v is the orbital speed of the satellite and is equal to
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 99
Substituting the value of v in (2) the kinetic energy of the satellite becomes,
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 100
Therefore the total energy of the satellite is
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 101
The negative sign in the total energy implies that the satellite is bound to the Earth and it cannot escape from the Earth.
Note: As h approaches ∞, the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth’s gravity and is not bound to Earth at large distance.

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 13.
Explain in detail the geostationary and polar satellites.
Answer:
The satellites orbiting the Earth have different time periods corresponding to different orbital radii. Kepler’s third law is used to find then radius of the orbit.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 102
Substituting for the time period (24 hours = 86400 seconds), mass, and radius of the Earth, h turns out to be 36,000 km. Such satellites are called “geo-stationary satellites”, since they appear to be stationary when seen from Earth.

India uses the INSAT group of satellites that are basically geo-stationary satellites for the purpose of telecommunication. Another type of satellite which is placed at a distance of 500 to 800 krn from the surface of the Earth orbits the Earth from north to south direction. This type of satellite that orbits Earth from North Pole to South Pole is called a polar satellite. The time period of a polar satellite is nearly 100 minutes and the satellite completes many revolutions in a day. A polar satellite covers a small strip of area from pole to pole during one revolution. In the next revolution it covers a different strip of area since the Earth would have moved by a small angle. In this way polar satellites cover the entire surface area of the Earth.

Question 14.
Explain how geocentric theory is replaced by heliocentric theory using the idea of retrograde motion of planets.
Answer:
When the motion of the planets are observed in the night sky by naked eyes over a period of a few months, it can be seen that the planets move eastwards and reverse their motion for a while and return to eastward motion again. This is called “retrograde motion of planets.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 103

Careful observation for a period of a year clearly shows that Mars initially moves eastwards (February to June), then reverses its path and moves backwards (July, August, September). It changes it direction of motion once again and continues its forward motion (October onwards). In olden days, astronomers recorded the retrograde motion of all visible planets and tried to explain the motion. According to Aristotle, the other planets and the Sun move around the Earth in the circular orbits. If it was really a circular orbit it was not known how the planet could reverse its motion for a brief interval. To explain this retrograde motion, Ptolemy introduced the concept of “epicycle” in his geocentric model. According to this theory, while the planet orbited the Earth, it also underwent another circular motion termed as “epicycle’. A combination of epicycle and circular motion around the Earth gave rise to retrograde motion of the planets with respect to Earth. Essentially Ptolemy retained the Earth centric idea of Aristotle and added the epicycle motion to it.

But Ptolemy’s model became more and more complex as every planet was found to undergo retrograde motion. In the 15th century, the Polish astronomer Copernicus proposed the heliocentric model to explain this problem in a simpler manner. According to this model, the Sun is at the centre of the solar system and all planets orbited the Sun. The retrograde motion of planets with respect to Earth is because of the relative motion of the planet with respect to Earth.
Earth orbits around the Sun faster than Mars. Because of the relative motion between Mars and Earth, Mars appears to move backwards from July to October. In the same way the retrograde motion of all other planets was explained successfully by the Copernicus model. It was because of its simplicity, the heliocentric model slowly replaced the geocentric model.

Question 15.
Explain in detail the Eratosthenes method of finding the radius of Earth.
Answer:
Eratosthenes observed that during noon time of summer solstice the Sun’s rays cast no shadow in the city Syne which was located 500 miles away from Alexandria. At the same day and same time he found that in Alexandria the Sun’s rays made 7.2 degree with local vertical. He realized that this difference of 7.2 degree was due to the curvature of the Earth.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 105
If S is the length of the arc between the cities of Syne and Alexandria, and if R is radius of Earth, then
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 106
1 mile is equal to 1.609 km. So, he measured the radius of the Earth to be equal to R = 6436 km, which is amazingly close to the correct value of 6378 km.

