Class 12

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 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 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 12th Commerce Solutions Chapter 18 Grievance Redressal Mechanism

Students can Download Commerce Chapter 18 Grievance Redressal Mechanism Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Commerce Book 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 Commerce Solutions Chapter 18 Grievance Redressal Mechanism

Samacheer Kalvi 12th Commerce Grievance Redressal Mechanism Textbook Exercise Questions and Answers

I. Choose the Correct Answer

Question 1.
The Chairman of the National Consumer Disputes Redressal Council is ________
(a) Serving or Retired Judge of the Supreme Court of India
(b) Prime Minister
(c) President of India
(d) None of the above
Answer:
(a) Serving or Retired Judge of the Supreme Court of India

Question 2.
The Chairman of the State Consumer Protection Council is ________
(a) Judge of a High Court
(b) Chief Minister
(c) Finance Minister
(d) None of the above
Answer:
(a) Judge of a High Court

Question 3.
The Chairman of the District Forum is ________
(a) District Judge
(b) High Court Judge
(c) Supreme Court Judge
(d) None of the above
Answer:
(a) District Judge

Samacheer Kalvi 12th Commerce Solutions Chapter 18 Grievance Redressal Mechanism

Question 4.
The State Commission can entertain complaints where the value of the goods or services and the compensation, if any claimed exceed ________
(a) ₹ 2 lakhs but does not exceed ₹ 5 lakhs
(b) ₹ 20 lakhs but does not exceed ₹ 1 crore
(c) ₹ 3 lakhs but does not exceed ₹ 5 lakhs
(d) ₹ 4 lakhs but does not exceed ₹ 20 lakhs
Answer:
(b) ₹ 20 lakhs but does not exceed ₹ 1 crore

Question 5.
The National Consumer Disputes Redressal Commission has jurisdiction to entertain complaints where the value of goods/services complained against and the compensation, if any claimed is ________
(a) Exceeding ₹ 1 crore
(b) Exceeding ₹ 10 lakhs
(c) Exceeding ₹ 5 lakhs
(d) Exceeding ₹ 12 lakhs
Answer:
(a) Exceeding ₹ 1 crore

Question 6.
The District Forum can entertain complaints where the value of goods or services and the compensation if any claimed is less than ________
(a) Below ₹ 10,00,000
(b) Below ₹ 20,00,000
(c) Below ₹ 40,00,000
(d) Below ₹ 50,00,000
Answer:
(b) Below ₹ 20,00,000

Question 7.
The International Organisation of Consumers Unions (IOCU) was first established in ________
(a) 1960
(b) 1965
(c) 1967
(d) 1987
Answer:
(a) 1960

Question 8.
Consumer awareness covers the following:
(а) Consumer awareness about Maximum Retail Price (MRP)
(b) Consumer awareness about Fair Price Shop
(c) Consumer awareness about price, quality, and expiry date of the product
(d) All of the above
Answer:
(d) All of the above

Question 9.
Complaints can also be filed by the ________
(a) Central Government
(b) State Government
(c) A group of consumers
(d) All of the above
Answer:
(d) All of the above

Samacheer Kalvi 12th Commerce Solutions Chapter 18 Grievance Redressal Mechanism

Question 10.
A consumer has to be protected against ________
(a) Defects of product
(b) Deficiencies of product
(c) Unfair and restrictive trade practices
(d) All of the above
Answer:
(d) All of the above

II. Very Short Answer Questions

Question 1.
What do you meant by Redressal Mechanism?
Answer:
Exploitation is common where consumers are unaware of their rights and privileges. Government has also taken necessary steps to save the Consumers. It is in this context grievance redressal mechanism becomes important.

Question 2.
What do you know about National Commission?
Answer:
The National Consumer Disputes Redressal Commission (NCDRC), India is a quasi-judicial commission in India which was set up in 1988 under the Consumer Protection Act of 1986. The National Consumer Disputes Redressal Commission (NCDRC) is also called as National Commission.

Question 3.
State the meaning of the term State Commission.
Answer:
The State Commission is to be appointed by the State Government in consultation with the Centre. The State Consumer Protection Council is also called State Commission.

Question 4.
What is an term District Forum?
Answer:
As per the Consumer Protection Act of 1986 and Section 9 thereof the establishment of a District Forum by the State Government in each district is necessary today to protect the interest of aggrieved consumers in that district. Complaints can be filed with the forum by a consumer.

Question 5.
How to register the complaints?
Answer:
A complaint can be filed by a complainant against the seller, manufacturer, or dealer of goods which are defective or against the provider of services.

Samacheer Kalvi 12th Commerce Solutions Chapter 18 Grievance Redressal Mechanism

III. Short Answer Questions

Question 1.
Is Consumer Protection necessary?
Answer:
Consumer is supposed to be the king in the business. Consumer satisfaction is compulsorily necessary. Because the business depends upon the consumer. If there is no consumer, no need of production or sales. So the consumer is to be fully satisfied and respected in the market.

Question 2.
Who are the members of the National Commission?
Answer:
Members: The National Consumer Disputes Redressal Commission has been constituted by a Notification.

  1. The National Commission should have five members.
  2. One should be from judiciary.
  3. Four other members of ability, knowledge and experience from any other fields.
  4. It should include a woman.

Question 3.
What is the Pecuniary Jurisdiction of the State Commission?
Answer:
The Jurisdiction of the State Commission is as follows:

  1. The State Commission can entertain complaints within the territory of entire state and the compensation, if any claimed exceed Rs. 20 lakhs and below Rupees One Crore.
  2. The State Commission also has the jurisdiction to entertain appeals against the orders of any District Forum within the State.

Samacheer Kalvi 12th Commerce Solutions Chapter 18 Grievance Redressal Mechanism

Question 4.
Does District Forum exceeds the claim limit of Rs 20 Fakhs? Explain the condition.
Answer:
If the value of the complaint exceeds this limit of Rs 20 Lakhs the complaint should be made direct to the State Commission. Further the District Forum also may pass orders against traders indulging in unfair trade practices, sales of defective goods or rendering deficient services, where the turnover of goods or value of services does not exceed Rs 20 Lakhs.

