Class 11

Students can Download Physics Chapter 4 Work, Energy and Power Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 4 Work, Energy and Power

Samacheer Kalvi 11th Physics Work, Energy and Power Textual Questions Solved

Samacheer Kalvi 11th Physics Work, Energy and Power Multiple Choice Questions
Question 1.
A uniform force of (\(2 \hat{i}+\hat{j}\)) + N acts on a particle of mass 1 kg. The particle displaces from position \((3 \hat{j}+\hat{k})\) m to \((5 \hat{i}+3 \hat{j})\) m. Th e work done by the force on the particle is
[AIPMT model 2013]
(a) 9 J
(b) 6 J
(c) 10 J
(d) 12 J
Answer:
(c) 10 J

Question 2.
A ball of mass 1 kg and another of mass 2 kg are dropped from a tall building whose height is 80 m. After, a fall of 40 m each towards Earth, their respective kinetic energies will be in the ratio of [AIPMT model 2004]
(a) \(\sqrt{2}\) : 1
(b) 1 : \(\sqrt{2}\)
(c) 2 : 1
(d) 1 : 23
Answer:
(d) 1 : 23

Question 3.
A body of mass 1 kg is thrown upwards with a velocity 20 m s^-1. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction?
(Take g = 10 ms^-2) [AIPMT 2009]
(a) 20 J
(b) 30 J
(c) 40 J
(d) 10 J
Answer:
(a) 20 J

Question 4.
An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water of the jet. What is the rate at which kinetic energy is imparted to water ? [AIPMT 2009]

Answer:
(a) \(\frac{1}{2} m v^{2}\)

Question 5.
A body of mass 4 m is lying in xv-plane at rest. It suddenly explodes into three pieces. Two pieces each of mass m move perpendicular to each other with equal speed v the total kinetic energy generated due to explosion is [AIPMT 2014]

Answer:
(b) \(\frac{3}{2} m v^{2}\)

The potential energy calculator to find how much energy is stored in an object raised off the ground.

Question 6.
The potential energy of a system increases, if work is done
(a) by the system against a conservative force
(b) by the system against a non-conservative force
(c) upon the system by a conservative force
(d) upon the system by a non-conservative force
Answer:
(a) by the system against a conservative force

Question 7.
What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

Answer:
(c) \(\sqrt{5 g R}\)

Question 8.
The work done by the conservative force for a closed path is
(a) always negative
(b) zero
(c) always positive
(d) not defined
Answer:
(b) zero

Question 9.
If the linear momentum of the obj ect is increased by 0.1 %, then the kinetic energy is increased by
(a) 0.1%
(b) 0.2%
(c) 0.4%
(d) 0.01%
Answer:
(b) 0.2%

Question 10.
If the potential energy of the particle is , then force experienced by the particle is

Answer:
(c) F = -βx

Question 11.
A wind-powered generator converts wind energy into electric energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to
(a) v
(b) v²
(c) v³
(d) v⁴
Answer:
(c) v⁴

Question 12.
Two equal masses m₁ and m₂ are moving along the same straight line with velocities 5 ms^-1 and -9 ms^-1 respectively. If the collision is elastic, then calculate the velocities after the collision of Wj and m2, respectively
(a) -4 ms^-1 and 10 ms^-1
(b) 10 ms^-1 and 0 ms^-1
(c) -9 ms^-1 and 5 ms^-1
(d) 5 ms^-1 and 1 ms^-1
Answer:
(c) -9 ms^-1 and 5 ms^-1

Question 13.
A particle is placed at the origin and a force F = kx is acting on it (where k is a positive constant). If U(0) = 0, the graph of U(x) versus x will be (where U is the potential energy function) [IIT 2004]

Answer:

Question 14.
A particle which is constrained to move along x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = -kx + ax³. Here, k and a are positive constants. For x ≥ 0, the functional form of the potential energy U(x) of the particle is [IIT 2002]

Answer:

Question 15.
A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then, the long piece will have a force constant of

Answer:
(b) \(\frac{3}{2} k\)

Samacheer Kalvi 11th Physics Work, Energy and Power Short Answer Questions

Question 1.
Explain how the definition of work in physics is different from general perception.
Answer:
The term work is used in diverse contexts in daily life. It refers to both physical as well as mental work. In fact, any activity can generally be called as work. But in Physics, the.term work is treated as a physical quantity with a precise definition. Work is said to be done by the force when the force applied on a body displaces it.