Question 16.
Describe the measurement of Earth’s shadow (umbra) radius during total lunar eclipse.
Answer:
Lunar eclipse and measurements of shadow of Earth: On January 31, 2018 there was a total lunar eclipse which was observed from various place including Tamil Nadu. It is possible to measure the radius of shadow of the Earth at the point where the Moon crosses.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 107
When the Moon is inside the umbra shadow, it appears red in colour. As soon as the Moon Schematic diagram of umbra disk radius exits from the umbra shadow, it appears in crescent shape.

By finding the apparent radii of the Earth’s umbra shadow and the Moon, the ratio of the these radii can be calculated.
The apparent radius of Earth’s umbra shadow = Rs = 13.2 cm
The apparent radius of the Moon = Rm = 5.15 cm
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 108
The radius of Moon Rm = 1737 km
The radius of the Earth’s umbra shadow is Rs = 2.56 × 1737 km ≅ 4446
The correct radius is 4610 km
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 109
The error will reduce if the pictures taken using a high quality telescope are used.

Samacheer Kalvi 11th Physics Gravitation Conceptual Questions

Question 1.
In the following what are the quantities which that are conserved?
(a) Linear momentum of planet
(b) Angular momentum of planet
(c) Total energy of planet
(d) Potential energy of a planet
Answer:
Since there is a net force acting on the planet, its velocity changes which means its linear momentum changes. In fact, the absolute value of linear momentum changes too since the planet’s speed is variable as it goes around in its elliptical orbit . So the linear momentum of planet is not conserved. But the angular momentum about the sun is conserved. Since the torque of gravitational force is zero.

The total mechanical energy remains constant for an isolated system objects that interact with conservative forces. So, the total energy of the system of planet is conserved. The single energy of the planet is not conserved.

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 2.
The work done by Sun on Earth in one year will be
(a) Zero
(b) None zero
(c) positive
(d) negative
Answer:
(a) Zero

Question 3.
The work done by Sun on Earth at any finite interval of time is
(a) positive, negative or zero
(b) Strictly positive
(c) Strictly negative
(d) It is always zero
Answer:
(d) It is always zero
2 & 3. No work is done on the earth revolving around it in perfectly circular orbit. The earth revolves around the sun due to gravitational force of attraction between the sun and the earth. This force around the sun. This centripetal force is always perpendicular to the linear displacement.
Work done = W = F.d cos θ Since θ = 90°
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 110

Question 4.
If a comet suddenly hits the Moon and imparts energy which is more than the total energy of the Moon, what will happen?
Answer:
A comet with small velocity and high mass, doesn’t trigger the moon much. It just makes a circular shaped impact. The moon is ment for the protection for life on earth and to attain stability for the earth rotation. But a comet with large mass and with large velocity may destroy the moon completely or its impact makes the moon, go out of its orbit.

Question 5.
If the Earth’s pull on the Moon suddenly disappears, what will happen to the Moon?
Answer:
Basically, the moon would become the third planet, however the orbit may change. The moon is in orbit around the earth, and what happens will depend on just when the earth disappears. The moon travels around the earth at about 1 km/sec and the Earth – moon pair travel around the sun at about 30 km/sec. This extra 1 km/sec movement will result in an orbit change.

Question 6.
If the Earth has no tilt, what happens to the seasons of the Earth?
Answer:
If the earth weren’t tilted on its axis, there would be no seasons. And humanity would suffer.

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 7.
A student was asked a question ‘why are there summer and winter for us? He replied as ‘since Earth is orbiting in an elliptical orbit, when the Earth is very faraway from .the Sun (aphelion) there will be winter, when the Earth is nearer to the Sun (perihelion) there will be winter}. Is this answer correct? If not, what is the correct explanation for the occurrence of summer and winter?
Answer:
Early astronomers proved that Earth is spherical in shape by looking at the shape of the shadow cast by Earth on the Moon during lunar eclipse.