Question 5.
Write a note on the Voluntary Consumer Organisation.
Answer:
Voluntary consumer organisations refer to the organisation formed voluntarily by the consumers to protect their rights and interests.

IV. Long Answer Questions

Question 1.
What are the Functions of the National Commission?
Answer:
The National Consumer Disputes Redressal Commission is also called as National commision. Its head office is New Delhi.
Functions:

  1. Compensation can be given foe the complaints to the value more than Rupees one crore.
  2. The orders passed in the state commission can be appealed.
  3. To call for the records and pass appropriate orders from the state commission and district
    forum.

Question 2.
Explain the overall performance of State Commission.
Answer:
The State Commission is to be appointed by the State Government. The person who is a judge or retired judge of high court is the president of the state commission.
Performances:

  1. The compensation of the value should not exceed Rs.20 lakhs and below Rs. 1 crore.
  2. The state commission has the power to call for the records and pass orders in the district forum
  3. To furnish the information which is required for the purpose of the Act to any officer.

Samacheer Kalvi 12th Commerce Solutions Chapter 18 Grievance Redressal Mechanism

Question 3.
Explain the term District Forum and explain the functions of District Forum.
Answer:
As per the Consumer Protection Act 1986, a district forum is established in each and every district to solve the problems in the concerned district.
Present or Retired district judge is the president of District forum.
Functions:

  1. The complaints relating to the district can be solved by district forum.
  2. Compensation that can be claimed is less than Rs. 20 lakhs.
  3. Sometimes the district forum also may pass orders against traders.

Question 4.
What is Voluntary Consumer Organisations? Explain its Functions.
Answer:
Voluntary consumer organisations refer to the organisation formed voluntarily by the consumers to protect their rights and interests.
Functions:

  1. Collecting Data on Different Products: These organizations collect samples of different products from time to time and test them.
  2. Filing Suit on Behalf of Consumers: If a consumer is not able to protest regarding his complaints, these organisations file case in the court, on behalf of a consumer.
  3. Protests against Adulteration: The consumer organisations play a significant role in eliminating the evil of adulteration, hoarding, black-marketing.
  4. Helping Educational Institutions: These organizations advice the educational institutions the way to prepare courses of study.
  5. Extending Support to Government: Consumer organisations keep informing the government agencies about adulteration, artificial scarcity, inferior quality products.

Samacheer Kalvi 12th Commerce Solutions Chapter 18 Grievance Redressal Mechanism

Question 5.
How to create consumer awareness?
Answer:
The first priority of a consumer organisation is to increase consumer awareness towards their rights.
The following are the points for awareness:

  1. To publish brochures, journals and monographs.
  2. To arrange conferences, seminars and workshops.
  3. To educate consumers to help themselves.
  4. To provide special education to women about consumerism.

Samacheer Kalvi 12th Commerce Grievance Redressal Mechanism Additional Questions and Answers

I. Choose the Correct Answer

Question 1.
The State consumer protection council is also called as
(a) State Commission
(b) National commission
(c) District commission
(d) City commission
Answer:
(a) State Commission

Question 2.
The National commission should have members.
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
(c) 5

II. Very Short Answer Questions

Question 1.
What did Mahatma Gandhi tell about the Customer?
Answer:
“A customer is the most important visitor on our premises. He is not dependent on us. We are dependent on him. He is not an interruption of our work. He is the purpose of it. He is not an outsider of our business.”

Samacheer Kalvi 12th Commerce Solutions Chapter 18 Grievance Redressal Mechanism

III. Short Answer Questions

Question 1.
Who can make complaint?
Answer:
There are certain persons eligible to make complaint.
They are as follows:

  1. A consumer as defined under Consumer Protection Act, 1986.
  2. A registered Voluntary Consumer Association.
  3. Central Government.
  4. State Govemment / Union Territory.
  5. Consumers having the common problem.

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10

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 6 Applications of Vector Algebra Ex 6.10

Choose the correct or the most suitable answer from the given four alternatives:
Question 1.
If \(\vec{a}\) and \(\vec{b}\) are parallel vector, then \([\vec{a}, \vec{c}, \vec{b}]\) is equal to ………………..
(a) 2
(b) -1
(c) 1
(d) 0
Solution:
(d) 0
Hint:
\(\vec{a}\) and \(\vec{b}\) are parallel vectors, so \(\vec{a} \times \vec{b}\) = 0
then \([\vec{a} \vec{c} \vec{b}]=-[\vec{a} \vec{b} \vec{c}]=-(\vec{a} \times \vec{b}) \cdot \vec{c}\) = 0

Question 2.
If a vector \(\vec{\alpha}\) lies in the plane of \(\vec{\beta}\) and \(\vec{\gamma}\), then …………
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 1
Solution:
(c) \([\vec{\alpha}, \vec{\beta}, \vec{\gamma}]\) = 0
Hint:
Since \([\vec{\alpha}, \vec{\beta}, \vec{\gamma}]\) are lie in the same plane
so \([\vec{\alpha}, \vec{\beta}, \vec{\gamma}]\) = 0

Question 3.
If \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}\)= 0, then the value of \(|[\vec{a}, \vec{b}, \vec{c}]|\) is ……………….
(a) \(|\vec{a}||\vec{b}||\vec{c}|\)
(b) \(\frac{1}{3}|\vec{a}||\vec{b}||\vec{c}|\)
(c) 1
(d) -1
Solution:
(a) \(|\vec{a}||\vec{b}||\vec{c}|\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 2

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10

Question 4.
If \(\vec{a}, \vec{b}, \vec{c}\) are three unit vectors such that \(\vec{a}\) is perpendicular to \(\vec{b}\), and is parallel to \(\vec{c}\) then \(\vec{a} \times(\vec{b} \times \vec{c})\)is equal to ………………….
(a) \(\vec{a}\)
(b) \(\vec{b}\)
(c) \(\vec{c}\)
(d) \(\vec{0}\)
Solution:
(b) \(\vec{b}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 3

Question 5.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 4
(a) 1
(b) -1
(c) 2
(d) 3
Solution:
(a) 1
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 5

Question 6.
The volume of the parallelepiped with its edges represented by the vectors \(\hat{i}+\hat{j}, i+2 \hat{j}\), \(\hat{i}+\hat{j}+\pi \hat{k}\) is ……………
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 6
Solution:
(c) π
Hint:
Volume = \([\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}{1} & {1} & {0} \\ {1} & {2} & {0} \\ {1} & {1} & {\pi}\end{array}\right|\)
= π(2 – 1) = π cubic units

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10

Question 7.
If \(\vec{a}\) and \(\vec{b}\) are unit vectors such that \([\vec{a}, \vec{b}, \vec{a} \times \vec{b}]=\frac{\pi}{4}\) then the angle between \(\vec{a}\) and \(\vec{b}\) is …………
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 7
Solution:
(a) \(\frac{\pi}{6}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 8

Question 8.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 9
(a) 0
(b) 1
(c) 6
(d) 3
Solution:
(a) 0
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 10
Equate corresponding coefficients on both sides
λ + µ = 0 and λ = -1 this gives µ = 1
∴ Then the value of λ + µ = 0.