Question 2.
Write the various types of potential energy. Explain the formulae.
Answer:
(a) U = mgh
U – Gravitational potential energy
m – Mass of the object,
g – acceleration due to gravity
h – Height from the ground,

u – Elastic potential energy
k – String constant; x-displacement.

U – electrostatic potential energy
\(\varepsilon_{0}\) = absolute permittivity
q₁, q₂ – electric charges

Question 3.
Write the differences between conservative and non-conservative forces. Give two examples each.
Answer:

Question 4.
Explain the characteristics of elastic and inelastic collision.
Answer:
In any collision process, the total linear momentum and total energy are always conserved whereas the total kinetic energy need not be conserved always. Some part of the initial kinetic energy is transformed to other forms of energy. This is because, the impact of collisions and deformation occurring due to collisions may in general, produce heat, sound, light etc. By taking these effects into account, we classify the types of collisions as follows:
(a) Elastic collision
(b) Inelastic collision
(a) Elastic collision: In a collision, the total initial kinetic energy of the bodies (before collision) is equal to the total final kinetic energy of the bodies (after collision) then, it is called as elastic collision, i.e.,
Total kinetic energy before collision = Total kinetic energy after collision
(b) Inelastic collision: In a collision, the total initial kinetic energy of the bodies (before collision) is not equal to the total final kinetic energy of the bodies (after collision) then, it is called as inelastic collision, i.e.,
Total kinetic energy before collision ≠ Total kinetic energy after collision

Even though kinetic energy is not conserved but the total energy is conserved. This is because the total energy contains the kinetic energy term and also a term ∆Q, which includes all the losses that take place during collision. Note that loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, etc. Further, if the two colliding bodies stick together after collision such collisions are known as completely inelastic collision or perfectly inelastic collision. Such a collision is found very often. For example when a clay putty is thrown on a moving vehicle, the clay putty (or Bubblegum) sticks to the moving vehicle and they move together with the same velocity.

Question 5.
Define the following
(a) Coefficient of restitution
(b) Power
(c) Law of conservation of energy
(d) Loss of kinetic energy in inelastic collision.
Answer:
(a) The ratio of velocity of separation after collision to the velocity of approach before collision

(b) Power is defined as the rate of work done or energy delivered

Its unit is watt.
(c) The law of conservation of energy states that energy can neither be created nor destroyed. It may be transformed from one form to another but the total energy of an isolated system remains constant.
(d) In perfectly inelastic collision, the loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, heat, light etc. Let KE_i be the total kinetic energy before collision and KE_f be the total kinetic energy after collision.
Total kinetic energy before collision,

Total kinetic energy after Collision,

Then the loss of kinetic energy is

Samacheer Kalvi 11th Physics Work, Energy and Power Long Answer Questions

Question 1.
Explain with graphs the difference between work done by a constant force and by a variable force.
Answer:
Work done by a constant force: When a constant force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation,
dW = (F cos θ) dr ..(1)
The total work done in producing a displacement from initial position r_i to final position r_f is,

The graphical representation of the work done by a constant force is shown in figure given below. The area under the graph shows the work done by the constant force.

Work done by a variable force: When the component of a variable force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation
dW = F cos θ dr [F cos θ is the component of the variable force F]
where, F and θ are variables. The total work done for a displacement from initial position r_i to final position r_f is given by the relation,

A graphical representation of the work done by a variable force is shown in figure given below. The area under the graph is the work done by the variable force.

Question 2.
State and explain work energy principle. Mention any three examples for it.
Answer:
(i) If the work done by the force on the body is positive then its kinetic energy increases.
(ii) If the work done by the force on the body is negative then its kinetic energy decreases.
(iii) If there is no work done by the force on the body then there is no change in its kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.
(iv) When a particle moves with constant speed in a circle, there is no change in the kinetic energy of the particle. So according to work energy principle, the work done by centripetal force is zero.