Question 8.
The following photographs are taken from the recent lunar eclipse which occurred on January 31, 2018. Is it possible to prove that Earth is a sphere from these photographs?
Answer:
Early astronomers proved that Earth is spherical in shape by looking at the shape of the shadow cast by Earth on the Moon during lunar eclipse.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 111

Samacheer Kalvi 11th Physics Gravitation Numerical Problems

Question 1.
An unknown a planet orbits the Sun with distance twice the semi major axis distance of the Earth’s orbit. If the Earth’s time period is Tp what is the time period of this unknown planet?
Answer:
By Kepler’s 3rd law T2 ∝ a3
Time period of unknown planet = T2
Time period of Earth = T1
Distance of unknown planet from the Sun = a2
Distance of the Earth from the Sun = a1
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 200

Question 2.
Assume that you are in another solar system and provided with the set of data given below consisting of the planets’ semi major axes and time periods. Can you infer the relation connecting semi major axis and time period?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 201
In a given datas tells us the relation connecting to the semi major axis is proportional to the two times of square of the time period.
a ∝ 2T2

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 3.
If the masses and mutual distance between the two objects are doubled, what is the change in the gravitational force between them?
Answer:
By Newton’s law of gravitation
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 202
Here, the masses and mutual distance between the two objects are doubled .
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 203
There is no change in the gravitational force between them.

Question 4.
Two bodies of masses m and 4m are placed at a distance r. Calculate the gravitational potential at a point on the line joining them where the gravitational field is zero.
Answer:
Let the point be the position when the gravitational field is zero.,
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 204
The point P is at a distance \(\frac{r}{3}\) from mass ‘m’ and \(\frac{2r}{3}\) from mass ‘4m’
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 205

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 5.
If the ratio of the orbital distance of two planets \(\frac{d_{1}}{d_{2}}\) = 2, what is the ratio of gravitational field experienced by these two planets?
Answer:
The gravitational field experienced by planets 1
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 206
The gravitational field experienced by planet 2
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 207

Question 6.
The moon Io orbits Jupiter once in 1.769 days. The orbital radius of the Moon I0 is 421700 km. Calculate the mass of Jupiter?
Answer:
Kepler’s third law is used to find the mass of the planet
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 208
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 209

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 7.
If the angular momentum of a planet is given by \(\overrightarrow{\mathbf{L}}=5 t^{2} \hat{i}-6 \hat{t} \hat{j}+3 \hat{k}\). What is the torque experienced by the planet? Will the torque be in the same direction as that of the angular momentum?
Answer:
The torque experienced by the planet
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 210

Question 8.
Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. Calculate the speed of each particle.
Answer:
The net gravitational force = Centripetal force
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 211
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 212
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 213

Question 9.
Suppose unknowingly you wrote the universal gravitational constant value as G = 6.67 × 1011 instead of the correct value G = 6.67 × 10-11, what is the acceleration due to gravity g’ for this incorrect G? According to this new acceleration due to gravity, what will be your weight W?
Answer:
Data: Incorrect Gravitational constant G = 6.67 × 1011 Nm2 kg-2
Mass of the Earth Me = 5.972 × 1024 kg
Radius of the earth Re = 6371 km (or) 6371 × 103 m
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 214

Question 10.
Calculate, the gravitational field at point O due to three masses m1, m2, and m3 whose positions are given by the following figure. If the masses m1 and m2 are equal what is the change in gravitational field at the point O?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 215
(Negative sign indicates field acting along negative x direction)
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 216
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 217
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 218

Question 11.
What is the gravitational potential energy of the Earth and Sun? The Earth to Sun distance is around 150 million km. The mass of the Earth is 5.9 × 1024 kg and mass of the Sun is 1.9 × 1030 kg.
Answer:
Mass of the Earth ME = 5.9 × 1024 kg
Mass of the Sun MS = 1.9 × 1030 kg
Distance between the Sun and Earth
r = 150 million km; r = 150 × 109 m
Gravitational constant G = 6.67 × 10-11 Nm2 kg-2
The gravitational potential energy
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 219