Question 9.
If \(\vec{a}, \vec{b}, \vec{c}\) are three non-coplanar vectors such that \(\vec{a} \times(\vec{b} \times \vec{c})=\frac{\vec{b}+\vec{c}}{\sqrt{2}}\), then the angle between \(\vec{a}\) and \(\vec{b}\) is …………
(a) 81
(b) 9
(c) 27
(d) 18
Solution:
(a) 81
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 11

Question 10.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 12
Solution:
(b) \(\frac{3 \pi}{4}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 13

Question 11.
If the volume of the parallelepiped with \(\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}\) as coterminous edges is 8 cubic units, then the volume of the parallelepiped with \((\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c}),(\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})\) and \((\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b})\) as coterminous edges is ………………
(a) 8 cubic units
(b) 512 cubic units
(c) 64 cubic units
(d) 24 cubic units
Solution:
(c) 64 cubic units
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 14

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10

Question 12.
Consider the vectors \(\vec{a}, \vec{b}, \vec{c}, \vec{d}\) such that \((\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\overrightarrow{0}\). Let P1 and P2 be the planes determined by the pairs of vectors \(\vec{a}, \vec{b}\) and \(\vec{c}, \vec{d}\) respectively. Then the angle between P1 and P2 is ……………..
(a) 0°
(b) 45°
(c) 60°
(d) 90°
Solution:
(a) 0°
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 15

Question 13.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 16
(a) perpendicular
(b) parallel
(c) inclined at an angle \(\frac{\pi}{3}\)
(d) inclined at an angle \(\frac{\pi}{6}\)
Solution:
(b) parallel
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 17

Question 14.
If Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 18, then a vector perpendicular to \(\vec{a}\) and lies in the plane containing \(\vec{b}\) and \(\vec{c}\) is
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 19
Solution:
(d) \(-17 \hat{i}-21 \hat{j}-97 \hat{k}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 20

Question 15.
The angle between the lines Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 21 is ………..
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 22
Solution:
(d) \(\frac{\pi}{2}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 23
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 24

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10

Question 16.
If the line Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 25 lies in the plane x + 3 + αz + β = 0, then (α, β) is …………..
(a) (-5, 5)
(b) (-6, 7)
(c) (5, -5)
(d) (6, -7)
Solution:
(d) (-6, 7)
Hint:
d.c.s of the first line = (3, -5, 2)
d.c.s of the line perpendicular to plane = (1, 3, -α)
a1a2 + b1b2 + c1c2 = 0
3 – 15 – 2α = 0 => -12 – 2α = 0
-2α =12 => α = -6
Plane passes through the point (2, 1, -2) so
2 + 3 + 6(-2) + β = 0 => β = 7
(α, β) = (-6, 7)

Question 17.
The angle between the line \(\vec{r}=(\hat{i}+2 \hat{j}-3 \hat{k})+t(2 \hat{i}+\hat{j}-2 \hat{k})\) and the plane \(\vec{r} \cdot(\hat{i}+\hat{j})+4\) = 0 is ……………
(a) 0°
(b) 30°
(c) 45°
(d) 90°
Solution:
(c) 45°
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 26

Question 18.
The coordinates of the point where the line \(\vec{r}=(6 \hat{i}-\hat{j}-3 \hat{k})+t(-\hat{i}+4 \hat{k})\) meets the plane \(\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})\) are
(a) (2, 1, 0)
(b) (7, -1, -7)
(c) (1, 2, -6)
(c) (5, -1, 1)
Solution:
(d) (5, -1, 1)
Hint:
Cartesian equation of the line
\(\frac{x-6}{-1}=\frac{y+1}{0}+\frac{z+3}{4}\) = λ
(-λ + 6, -1, 4λ – 3)
This meets the plane x + y – z = 3
-λ + 6 – 1 – 41 + 3 = 3 ⇒ -5λ = -5
λ = 1
The required point (5, -1, 1).

Question 19.
Distance from the origin to the plane 3x – 6y + 2z + 7 = 0 is ………………
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 1
Hint:
Distance from the origin (0, 0, 0) to the plane
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 27

Question 20.
The distance between the planes x + 2y + 3z + 7 = 0 and 2x + 4y + 6z + 7 = 0 is ……………
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 28
Solution:
(a) \(\frac{\sqrt{7}}{2 \sqrt{2}}\)
Hint:
x + 2y + 3z + 1 = 0; 2x + 4y + 6z + 7 = 0
Multiplying 2 on both sides
2x + 4y + 6z + 14 = 0 .
a = 2, b = 4, c = 6, d1 = 14, d2 = ?
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 29

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10

Question 21.
If the direction cosines of a line are \(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\), then ……………….
(a) c = ±3
(b) c = ± \(\sqrt{3}\)
(c) c > 0
(d) 0 < c < 1
Solution:
(b) c = ± \(\sqrt{3}\)
Hint:
We know that sum of the squares of direction cosines = 1
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 30

Question 22.
The vector equation \(\vec{r}=(\hat{i}-2 \hat{j}-\hat{k})+t(6 \vec{j}-\hat{k})\) represents a straight line passing through the points ……………. (a) (0, 6, -1) and (1, -2, -1) (b) (0, 6, -1) and (-1, -4, -2) (c) (1, -2, -1) and (1, 4, -2) (d) (1, -2, -1) and (0, -6, 1)
Solution:
(c) (1, -2, -1) and (1, 4, -2)
Hint:
The required vector equation is \(\vec{r}=\vec{a}+t(\vec{b}-\vec{a})\)
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 31
From (1) and (2) The points are (1, -2, -1) and (1, 4, -2)