Question 3.
Arrive at an expression for power and velocity. Give some examples for the same.
Answer:
The work done by a force \(\overrightarrow{\mathrm{F}}\) for a displacement \(d \vec{r}\) is

Left hand side of the equation (i) can be written as

Since, velocity is . Right hand side of the equation (i) can be written as dt

Substituting equation (ii) and equation (iii) in equation (i), we get

This relation is true for any arbitrary value of dt. This implies that the term within the bracket must be equal to zero, i.e.,

Hence power \(\mathrm{P}=\overrightarrow{\mathrm{F}} \cdot \vec{v}\)

Question 4.
Arrive at an expression for elastic collision in one dimension and discuss various cases.
Answer:
Consider two elastic bodies of masses m₁ and m₂ moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure given below.

In order to have collision, we assume that the mass m] moves faster than mass m₂ i.e., u₁ > u₂. For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same.

From the law of conservation of linear momentum,
Total momentum before collision (p_i) = Total momentum after collision (p_f)
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ …(i)


This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for v₁ and v₂,
v₁ = v₂ + u₂ – u₂ …(vi)
Or v₂ = u₁ + v₁ – u₂ …(vii)
To find the final velocities v₁ and v₂:
Substituting equation (vii) in equation (ii) gives the velocity of as m₁ as
m₁ (u₁ – v₁) = m₂(u₁ + v₁ – u₂ – u₂)
m₁ (u₁ – y₁) = m₂ (u₁ + + v₁  – 2u₂)
m₁u₁ – m₁v₁ = m₂u₁ + m₂v₁ + 2m₂u₂
m₁u₁ – m₂u₁ + 2m₂u₂ = m₁v₁ + m₂v₁
(m₁– m₂) u₁ + 2m₂u₂ = (m₁ + m₂) v₁

Similarly, by substituting (vi) in equation (ii) or substituting equation (viii) in equation (vii), we get the final velocity of m₂ as

Case 1: When bodies has the same mass i.e., m₁ = m₂,

The equations (x) and (xi) show that in one dimensional elastic collision, when two bodies of equal mass collide after the collision their velocities are exchanged.
Case 2: When bodies have the same mass i.e., m₁ = m₂ and second body (usually called target) is at rest (u₂ = 0),
By substituting m₁ = m₂ = and u₂ = 0 in equations (viii) and equations (ix) we get,
from equation (viii) ⇒ v₁ = 0 …(xii)
from equation (ix) ⇒ v₂ = u₁ ….. (xiii)
Equations (xii) and (xiii) show that when the first body comes to rest the second body moves with the initial velocity of the first body.
Case 3: The first body is very much lighter than the second body

Dividing numerator and denominator of equation (viii) by m₂, we get

Similarly the numerator and denominator of equation (ix) by m₂, we get

The equation (xiv) implies that the first body which is lighter returns back (rebounds) in the opposite direction with the same initial velocity as it has a negative sign. The equation (xv) implies that the second body which is heavier in mass continues to remain at rest even after collision. For example, if a ball is thrown at a fixed wall, the ball will bounce back from the wall with the same velocity with which it was thrown but in opposite direction.

Dividing numerator and denominator of equation (xiii) by m₁, we get

The equation (xvi) implies that the first body which is heavier continues to move with the same initial velocity. The equation (xvii) suggests that the second body which is lighter will move with twice the initial velocity of the first body. It means that the lighter body is thrown away from the point of collision.