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 12.
Earth revolved around the Sun at 30 km s-1. Calculate the kinetic energy of the Earth. In the previous example you calculated the potential energy of the Earth. What is the total energy of the Earth in that case? Is the total energy positive? Give reasons.
Answer:
Mass of the Earth ME = 5.9 × 1024 kg
Speed of the Earth rovolves around the Sun
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 220
= 2.655 × 1033 + (- 4.985 × 1033)
= 2.655 × 1033 – 4.985 × 1033
E = -2.33 × 1033 joule (or) J
‘-Ve’ implies that Earth is bounded with Sun.

Question 13.
An object is thrown from Earth in such a way that it reaches a point at infinity with non-zero kinetic energy Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 221 with that velocity should the object be thrown from Earth?
Answer:
An object is thrown up with an initial velocity is vi, So Total energy of the object is
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 223
Now, the object reaches a height with a non-zero K.E.
K.E becomes infinity. P.E becomes zero.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 222

Question 14.
Suppose we go 200 km above and below the surface of the Earth, what are the g values at these two points? In which case, is the value of g small?
Answer:
Variation of g’ with depth
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 1390
Variation of g’ with altitude
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 224

Question 15.
Calculate the change in g value in your district of Tamil Nadu. (Hint: Get the latitude of your district of Tamilnadu from the Google). What is the difference in g values at ‘ Chennai and Kanyakumari?
Answer:
Variation of ‘g’ value in the latitude to chennai
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 225
Period of revolution (T) = 1 day = 86400 sec
Radius of the Earth (R) = 6400 × 103 m
Latitude of Chennai (λ) = 13° = 0.2268 rad
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 226
Variation of ‘g’ value in the latitude of Kanyakumari
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 27

Samacheer Kalvi 11th Physics Gravitation Additional Questions

I. Choose the correct answer from the following:

Question 1.
According to Kepler, planet move in
(a) Circular orbits around the Sun
(b) Elliptical orbits around the Sim with Sun at exact centre
(c) Straight lines with constant velocity ‘
(d) Elliptical orbits around the Sun with Sun at one of its foci.
Answer:
(d) Elliptical orbits around the Sun with Sun at one of its foci.

Question 2.
Kepler’s second law regarding constancy of aerial velocity of a planet is consequence of the law of conservation of ………
(a) energy
(b) angular momentum
(c) linear momentum
(d) None of these
Answer:
(b) angular momentum
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 120

Question 3.
According to Kepler, the period of revolution of a planet (T) and its mean distance from the Sun (a) are related by the equation
(a) T3 a3 = constant
(b) T2 a-3 = constant
(c) Ta3 = constant
(d) T2a = constant
Answer:
(b) T2 a-3 = constant
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 121

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 4.
The period of Moon’s rotation around the Earth is nearly 29 days. If Moon’s mass were 2 fold
its present value and all other things remained unchanged the period of Moon’s rotation would be nearly ….. days.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 122
Answer:
(d) 29
Hint:
Time period does not depends upon the mass of satellite.

Question 5.
The period of revolution of planet A around the Sun is 8 times that of B. The distance of A from the Sun is how many times greater than that of B from the Sun.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 123

Question 6.
The radius of orbit of a planet is two times that of Earth. The time period of planet is years …….
(a) 4.2
(b) 2.8
(c) 5.6
(d) 8.4
Answer:
(b) 2.8
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 124

Question 7.
A geostationary satellite orbits around the earth in a circular orbit of radius 3600 km the time period of a satellite orbiting a few hundred kilometers above the earth’s surface (RE = 6400 km) will be approximately be …… hours.
(a) 1/2
(b) 1
(c) 2
(d) 4
Answer:
(c) 2
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 1256

Question 8.
What does not change in the field of central force?
(a) Potential energy
(b) kinetic energy
(c) linear momentum
(d) Angular momentum
Answer:
(d) Angular momentum
Hint:
For central force torque is zero.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 126