Question 23.
If the distance of the point (1, 1, 1) from the origin is half of its distance from the plane x + y + z + k= 0, then the values of k are ………….. (a) ±3 (b) ±6 (c) -3, 9 (d) 3, -9
Solution:
(d) 3, -9
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 32

Question 24.
If the planes \(\vec{r} \cdot(2 \hat{i}-\lambda \hat{j}+\hat{k})=3 \text { and } \vec{r} \cdot(4 \hat{i}+\hat{j}-\mu \hat{k})\) = 5 are parallel, then the value of λ and µ are ……………
Solution:
(c) \(-\frac{1}{2}\), -2
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 33

Question 25.
If the length of the perpendicular from the origin to the plane 2x + 3y + λz = 1, λ > 0 is \(-\frac{1}{5}\), then the value of λ is …………..
(a) \(2 \sqrt{3}\)
(b) \(3 \sqrt{2}\)
(c) 0
(d) 1
Solution:
(a) \(2 \sqrt{3}\)
Hint:
Given length of perpendicular from origin to the plane = \(-\frac{1}{5}\)
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 34

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 Additional Problems

Question 1.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 35
(a) 6
(b) 10
(c) 12
(d) 24
Solution:
(c) 12
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 36

Question 2.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 37
(a) only x
(b) only y
(c) Neither x nor y
(d) Both x and y
Solution:
(c) Neither x nor y
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 38

Question 3.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 39
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 3
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 40

Question 4.
The value of Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 41 = ………………
(a) 1
(b) 3
(c) -3
(d) 0
Solution:
(b) 3
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 42

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10

Question 5.
Let a, b, c be distinct non-negative numbers. If the vectors Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 43 lie in a plane, then c is ……………..
(a) the A.M. of a and b
(b) the G.M. of a and b
(c) the H.M. of a and b
(d) equal to zero.
Solution:
(b) the G.M. of a and b
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 44

Question 6.
The value of \(\hat{i} \cdot(\hat{j} \times \hat{k})+(\hat{i} \times \hat{k}) \cdot \hat{j}\) ………….
(a) 1
(b) -1
(c) 0
(d) \(\hat{j}\)
Solution:
(c) 0
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 45

Question 7.
The value of \((\hat{i}-\hat{j}, \hat{j}-\hat{k}, \hat{k}-\hat{i})\) is …………..
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 46
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 47

Question 8.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 48
Solution:
(c) \(\vec{u}=\overrightarrow{0}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 49

Question 9.
The area of the parallelogram having a diagonal \(3 \vec{i}+\vec{j}-\vec{k}\) and a side \(\vec{i}-3 \vec{j}+4 \vec{k}\) is ………………
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 50
Solution:
(d) \(3 \sqrt{30}\)
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 51

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10

Question 10.
If Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 52, then ……………….
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 53
Solution:
(d) \(\vec{x}=\overrightarrow{0} \text { or } \vec{y}=\overrightarrow{0} \text { or } \vec{x} \text { and } \vec{y}\) are parallel
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 54

Question 11.
If \(\overrightarrow{\mathrm{PR}}=2 \vec{i}+\vec{j}+\vec{k}, \overrightarrow{\mathrm{Question}}=-\vec{i}+3 \vec{j}+2 \vec{k}\), then the area of the quadrilateral PQRS is ………………..
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 55
Solution:
(c) \(\frac{5 \sqrt{3}}{2}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 56

Question 12.
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 57
Solution:
(c) \(\vec{c}\) parallel to \(\vec{a}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 58

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6

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 7.6

Question 1.
Find the absolute extrema of the following functions on the given closed interval.
(i) f(x) = x3 – 12x + 10; [1, 2]
(ii) f(x) = 3x4 – 4x3 ; [-1, 2]
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 1
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 2
∴ Absolute maximum is – 1 and absolute minimum is – 6

(ii) f(x) = 3x4 – 4x³; [-1, 2]
f'(x) = 12x³ – 12x²
f'(x) = 0 ⇒ 12x²(x – 1) = 0
x = 0, 1 ∈ (-1, 2)
Critical points x = 0, 1 and end points of the interval x = -1, 2
f(-1) = 3 + 4 = 7
f(0) = 0 – 0 = 0
f(1) = 3 – 4 = -1
f(2) = 48 – 32 = 16
Absolute maximum f(2) = 16
Absolute minimum f(l) = -1

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6

(iii)
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 3

(iv) f(x) = 2 cos x + sin 2x
f'(x) = -2 sin x + 2 cos 2x
f'(x) = 0 ⇒ cos 2x = sin x
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 4

Question 2.
Find the intervals of monotonicities and hence find the local extremum for the following functions
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 5
Solution:
(i) f(x) = 2x3 + 3x2 – 12x
f'(x) = 6x2 + 6x – 12
f'(x) = 0 ⇒ 6(x2 + x – 2) = 0
(i.e.,) 6(x + 2)(x – 1) = 0
⇒ x = -2 or 1
Taking the points in the number line
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 6
The intervals are (-∞, -2), (-2, 1), (1, ∞)
when x ∈ (-∞, -2), f'(x) = 6 (-1) (-4) = +ve
say x = – 3
⇒ f(x) is strictly increasing in the interval (-∞, -2) when x ∈ (-2, 1), f’ (x) = 6 (2) (-1) = -ve
say x = 0
⇒ f(x) is strictly decreasing in the interval (-2, 1)
when x ∈ (1, ∞), f'(x) = 6 (4) (+1) = + ve say x = 2
⇒ f(x) is strictly increasing in (1, ∞)
Since f(x) changes from +ve to – ve when passing through -2, the first derivative, test tells us there is a local maximum at x = -2 and the local maximum value is f(-2) = 20.
Again f'(x) changes from – ve to +ve when passing through 1 ⇒ there is a local minimum at x = 1 and the local minimum value is f( 1) = -7. So (1 )f(x) is strictly increasing on (-∞, -2) and (1, ∞). And (2)f(x) is strictly decreasing on (-2, 1)
The local maximum = 20 and the local minimum = -7
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 7
f(x) is strictly decreasing on (-∞, 5) and (5, ∞)
And there is no local extremum
(iii)
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 8
For all x values, so f(x) is strictly increasing in (-∞, ∞) and there is no local extremum.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 9
⇒ f(x) is strictly decreasing in (0, 1)
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 11
say x = 2
⇒ f(x) is strictly increasing in (1, ∞)
since f'(x) changes from -ve to +ve at x = 1, there is a local minimum at x = 1 and the local minimum values is f(1) = \(\frac{1}{3}-0=\frac{1}{3}\)
So the function is strictly decreasing on (0, 1) and strictly increasing on (1, ∞) and the local minimum value is \(\frac{1}{3}\)
(v)
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 111
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 12
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 13

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 Additional Problems

Find the absolute maximum and absolute minimum values of f on the given interval.