Question 5.
What is inelastic collision? In which way it is different from elastic collision. Mention few examples in day to day life for inelastic collision.
Answer:
Inelastic collision: In a collision, the total initial kinetic energy of the bodies (before collision) is not equal to the total final kinetic energy of the bodies (after collision) then, it is called as inelastic collision, i.e.,
Total kinetic energy before collision ≠ Total kinetic energy after collision

Even though kinetic energy is not conserved but the total energy is conserved. This is because the total energy contains the kinetic energy term and also a term ∆Q, which includes all the losses that take place during collision. Note that loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, etc. Further, if the two colliding bodies stick together after collision such collisions are known as completely inelastic collision or perfectly inelastic collision. Such a collision is found very often. For example when a clay putty is thrown on a moving vehicle, the clay putty (or Bubblegum) sticks to the moving vehicle and they move together with the same velocity.
Difference between Elastic & in elastic collision

Samacheer Kalvi 11th Physics Numerical Problems

Question 1.
Calculate the work done by a force of 30N in lifting a load of 2 Kg to a height of 10m(g = 10 ms^-1)
Answer:
Given: F = 30 N, load (m) = 2 kg; height = 10 m, g = 10 ms^-2
Gravitational force F = mg = 30 N
The distance moved h = 10 m
Work done on the object W = Fh = 30 × 10 = 300 J.

Question 2.
A ball with a velocity of 5 ms^-1 impinges at angle of 60° with the vertical on a smooth horizontal plane. If the coefficient of restitution is 0.5, find the velocity and direction after the impact.
Answer:
Given: Velocity of ball: 5 ms^-1
Angle of inclination with vertical: 60°
Coefficient of restitution = 0.5.
Note: Let the angle reflection is θ’ and the speed after collision is v’. The floor exerts a force on the ball along the normal during the collision. There is no force
parallel to the surface. Thus, the parallel component of the velocity of the ball remains unchanged. This gives
v’ sin θ’ = v sin θ …… (i)
Vertical component with respect to floor = v’ cos θ’ (velocity of separation)
Velocity of approach = v cos θ


from (i) and (ii)

Question 3.
A bob of mass m is attached to one end of the rod of negligible mass and length r, the other end of which is pivoted freely at a fixed center O as shown in the figure.
What initial speed must be given to the object to reach the top of the circle?
(Hint: Use law of conservation of energy). Is this speed. less or greater than speed obtained in the section 4.2.9?

Answer:
To get the vertical speed given to the object to reach the top of the circle, law of conservation of energy can be used at a points (1) and (2)
Total energy at 1 = Total energy at 2
∴ Potential energy at point 1 = 0

from eqn (i)

In this case bob of mass m is connected with a rod of negligible mass, so the velocity of bob at highest point can be equal to zero i.e. v₂ = 0

The speed of bob obtained here is lesser than the speed obtained in section 4.2.9. It is only because of string is replaced by a massless rod here.

Question 4.
Two different unknown masses A and B collide. A is initially at rest when B has a speed v. After collision B has a speed v/2 and moves at right angles to its original direction of motion. Find the direction in which A moves after collision.
Answer:

Question 5.
A bullet of mass 20 g strikes a pendulum of mass 5 kg. The centre of mass of pendulum rises a vertical distance of 10 cm. If the bullet gets embedded into the pendulum, calculate its initial speed.
Answer:
Given: m₁ = 20 g = 20 × 10^-3 kg; m₂ = 5 kg; s = 10 × 10^-2 m.
Let the speed of the bullet be v. The common velocity of bullet and pendulum bob is V. According to law of conservation of linear momentum.

The bob with bullet go up with a deceleration of g = 9.8 ms^-2. Bob and bullet come to rest at a height of 10 × 10^-2 m.
from III rd equation of motion

Samacheer Kalvi 11th Physics Conceptual Questions

Question 1.
A spring which in initially in un-stretched condition, is first stretched by a length x and again by a further length x. The work done in the first case W₁ is one third of the work done in second case W₂. True or false?
Answer:
The amount of work done to stretching distance x

Total work done in stretching the spring through a distance 2x is

Extra work required to stretch the additional x distance is
W = W₂ – W₁ = 4W₁ – W₁ = 3W₁

Hence it is true

Question 2.
Which is conserved in inelastic collision? Total energy (or) Kinetic energy?
Answer:
In inelastic collision total energy is only conserved but kinetic energy is not conserved. A part of kinetic energy is converted into some other form of energy such as sound, heat energy.
Note: The linear momentum is also conserved.

Question 3.
Is there any net work done by external forces on a car moving with a constant speed along a straight road?
Answer:
If the car moves at constant speed, then there is no change in its kinetic energy. It implies that if there is no change in kinetic energy then there is no work done by the force on the body provided its mass remains constant.