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 9.
A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the center of earth in new orbit is two times of the earlier orbit. The time period in second orbit is …… hours.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 127
Answer:
(b) \(48 \sqrt{2}\)
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 128

Question 10.
If the Earth is at one-fourth of its present distance from the sun the duration of year will be:
(a) half the present year
(b) one-eight the present year
(c) one-fourth the present year
(d) one-sixth the present year
Answer:
(b) one-eight the present year
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 129

Question 11.
The Earth E moves in an elliptical orbit with the Sun S at one of the foci as shown in figure.
Its speed of motion will be maximum at a point ……..
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 130
(a) C
(b) A
(c) B
(d) D
Answer:
(b) A
Hint: Speed at the Earth will be maximum when its distance from the Sun is minimum because
mvr = constant

Question 12.
Rockets are launched in eastward direction to take advantage of …..
(a) the clear sky on eastern side
(b) Earth’s rotation
(c) the thinner atmosphere on this side
(d) Earth’s tilt
Answer:
(b) Earth’s rotation
Hint:
Because Earth rotation from west to east direction.

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 13.
Two sphere of mass M1 and M2 are situated in air and the gravitational force between them is F. The space around the masses is now filled with liquid of specific gravity 3. The gravitational force will now be ……
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 1300
Answer:
(a) F
Hint:
Gravitational force does not depend upon the medium.

Question 14.
Which of the following statement about the gravitational constant is true?
(a) It is a force
(b) It has same value in all system of unit
(c) It has not unit
(d) It depends on the value of the masses
Answer:
(a) It is a force

Question 15.
Energy required to move a body of mass ‘M’ from an orbit of radius 2R to 3R is …….
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 131
Answer:
(d) \(\frac{\mathrm{GMm}}{6 \mathrm{R}}\)
Hint:
Change in P.E. in displacing a body from r1 and r2 is given by:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 132

Question 16.
The mass of the earth is 6 × 1024 kg and that of the Moon is 7.4 × 1022 kg. The constant of gravitation G is 6.67 × 10-11 Nm2 kg-2. The potential energy of the system is – 7.79 × 1028J. The mean distance between the Earth and Moon is …… metre.
(a) 3.80 × 108
(b) 3.37 × 108
(c) 7.60 × 108
(d) 1.90 × 102
Answer:
(a) 3.80 × 108
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 133

Question 17.
What is the intensity of gravitational field at the center of spherical shell?
(a) \(\frac{\mathrm{G} m}{r^{2}}\)
(b) g
(c) zero
(d) None of these
Answer:
(c) zero

Question 18.
A body of mass m is taken from the Earth’s surface to a height equal to the radius R of the earth. If g is the acceleration to gravity at the surface of the Earth, then find the change in the potential energy of the body ……
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 134
Answer:
(b) \(\frac{1}{2} m g R\)
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 135

Question 19.
A satellite is orbiting around the Earth in a circular orbit with velocity v. If m is the mass of the satellite, its total energy is ……
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 136
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 13934
Hint:
The total energy is negative of the kinetic energy.

Question 20.
Escape velocity of a body of 1 kg. On a planet is 100 ms-1. Gravitational potential energy of the body at the planet is ……
(a) -5000 J
(b) – 1000 J
(c) – 2400 J
(d) 4000 J
Answer:
(a) – 5000 J
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 137

Question 21.
A particle falls towards earth from infinity. It’s velocity reaching the Earth would be ……
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 138
Answer:
(b) \(\sqrt{2 g \mathrm{R}}\)
Hint:
This should be equal to escape velocity is = \(\sqrt{2 g \mathrm{R}}\)

Question 22.
An artificial satellite is revolving round the Earth in a circular orbit, its velocity is half the escape velocity. Its height from the Earth surface is …. km.
(a) 6400
(b) 12800
(c) 3200
(d) 1600
Answer:
(a) 6400

Question 23.
The escape velocity of a body on the surface of the Earth is 11.2 km/s. If the mass of the Earth is increase to twice its present value and the radius of the earth becomes half, the escape velocity becomes = …… kms-1
(a) 5.6
(b) 11.2
(c) 22.4
(d) 494.8
Answer:
(c) 22.4
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 1388
If M becomes double and R becomes half, then escape velocity becomes two times.