Question 1.
f(x) = 1 – 2x – x2, [-4, 1]
Solution:
f(x) = 1 – 2x – x2
f'(x) = -2 – 2x
f'(x) = 0 ⇒ 2x = -2 ⇒ x = -1; The points are -4, -1, 1
At x = -4, f(x) = 1 – 2(-4) – (-4)2 = 1 + 8 – 16 = -7
At x = -1, f(x) = 1 – 2 (-1) – (-1)2 = 1 + 2 – 1 = 2
At x = 1, f(x) = 1 – 2 (1) – (1)2 = 1 – 2 – 1 = – 2
Therefore, the absolute minimum is -7 and the absolute maximum is 2.

Question 2.
f(x) = x3 – 12x + 1, [-3, 5]
Solution:
f(x) = x3 – 12x + 1
f'(x) = 3x2 – 12
f'(x) = 0 ⇒ 3x2 – 12 = 0
3x2 = 12 ⇒ x2 = 4 ⇒ x = ±2
The x values are -3, -2, 2, 5
f(x) (at x = -3) = (-3)3 – 12 (-3) + 1 = -27 + 36 + 1 = 10
f(x) (at x = -2) = (-2)3 – (12)(-2) + 1 = -8 + 24 + 1 = 17
f(x) (at x = 2) = 23 – 12 (2) + 1 = 8 – 24 + 1 = -15
f(x) (at x = 5) = 53 – 12 (5) + 1 = 125 – 60 + 1 = 66
From the above four values 10,17,-15 and 66, we see that absolute maximum is 66 and the absolute minimum is -15.

Question 3.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 14
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 15
So, the absolute maximum is \(\frac{2}{3}\) and the absolute minimum is \(\frac{1}{2}\)

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6

Question 4.
f(x) = sin x + cos x, [0, π/3]
f'(x) = sin x + cos x
f'(x) = cos x – sin x
f’ (x) = 0 ⇒ cos x – sin x = 0
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 17

Question 5.
f(x) = x – 2 cos x, [-π, π]
Solution:
f(x) = x – 2 cos x
f'(x) = 1 + 2 sin x
f'(x) = 0 ⇒ 1 + 2 sin x = 0
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 18
From the above values, absolute maximum is π + 2 and the absolute minimum is \(\frac{-\pi}{6}-\sqrt{3}\)
From the local maximum and minimum values of the following functions.

Question 6.
2x3 + 5 x2 – 4x
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 19
Local maximum = 12

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6

Question 7.
t + cos t
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.6 20
t = \(\frac{\pi}{2}\) is neither a maximum point nor a minimum point. So there is no maximum or minimum.

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

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 9 Applications of Integration Ex 9.10

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 1
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 2
Solution:
(a) \(\frac{\pi}{6}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 3

Question 2.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 4
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 5
Solution:
(c) \(\frac{5}{2}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 7

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

Question 3.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 8
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 9
Solution:
(c) 0
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 10

Question 4.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 11
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 12
Solution:
(d) \(\frac{2}{3}\)
Hint:
It is an even function
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 13

Question 5.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 14
(b) 2π
(c) 3π
(d) 4π
Solution:
(d) 4π
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 144

Question 6.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 15
(a) 4
(b) 3
(c) 2
(d) 0
Solution:
(c) 2
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 155

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

Question 7.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 16
(a) cos x – x sin x
(b) sin x + x cos x
(c) x cos x
(d) x sin x
Solution:
(c) x cos x
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 166

Question 8.
The area between y2 = 4x and its latus rectum is ………
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 17
Solution:
(c) \(\frac{8}{3}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 18
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 19

Question 9.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 20
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 21.
Solution:
(b) \(\frac{1}{10100}\)
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 22

Question 10.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 23
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 24
Solution:
(a) \(\frac{\pi}{2}\)
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 25

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

Question 11.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 26
(a) 10
(b) 5
(c) 8
(d) 9
Solution:
(d) 9
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 27

Question 12.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 28
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 29
Solution:
(b) \(\frac{2}{9}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 30

Question 13.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 31
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 32
Solution:
\(\frac{3 \pi}{8}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 33

Question 14.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 34
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 35
Solution:
(d) \(\frac{2}{27}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 36

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

Question 15.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 37
(a) 4
(b) 1
(c) 3
(d) 2
Solution:
(d) 2
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 38

Question 16.
The volume of solid of revolution of the region bounded by y2 = x(a – x) about x-axis is ……..
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 39
Solution:
(d) \(\frac{\pi a^{3}}{6}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 40

Question 17.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 41
(a) 3
(b) 6
(c) 9
(d) 5
Solution:
(c) 9
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 42

Question 18.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 43
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 44
Solution:
(d) \(\frac{\pi^{2}}{4}-2\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 45

Question 19.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 46
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 47
Solution:
(b) \(\frac{3 \pi a^{4}}{16}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 48

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

Question 20.
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 49
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 50
Solution:
(a) \(\frac{1}{2}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 51

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 Additional Problems

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.
The area bounded by the line y = x, the x – axis, the ordinates x = 1,x = 2 is …….
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 52
Solution:
(a) \(\frac{3}{2}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 53

Question 2.
The area of the region bounded by the graph of y = sin x and y = cos x between x = 0 and x = \(\frac{\pi}{4}\) is ……..
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 54
Solution:
(b) \(\sqrt{2}-1\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 55