Question 4.
A car starts from rest and moves on a surface with uniform acceleration. Draw the graph of kinetic energy versus displacement. What information you can get from that graph?
Answer:
A car starts from rest and moves with uniform acceleration. The graph between kinetic energy and displacement, is a straight line.
The slope of KE and displacement graph gives net force acting on the car to keep the car with uniform acceleration.

Question 5.
A charged particle moves towards another charged particle. Under what conditions the total momentum and the total energy of the system conserved?
Answer:
Coulomb force is acting in between the charged particles Internal force is a conservative force. If no external forces act or the work done by external forces is zero then the mechanical energy of the system and also total linear momentum also remains constant.

Samacheer Kalvi 11th Physics Work, Energy and Power Additional Questions Solved

Samacheer Kalvi 11th Physics Multiple Choice Questions

Question 1.
Thrust and linear momentum
(a) Thrust and linear momentum
(b) Work and energy
(c) Work and power
(d) Power and energy
Answer:
(b) Work and energy

Question 2.
The rate of work done is called as
(a) energy
(b) power
(c) force
(d) mechanical energy
Answer:
(b) power

Question 3.
Unit of work done
(a) Nm
(b) joule
(c) either a or b
(d) none
Answer:
(c) either a or b

Question 4.
Dimensional formula for work done is
(a) MLT^-1
(b) ML²T²
(c) M^-1L^-1T²
(d) ML²T^-2
Answer:
(d) ML²T^-2

Question 5.
When a body moves on a horizontal direction, the amount of work done by the gravitational force is
(a) positive
(b) negative
(c) zero
(d) infinity
Answer:
(c) zero

Question 6.
The amount of work done by centripetal force on the object moving in a circular path is
(a) zero
(b) infinity
(c) positive
(d) negative
Answer:
(a) zero

Question 7.
The work done by the goal keeper catches the ball coming towards him by applying a force is
(a) positive
(b) negative
(c) zero
(d) infinity
Answer:
(b) negative

Question 8.
If the angle between force and displacement is acute then the work done is
(a) positive
(b) negative
(c) zero
(d) maximum
Answer:
(a) positive

Question 9.
If the force and displacement are perpendicular to each other, then the work done is
(a) positive
(b) negative
(c) zero
(d) maximum
Answer:
(c) zero

Question 10.
If the angle between force and displacement is obtuse, then the work done is
(a) positive
(b) negative
(c) zero
(d) minimum
Answer:
(b) negative

Question 11.
The area covered under force and displacement graph is
(a) work done
(b) acceleration
(c) power
(d) kinetic energy
Answer:
(a) work done

Question 12.
The capacity to do work is
(a) force
(b) energy
(c) work done
(d) power
Answer:
(b) energy

Question 13.
The energy possessed by a body due to its motion is called as
(a) potential energy
(b) kinetic energy
(c) mechanical energy
(d) none
Answer:
(b) kinetic energy

Question 14.
The energy possessed by the body by virtue of its position is called as
(a) potential energy
(b) kinetic energy
(c) mechanical energy
(d) none
Answer:
(a) potential energy

Question 15.
1 erg is equivalent to
(a) 10^-7 J
(b) 1.6 × 10^-19 J
(c) 4.186 J
(d) 3.6 × 10^-6 J
Answer:
(a) 10^-7 J

Question 16.
1 electron volt is equivalent to
(a) 10^-7 J
(b) 1.6 × 10^-19 J
(c) 4.186 J
(d) 3.6 × 10^-6 J
Answer:
(b) 1.6 × 10^-19 J

Question 17.
1 kilowatt hour is equivalent to
(a) 10^-7 J
(b) 1.6 × 10^-19 J
(c) 4.186 J
(d) 3.6 × 10^-6 J
Answer:
(d) 3.6 × 10^-6 J

Question 18.
1 calorie is equivalent to
(a) 10^-7 J
(b) 1.6 × 10^-19 J
(c) 4.186 J
(d) 3.6 × 10⁶ J
Answer:
(c) 4.186 J