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 24.
The velocity with which a projectile must be fired so that it escapes Earth’s gravitational does
not depend on ……..
(a) Mass of Earth
(b) Radius of the projectile’s orbit
(c) Mass of the projectile
(d) Gravitational constant
Answer:
(c) Mass of the projectile

Question 25.
The escape velocity for a body projected vertically upwards from the surface of Earth is 11 kms-1. If the body is projected at an angle of 45° with the vertical, the escape velocity will be …. kms-1
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 140
Answer:
(d) 11
Hint:
Escape velocity does not depends upon the angle of projection.

Question 26.
Two satellites of mass ml and m2(m1> m2) are revolving round the earth in circular orbits of r1 and r2 (r1 > r2) respectively. Which of the following statement is true regarding their speeds v1 and v2 ……..
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 141
Answer:
(b) v1 < v2
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 13935

Question 27.
As astronaut orbiting the earth in a circular orbit 120 km above the surface of Earth, gently drops a spoon out of space-ship. The spoon will ……….
(a) fall vertically down to the Earth
(b) move towards the moon
(c) will move along with space-ship
(d) will move in an irregular way then fall down to Earth
Answer:
(c) will move along with space-ship
Hint:
The velocity of the spoon will be equal to the orbital velocity when dropped out of the space-ship

Question 28.
A satellite revolves around the Earth in an elliptical orbit. Its speed.
(a) is the same at all point in the orbit
(b) is greatest when it is closest to the Earth
(c) is greatest when it is farthest to the Earth
(d) goes on increasing or decreasing continuously depending upon the mass of the satellite. Answer:
(b) is greatest when it is closest to the Earth

Question 29.
A satellite is moving around the Earth with speed v in a circular orbit of radius r. If the orbit
radius is decreased by 1% its speed will …….
(a) increase by 1%
(b) increase by 0.5%
(c) decrease by 1%
(d) decrease by 0.5%
Answer:
(b) increase by 0.5%
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 145

Question 30.
Orbital velocity of an artificial satellite does not depend upon …….
(a) mass of Earth
(b) mass of satellite
(c) radius of Earth
(d) acceleration due to gravity
Answer:
(b) mass of satellite

Question 31.
The orbital speed of Jupiter is …….
(a) greater than the orbital speed of Earth
(b) less then the orbital speed of Earth
(c) zero
(d) equal to the orbital speed of Earth
Answer:
(b) less then the orbital speed of Earth
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 148

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 32.
As we go grom the equator to the poles, the value of g …..
(a) remains constant
(b) decreases
(c) increases
(d) decreases upto latitude of 45°
Answer:
(c) increases

Question 33.
The value of g on the Earth surface is 980 cm/sec2. Its value at a height of 64 km from the Earth surface is ……. cms2
(a) 960.40
(b) 984.90
(c) 982.45
(d) 977.55
Answer:
(a) 960.40
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 149

Question 34.
The Moon s radius is \(\frac{1}{4}\) that of earth and its mass is \(\frac{1}{80}\) times that of the Earth. If g represents the acceleration due to gravity on the surface of Earth, that on the surface of the Moon is ……
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 150
Answer:
(b) \(\frac{g}{5}\)
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 151

Question 35.
If the density of small planet is that of the same as that of the earth while the radius of the
planet is 0.2 times that of the Earth, the gravitational acceleration on the surface for the planet is ……
(a) 0.2g
(b) 0.4g
(c) 2g
(d) 4g
Answer:
(a) 0.2g
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 152

Question 36.
Assuming Earth to be a sphere of a uniform density, what is value of gravitational acceleration in mine 100 km below the Earth surface = ….. ms-2
(a) 9.66
(b) 7.64
(c) 5.00
(d) 3.1
Answer:
(a) 9.66
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 153