Question 3.
The area between the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 and its auxiliary circle is …….
(a) πb(a – b)
(b) 2πa(a – b)
(c) πa(a – b)
(d) 2πb(a – b)
Solution:
(c) πa(a – b)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 56

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

Question 4.
The area bounded by the parabola y2 = x and its latus rectum is ……..
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 57
Solution:
(b) \(\frac{1}{6}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 58

Question 5.
The volume of the solid obtained by revolving \(\frac{x^{2}}{9}+\frac{y^{2}}{16}\) = 1 about the minor axis is …….
(a) 48π
(b) 64π
(c) 32π
(d) 128π
Solution:
(b) 64π
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 59
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 60

Question 6.
The volume, when the curve y = \(\sqrt{3+x^{2}}\) from x = 0 to x = 4 is rotated about x – axis is ……
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 61
Solution:
(c) \(\frac{100}{3} \pi\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 62

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

Question 7.
The volume generated when the region bounded by y = x, y = 1, x = 0 is rotated about y – axis is ……….
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 63

Solution:
(c) \(\frac{\pi}{3}\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 64

Question 8.
Volume of solid obtained by revolving the area of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 about major and minor axes are in the ratio …….
(a) b2 : a2
(b) a2 : b2
(c) a : b
(d) b : a
Solution:
(d) b : a
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 65

Question 9.
The volume generated by rotating the triangle with vertices at (0, 0), (3, 0) and (3, 3) about x-axis is …….
(a) 18π
(b) 2π
(c) 36π
(d) 9π
Solution:
(d) 9π
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 66

Question 10.
The length of the arc of the curve Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 611 is …….
(a) 48
(b) 24
(c) 12
(d) 96
Solution:
(a) 48
Hint:
Length of the arc of the curve = 6a
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 67
∴ Required length = 6a = 6 × 8 = 48 units.

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

Question 11.
The surface area of the solid of revolution of the region bounded by y = 2x, x = 0 and x = 2 about x-axis is ……
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 68
Solution:
(a) \(8 \sqrt{5} \pi\)
Hint:
Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 69

Question 12.
The curved surface area of a sphere of radius 5, intercepted between two parallel planes of distance 2 and 4 from the centre is ……
(a) 20π
(b) 40π
(c) 10π
(d) 30π
Solution:
(a) 20π
Hint:
The curved surface area of a sphere of radius r intercepted between two parallel planes at a distance a and b from the centre of the sphere is 2πr (b – a)
Given radius, r = 5; a = 2; b = 4
Required surface area = 2πr (b – a)
= 2π × 5 × (4 – 2)
= 20π sq. units

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7

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 7.7

Question 1.
Find intervals of concavity and points of inflexion for the following functions:
(i) f(x) = x (x – 4)3
(ii) f(x) = sin x + cos x, 0 < x < 2π
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 1
Solution:
(i) f(x) = x(x – 4)3
f(x) = x[x3 – 12x2 + 48x – 64]
(i.e.,) f(x) = x4 – 12x3 + 48x2 – 64x
f'(x) = 4x3 – 36x2 + 96x – 64
f”(x) = 12x2 – 72x + 96
f”(x) = 0 ⇒ 12(x2 – 6x + 8) = 0
⇒ 12 (x – 2) (x – 4) = 0 ⇒ x = 2 or 4
Marking the points in the number line
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 2
in the interval (-∞, 2) the curve concave upwards
when x ∈ (2, 4), f”(x) = (3 – 2) (3 – 4)
[say x = 3] = -ve
⇒ The curve concave upwards in (2,4)
when x ∈ (4, ∞), f”(x) = (5 – 2) (5 – 4)
[say x = 5] = (+) (+) = +ve ⇒ The curve concave downwards is (4, ∞) in (-∞, 2) concave upwards and in (2, 4) the concave downwards
⇒ x = 2 is a point of inflection at x = 2, f(2) = -16
So (2, -16) is a point of inflection
Again in (2, 4) the curve concave downwards and in (4, ∞) the curve concave upwards ⇒ x = 4 is a point of inflection f(4) = 0
∴ (4, 0) is a point of inflection
So points of inflection are (2, -16) and (4, 0)

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7

(ii) f(x) = sin x + cos x
f'(x) = cos x – sin x
f”(x) = – sin x – cos x
f”(x) = 0 ⇒ – sin x = cos x
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 3
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 4
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 5
So the intervals are (-∞, 0), (0, ∞) when x ∈ (-∞, 0) f”(x) is negative
⇒ the curve concave downwards and when x ∈ (0, ∞) f”(x) is positive
⇒ the curve concave upwards
⇒ x = 0 is a point of inflection f(0) = \(\frac{1}{2}\)(1 – 1) = 0
So (0, 0) is the point of inflection and the curve concave upwards in (0, ∞) and curve concave downwards in (-∞, 0) and (0, 0) is the point of inflection.

Question 2.
Find the local extrema for the following functions using second derivative test:
(i) f(x) = – 3x5 + 5x3
(ii) f(x) = x log x
(iii) f(x) = x2 e-2x
Solution:
(i) f(x) – 3x5 + 5x3
f'(x) = -15x4 + 15x²
For maximum or minimum f'(x) = 0
⇒ -15x²(x² – 1) = 0
x = 0, 1, -1 [x = 0 is not possible as it gives no extremum.
f'(x) = -60x³ + 30x
f”(-1) > 0 ⇒ f(x) attains minimum
f”(1) < 0 ⇒ f(x) attains maximum
∴ Local minimum f(-1) = -3(-1) + 5(-1) = -2
Local maximum f(1) = -3(1) + 5(1) = 2

(ii) f(x) = x log x
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 6

(iii) f(x) = x² e-2x
f'(x) = -2x² e-2x + 2x e-2x
2x For maximum or minimum, f'(x) = 0
⇒ -2x e-2x(x – 1) = 0
x = 0, 1
f”(x) = -2 [- 2x² e-2x + 4x e-2x – e-2x]
at x = 0, f”(x) > 0 ⇒ f(x) attains minimum
at x = 1, f”(x) < 0 ⇒ f(x) attains maximum
∴ Local minimum f(0) = 0
Local maximum f(1) = \(\frac { 1 }{ e^2 }\)