Question 19.
The amount of work done by a moving body depends on the
(a) mass of the body
(b) velocity
(c) both (a) and (b)
(d) time
Answer:
(c) both (a) and (b)

Question 20.
The kinetic energy of a body is given by

Answer:
(a) \(\frac{1}{2} m v^{2}\)

Question 21.
Kinetic energy of the body is always
(a) zero
(b) infinity
(c) negative
(d) positive
Answer:
(d) positive

Question 22.
If the work done by the force on the body is positive then its kinetic energy
(a) increases
(b) decreases
(c) zero
(d) either increases or decreases
Answer:
(a) increases

Question 23.
If p is the momentum of the particle then its kinetic energy is

Answer:
(c) \(\frac{\mathbf{p}^{2}}{2 \mathbf{m}}\)

Question 24.
If two objects of masses m₁ and m₂ (m₁ > m₂) are moving with the same momentum then the kinetic energy will be greater for
(a) m₁
(b) m₂
(c) m₁ or m₂
(d) both will have equal kinetic energy
Answer:
(b) m₂

Question 25.
For a given momentum, the kinetic energy is proportional to

Answer:
(b) \(\frac{1}{\mathrm{m}}\)

Question 26.
Elastic potential energy possessed by a spring is

Answer:
(c) \(\frac{1}{2}\)kx²

Question 27.
Potential energy stored in the spring depends on
(a) spring constant
(b) mass
(c) gravity
(d) length
Answer:
(b) mass

Question 28.
Two springs of spring constants k₁ and k₂ (k₁ > k₂). If they are stretched by the same force then (u₁, u₂ are potential energy of the springs) is
(a) u₁ > u₂
(b) u₂ > u₁
(c) u₁ = u₂
(d) u₁ ≥ u₂
Answer:
(b) u₂ > u₁

Question 29.
Conservative force is
(a) electrostatic force
(b) magnetic force
(c) gravitational force
(d) all the above
Answer:
(d) all the above

Question 30.
Non conservative force is
(a) frictional force
(b) viscous force
(c) air resistance
(d) all the above
Answer:
(d) all the above

Question 31.
If the work done is completely recoverable, then the force is
(a) conservative
(b) non-conservative
(c) both (a) and (b)
(d) frictional in nature
Answer:
(b) non-conservative

Question 32.
The work done by the conservative forces in a cycle is
(a) zero
(b) one
(c) infinity
(d) having negative value
Answer:
(a) zero

Question 33.
Negative gradient of potential energy gives
(a) conservative force
(b) non conservative force
(c) kinetic energy
(d) frictional force
Answer:
(a) conservative force

Question 34.
When a particle moving in a vertical circle, the variable is/are
(a) velocity of the particle
(b) tension of the string
(c) both (a) and (b)
(d) mass of the particle
Answer:
(c) both (a) and (b)

Question 35.
Which of the following is zero at the highest point in vertical circular motion?
(a) velocity of the particle
(b) tension of the spring
(c) potential energy
(d) none
Answer:
(a) velocity of the particle

Question 36.
The body must have a speed at highest point in vertical circular motion to stay in the circular path

Answer:
(a) \(\geq \sqrt{\mathbf{g r}}\)

Question 37.
The body must have a minimum speed of lowermost point in vertical circular motion to complete the circle

Answer:
(c) \(\geq \sqrt{5 \mathrm{gr}}\)

Question 38.
The rate of work done is
(a) energy
(b) force
(c) power
(d) energy flow
Answer:
(c) power

Question 39.
The unit of power is
(a) J
(b) W
(c) J s^-1
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question 40.
One horse power (1 hp) is
(a) 476 W
(b) 674 W
(c) 746 W
(d) 764 W
Answer:
(c) 746 W

Question 41.
The dimension of power is
(a) ML²T^-2
(b) ML²T^-3
(c) ML^-2T²
(d) ML^-2T³
Answer:
(b) ML²T^-3

Question 42.
kWh is the practical unit of
(a) energy
(b) power
(c) electrical energy
(d) none
Answer:
(a) energy