Question 37.
The radii of two planets are respectively R1 and R2 and their densities are respectively ρ1 and ρ2 the ratio of the accelerations due to gravity at their surface is ……
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 154
Answer:
(d) \(g_{1}: g_{2}=\mathrm{R}_{1} \rho_{1}: \mathrm{R}_{2} \rho_{2}\)
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 155

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 38.
The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 156
Answer:
(c) dR
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 157

Question 39.
The acceleration of a body due to the attraction of the Earth (radius R) at a distance 2R from the surface of the Earth is …….
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 158
Answer:
\(\frac{g}{9}\)
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 159

Question 40.
If density of Earth increased 4 times and its radius becomes half of then out weight will be ……
(a) four times its present value
(b) doubled
(c) remains same
(d) halved
Answer:
(b) doubled
Hint:
g ∝ ρR

Question 41.
The radius of the Earth is 6400 km and g = 10 ms-2 in order that a body of 5 kg weights zero at the equator, the angular speed of the Earth is ….. rad s-1
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 160
Answer:
(c) \(\frac{1}{800}\)
Hint:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 161

Question 42.
Weight of a body is maximum at …..
(a) Moon
(b) poles of Earth
(c) equator of Earth
(d) centre of Earth
Answer:
(b) poles of Earth

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 43.
The weight of an astronaut, in an artificial satellite revolving around the Earth is:
(a) zero
(b) equal to that on the Earth
(c) more than that on Earth
(d) less than that on Earth
Answer:
(a) zero

Samacheer Kalvi 11th Physics Gravitation 2 Marks Questions

Question 1.
Distinguish between the terms gravitation and gravity.
Answer:
Gravitation: It is the force of attraction between any two bodies in the universe.
Gravity: It is the force of attraction between the earth and any object lying on or near its surface.

Question 2.
Why is G called the universal gravitational constant?
Answer:
The value of G does not depend on the nature and size of the bodies. It also does not depend on the nature of the medium between the two bodies. That is why G is called universal gravitational constant.

Question 3.
What is meant by the term free fall?
Answer:
The motion of a body under the influence of gravity alone is called a free fall.

Question 4.
What is meant by acceleration due to gravity? Is is a scalar or a vector?
Answer:
The acceleration produced in a freely falling body under the gravitational pull of the earth. It is a vector having direction towards the centre of the earth.

Question 5.
What do you mean by weight of a body? Is it a scalar or vector?
Answer:
Weight of a body is defined as the gravitational force with which a body is attracted towards the centre of the earth. Hence the weight of a body is given by w = mg (or) \(\overrightarrow{\mathrm{W}}=m \vec{g}\)

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 6.
Define orbital velocity.
Answer:
Orbital velocity is the velocity required to put the satellite into its orbit around the earth.

Question 7.
Give some uses of geostationary satellites.
Answer:

  • In communicating radio, T.V and telephone signals across the world.
  • In studying upper regions of the atmosphere.
  • In forecasting weather.
  • In deter ming the exact shape and dimensions of the earth.
  • In studying solar radiations and cosmic rays.

Question 8.
Give the uses of polar satellites.
Answer:

  • Polar satellites are uses in weather and environment monitoring.
  • They are used in spying work for military purposes.
  • They are used to study topography of Moon, Venus and Mars.