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7

Question 3.
For the function f(x) = 4x3 + 3x2 – 6x + 1 find the intervals of monotonicity, local extrema, intervals of concavity and points of inflection.
Solution:
4x3 + 3x2 – 6x + 1
f'(x) = 12x2 + 6x – 6
f”(x) = 24x + 6
f'(x) = 0 ⇒ 6(2x2 + x – 1) = 0
6(x + 1)(2x – 1) = 0
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 25
From (1) and (2)
x = -1 is a maximum point,
and f(-1)= -4 + 3 + 6 + 1 = 6
So local maximum is 6
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 8
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 9
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 10

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 Additional Problems

Question 1.
Discuss the curves y = x4 – 4x3 with respect to concavity and points of inflection.
Solution:
f(x) = x4 – 4x3 ⇒ f'(x) = 4x3 – 12x2
f”(x) = 12x2 – 24x = 12x (x – 2)
Since f”(x) = 0 when x = 0 or 2, we divide the real line into three intervals.
(-∞, 0) , (0, 2), (2, ∞) and complete the following chart.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 11
The point (0, f(0) i.e., (0, 0) is an inflection point since the curve changes from concave upward to concave downward there. Also (2, f(2)) i.e., (2, -16) is an inflection point since the curve changes from concave downward to concave upward there.

Question 2.
Find the intervals of concavity and the points of inflection of the following functions.
f(x) = 2x3 + 5x2 – 4x
Solution:
y = f (x) = 2x3 + 5x2 – 4x
f'(x) = 6x2 + 10x – 4
f”(x) = 12x + 10; f'”(x) = 12
f”(x) = 0 ⇒ 12x + 10 = 0
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 88
Consider x in (-∞, -5/6) say x = -1
f”(x)= -12 + 10 = -2 < 0
⇒ the curve convex upward in the interval (-∞, -5/6)
Consider x in (-5/6, ∞) say x = 0 f”(x) = 0 + 10 = 10 > 0
⇒ the curve is concave upward in (-5/6, -∞)
Thus, the curve is concave upward in (-5/6, ∞) and convex upward in (-∞, – 5/6)
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 12

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7

Question 3.
f(x) = x4 – 6x2
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 13
So, the points of inflection are at x = ± 1.
f”(x) = 0 ⇒ x2 = 1 ⇒ x = ± 1
So, the intervals are (-∞, -1), (-1, 1), (1, ∞)
When x ∈ (-∞, -1), say x = -2
f”(x) = 12(-2)2 – 12 = 48 – 12 = 36 > 0
⇒ f(x) is concave upward in (-∞, -1) when x ∈ (-1, 1) say x = 0.
f”(x) = 0 – 12 = -12 < 0
⇒ f(x) convex upward.
⇒ x = -1 is a point of inflection.
Again when x ∈ (-1, 1), the curve is convex upward and when x ∈ (1, ∞) say x = 2.
f”(x) = (1, ∞) say x = 2.
f”(x) = 12 (4) – 12 > 0
⇒ the curve is concave upward.
⇒ x = 1 is a point of inflection.
Thus x = ± 1 are the points of inflection.
At x = 1, f(x) = 1 – 6 = -5
At x = -1, f(x) = (-1)4 – 6(-1)2 = 1 – 6 = -5
The curve concave upward in (-∞, -1) u (1, ∞) and convex upward in (-1, 1) and the points of inflection are (1, -5) and (-1, -5).

Question 4.
f(θ) = sin 2θ in (θ, π)
Solution:
f'(θ) = sin 2θ
f’ (θ) = (cos 2θ) (2) = 2 cos 2θ
f”(θ)= 2[-sin 2θ] (2) = – 4 sin 2θ
f”(θ) = 0
⇒ -4 sin 2θ = 0
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 14

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7

Question 5.
y = 12x2 – 2x3 – x4
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 15
So the intervals are (-∞, -2), (-2, 1)(1, ∞)
When x ∈ (-∞, -2) say x = -3
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 16
The curve is concave upward.
When x ∈ (-2, 1) say x = 0.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 17
⇒ The curve is concave upward.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 18
⇒ The curve is convex upward. So, x = 1 is a point of inflection.
At x = – 2, y = 12(4) – 2(-8) – (16) = 48 + 16 – 16 = 48
At x = 1, y = 12(1) – 2(1) – 1 = 12 – 2 – 1 = 9
So, the points of inflection are (1, 9) and (-2, 48)

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8

Question 1.
Find two positive numbers whose sum is 12 and their product is maximum.
Solution:
Let the two positive numbers be ‘x’ and ‘y’
Given sum is 12 ⇒ x + y = 12
y = 12 – x
Product P = xy
P = x(12 – x)
P = 12x – x²
\(\frac { dP }{ dx }\) = 12 – 2x
For maximum or minimum,
\(\frac { dP }{ dx }\) = 0 ⇒ 12 – 2x = 0
x = 6
\(\frac { d^2P }{ dx^2 }\) = -2
at x = 6, \(\frac { d^2P }{ dx^2 }\) = -2 < 0
∴ Product ‘P’ is maximum when x = 6
∴ y = 12 – 6 = 6
Hence, the positive numbers are 6 and 6 and their product is 36.

Question 2.
Find two positive numbers whose product is 20 and their sum is minimum.
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 1
but x is positive (given)
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 2

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8

Question 3.
Find the smallest possible value of x2 + y2 given that x + y = 10.
Solution:
Given x + y = 10 ⇒ 7 = 10 – x
To find the smallest value of x2 + y2
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 3
at x = 5, f”(x) = 4 at x = 5, y = 10 – 5 = 5 = +ve
x = 5 is a minimum point.
So the minimum value of x2 + y2 = 52 + 52 = 50

Question 4.
A garden is to be laid out in a rectangular area and protected by a wire fence. What is the largest possible area of the fenced garden with 40 meters of wire?
Solution:
Let the length of the garden be ‘x’ m
Let the breadth of the garden be ‘y’ m
The Garden is fenced with 40 m wire
i.e., Perimeter = 2 (x + y) = 40
x + y = 20
y = 20 – x
Area of the Garden A = xy
A = x (20 – x)
A = 20x – x²
\(\frac { dA }{ dx }\) = 20 – 2x
For maximum or minimum,
\(\frac { dA }{ dx }\) = 0 ⇒ 20 – 2x = 0
x = 10
\(\frac { d^2A }{ dx^2 }\) = -2
at x = 10, \(\frac { d^2A }{ dx^2 }\) < 0 Now, y = 20 -10 = 10
Maximum area = 10 × 10 = 100 sq. m;