Samacheer Kalvi 11th Physics Gravitation Numerical Problems

Question 1.
A geo-stationary satellite is orbiting the Earth of a height of 6R above the surface of Earth R being the radius of the Earth calculate the time period of another satellite at a height of 2.5R from the surface of Earth.
Distance of satellite from the center are 7R and 3.5 R respectively.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 1

Question 2.
The time period of a satellite of Earth is 5 hours. If the separation between the Earth and the satellite is increased to four times the previous value, the new time period will become.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 2

Question 3.
The figure shows elliptical orbit of a planet ‘M’ about the Sun ‘S’, the shaded area SCD is twice the shaded area SAB. If t1 is the time for the planet to move from C and D and t2 is the time to move from A to B then.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 3

Question 4.
A satellite moves in a circle around the Earth, the radius of this circle is equal to one half of the radius of the Moon’s orbit. The satellite completes one revolution in…… lunar month.
Answer:
Time period of revolution of moon around the Earth Tm = 1 lunar month
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 4

Question 5.
Two identical solid copper spheres of radius R are placed in contact with each other. The gravitational force between them is proportional to.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 5
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 6
∴ The gravitational force between them is proportional to 4th power of radius.

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 6.
Two satellites A and B of the same mass are revolving around the Earth in circular orbits such that the distance of B from the centre of the Earth is thrice as compared to the distance of A from the centre. What will be the ratio of centripetal force on B to that on A.
Answer:
The necessary centripetal force is provided by the gravitational force of attraction, for circular orbit
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 7
The ratio of centripetal force,
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 8

Question 7.
An infinite number of bodies, each of mass 2 kg, are situated on x-axis at distances lm, 2m, 4m, 8m from the origin. What will be the resultant gravitational potential due to this system at the origin.
Answer:
Gravitational potential at the origin is
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 9

Question 8.
A body of mass ‘m’ kg starts falling from a point 2R above the Earth’s surface. What is its K.E. When it has fallen to a point ‘R’ above the Earth’s surface.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 10

Question 9.
A particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to done against the gravitational force between them to take the particle is away from the sphere.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 11
So, the amount of work done to take the particle upto infinite will be 6.67 × 10-10 J

Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation

Question 10.
The mass of a space ship is 1000 kg. It is to be launched from Earth ’s surface out into free space the value of g and R (radius of Earth) are 10 ms-2 and 6400 km respectively. The required energy for this work will be.
Answer:
Potential energy U = – mgRe + mgh
(The first term is independent of the height, so it can be taken to zero.)
W = U = mgh [h ≈ R]
= 1000 × 10 × 6400 × 103 = 64 × 109
W = 6.4 × 1010 J

Question 11.
If the mean radius of the Earth is R, its angular velocity is ω, and the acceleration due to gravity at the surface of the Earth is g, then what will be the cube of the radius of the orbit of a geostationary satellite.
Answer:
Let r be the radius of the geostationary orbit. Angular velocity of revolution of a geostationary satellite is same as the angular velocity of rotation of the Earth.
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 12

Question 12.
The escape velocity of a body from Earth’s surface is ve. What will be the escape velocity of the same body from a height equal to 7R from Earth’s surface.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 13

Question 13.
If R is the radius of the Earth and g is the acceleration due to gravity on the Earth’s surface. Find the mean density of the Earth.
Acceleration due to gravity g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 14

Question 14.
If the mass of Earth is 80 times of that of a planet and diameter is double that of planet and ‘g’ on the Earth is 9.8 ms-2. Calculate the value of ‘g’ on that planet?
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 15

Question 15.
At what distance from the centre of Earth, the value acceleration due to gravity ‘g’ will be half that of the surface?
Answer:
According to acceleration due to gravity
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 16

Question 16.
A body weight 700 g on the surface of Earth. How much it weight on the surface of planet whose mass is \(\frac{1}{7}\) and radius is half that of the Earth.
Answer:
Acceleration due to gravity g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 18

Question 17.
An object weight 72 N on the Earth. What its weight at a height \(\frac{\mathbf{R}}{2}\) from Earth.
Answer:
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 19

Question 18.
A body weight 500 N on the surface of the Earth. How much would it weight half way below the surface of Earth.
Answer:
Weight on surface of Earth, mg = 500 N
Weight below the surface of Earth at d = \(\frac{\mathrm{R}}{2}\)
From variation of ‘g’ with depth
Samacheer Kalvi 11th Physics Solutions Chapter 6 Gravitation 20