Question 5.
A rectangular page is to contain 24 cm2 of print. The margins at the top and bottom of the page are 1.5 cm and the margins at other sides of the page is 1 cm. What should be the dimensions of the page so that the area of the paper used is minimum.
Solution:
Let the length of the printed page be = x cm
and breadth = y cm
Now xy = 24
⇒ y = \(\frac{24}{x}\)
The length of the paper = y + 3
Area A = (x + 2)(y + 3)
= xy + 3x + 2y + 6
= 24 + 3x + 2y + 6
= 3x + 2y + 30 ……. (2)
Substituting (1) in (2) we get
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 614
∴ Dimensions of the paper are
x + 2 = 4 + 2 = 6 cm
and y + 3 = 6 + 3 = 9 cm
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 26

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8

Question 6.
A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 1,80,000 sq. mtrs in order to provide enough grass for herds. No fencing is needed along the river. What is the length of the minimum needed fencing material?
Solution:
Given Area = 180000 sq. meters
Let length be = x
and breadth be = y
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 27
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 28

Question 7.
Find the dimensions of the rectangle with the maximum area that can be inscribed in a circle of a radius of 10 cm.
Solution:
P is a point on the circumference of a circle of radius 10 cm P = (10 cos α, 10 sin α)
∴ PQ = 20 sinα and
PS = 20 cos α
A = area of PQRS = (20 sin α) (20 cos α)
= 400 sin α cos α
= (200) (2 sin α cos α)
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 30
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 300

Question 8.
Prove that among all the rectangles of the given perimeter, the square has the maximum area.
Solution:
Let the length and breadth of the rectangle be x and y respectively.
P = 2(x + y) [given]
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 31
Substitute (3) in (2) we get
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 32
so x = 6 cm and y = 3 cm

Question 9.
Find the dimensions of the largest rectangle that can be inscribed in a semi-circle of radius r cm.
Solution:
Let θ be the angle made by OP with the positive direction of the x-axis.
Then the area of rectangle A is
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 35
A(θ) = (2 r cos θ)(r sin θ)
= r2 2 sin θ cos θ = r2 sin 2θ
Now A(θ) is maximum when sin 2θ is maximum.
The maximum value of
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 36

Question 10.
A manufacturer wants to design an open box having a square base and a surface area of 108 sq. cm. Determine the dimensions of the box for the maximum volume.
Solution:
Let the side of the square base be = x cm and the height be = y cm
Surface area = 108 sq cm
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 37
Substituting (3) in (2) we get
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 377
so x = 6 cm and y = 3 cm

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8

Question 11.
The volume of a cylinder is given by the formula V = πr2h. Find the greatest and least values of V if r + h = 6.
Solution:
V = πr2h
Given r + h = 6 ⇒ h = 6 – r
V = πr2(6 – r) = 6πr2 – πr3
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 38
3πr(4 – r) = 0 ⇒ r = 0 or 4
when r = 4, h = 2
So v = π(16)(2) = 32π
when r = 0, V = 0
So the maximum volume = 32π and the minimum volume = 0

Question 12.
A hollow cone with a base radius of a cm and a height of b cm is placed on a table. Show that the volume of the largest cylinder that can be hidden underneath is \(\frac{4}{9}\) times the volume of the cone.
Solution:
The height of cone = h = b
The base radius = r = a
The base radius of cylinder = r
The height of cylinder = h
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 39
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 399

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 Additional Problems

Question 1.
The top and bottom margins of a poster are each 6 cms and the side margins are each 4 cms. If the area of the printed material on the poster is fixed at 384 cms2, find the dimension of the poster with the smallest area.
Solution:
Let x and y be the length and breadth of the printed area, then the area xy = 384
Dimensions of the poster area are (x + 8) and (y + 12) respectively.
Poster area A = (x + 8) (y + 12)
= xy + 12x + 8y + 96
= 12x + 8y + 480
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 60
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 600
But x > 0
∴ x = 16
when x = 16, A” > 0
when x = 16, the area is minimum
y = 24
∴ x + 8 = 24,
y + 12 = 36
Hence the dimensions are 24 cm and 36 cm

Question 2.
Show that the volume of the largest right circular cone that can be inscribed in a sphere of radius a is \(\frac{8}{27}\) (volume of the sphere).
Solution:
Given that a is the radius of the sphere and let x be the base radius of the cone.
If h is the height of the cone, then its volume is
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 61
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 611
where OC = y so that height h = a + y
From the diagram x2 + y2 = a2
Using (2) in (1) we have
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 62
For the volume to be maximum:
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 63
⇒ 3y = + a or y = -a
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 64

Question 3.
A closed (cuboid) box with a square base is to have a volume of 2000 c.c. The material for the top and bottom of the box is to cost Rs. 3 per square cm and the material for the sides is to cost Rs. 1.50 per square cm. If the cost of the materials is to be the least, find the dimensions of the box.
Solution:
Let x, y respectively denote the length of the side of the square base and the depth of the box.
Let C be the cost of the material
Area of the bottom = x2
Area of the top = x2
The combined area of the top and bottom = 2x2
Area of the four sides = 4xy
Cost of the material for the top and bottom = 3(2x)2
Cost of the material for the sides = (1.5)(4xy) = 6xy
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 65

Question 4.
Find two numbers whose sum is 100 and whose product is a maximum.
Solution:
Let the two numbers be x and y.
x + y = 100
⇒ y = 100 – x
Product = xy = x(100 – x)
f = x(100 – x) = 100x – x2
We have to find x at which f is maximum.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 66
∴ f is maximum at x = 50
So, y = 100 – x = 100 – 50 = 50
So, the two numbers are 50, 50.

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8

Question 5.
Find two positive numbers whose product is 100 and whose sum is minimum.
Solution:
Let the two numbers be x and y.
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 67
To find x at which f is maximum
Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 68
So the two numbers are 10, 10.