Question 43.
If a force F is applied on a body and the body moves with velocity v, the power will be
(a) F.V
(b) F/V
(c) FV²
(d) FW²
Answer:
(a) F.V

Question 44.
A body of mass m is thrown vertically upward with a velocity v. The height at which the kinetic energy of the body is one third of its initial value is given by

Answer:
(c) \(\frac{v^{2}}{6 g}\)
Solution:
Initial . The loss in K.E will be the gain in potential energy

Question 45.
A body of mass 5 kg is initially at rest. By applying a force of 20 N at an angle of 60° with horizontal the body is moved to a distance of 4 m. The kinetic energy acquired by the body is
(a) 80 J
(b) 60 J
(c) 40 J
(d) 17.2 J
Answer:
(c) 40 J
Solution:
The work done is equal to its kinetic energy
∴ K.E gained = Fs cos θ = 20 × 4 cos 60° = 40 J.

Question 46.
A bullet is fired normally on an immovable wooden plank of thickness 2 m. It loses 20% of its kinetic energy in penetrating a thickness 0.2 m of the plank. The distance penetrated by the bullet inside the wooden plank is
(a) 0.2 m
(b) 0.8 m
(c) 1 m
(d) 1.5 m
Answer:
(c) 1 m
Solution:
The wood offers a constant retardation. If the bullet loses 20% of its kinetic energy by penetrating 0.2m. it can penetrate further into 4 × 0.2 = 0.8 m with the remaining kinetic energy. So the total distance penetrated by the bullet is 0.2 + 0.8 = 1 m.

Question 47.
Which of the following quantity is conserved in all collision process?
(a) kinetic energy
(b) linear momentum
(c) both (a) and (b)
(d) none.
Answer:
(b) linear momentum

Question 48.
The kinetic energy is conserved in
(a) elastic collision
(b) inelastic collision
(c) both (a) and (b)
(d) none
Answer:
(a) Elastic collision

Question 49.
The kinetic energy is not conserved in
(a) Elastic collision
(b) In elastic collision
(c) both (a) and (b)
(d) none
Answer:
(b) In elastic collision

Question 50.
In inelastic collision, which is conserved
(a) linear momentum
(b) total energy
(c) both (a) and (b)
(d) none
Answer:
(c) both (a) and (b)

Question 51.
If the two colliding bodies stick together after collision such collisions are
(a) elastic collision
(b) inelastic collision
(c) perfectly inelastic collision
(d) head on collision
Answer:
(c) perfectly inelastic collision

Question 52.
When bubblegum is thrown on a moving vehicle, it sticks is an example for
(a) elastic collision
(b) inelastic collision
(c) perfectly inelastic collision
(d) none
Answer:
(c) perfectly inelastic collision

Question 53.
Elastic collision is due to
(a) conservative force
(b) non conservative force
(c) gravitational force
(d) electrostatic force
Answer:
(b) non conservative force

Question 54.
Inelastic collision is due to
(a) conservative force
(b) non conservative force
(c) gravitational force
(d) electrostatic force
Answer:
(b) non conservative force

Question 55.
If the velocity of separation is equal to the velocity of approach, then the collision is
(a) conservative force
(b) non conservative force
(c) gravitational force
(d) electrostatic force
Answer:
(a) conservative force

Question 56.
For elastic collision, coefficient of restitution is
(a) 0
(b) 1
(c) 0 < e < 1
(d) ∞
Answer:
(b) 1

Question 57.
For inelastic collision co-efficient of restitution is
(a) 0
(b) 1
(c) 0 < e < 1
(d) ∞
Answer:
(c) 0 < e < 1

Question 58.
For perfectly inelastic collision, coefficient of restitution is
(a) 0
(b) 1
(c) 0 < e < 1
(d) ∞
Answer:
(a) 0

Question 59.
The ratio of velocities of equal masses in an inelastic collision with one of the masses is stationary is
60. A box is dragged across a surface by a rope which makes an angle 45° with the horizontal. The tension in the rope is 100 N when the box is dragged 10 m. The work done is

Answer:
(a) \(\frac{1-e}{1+e}\)

Question 60.
A box is dragged across a surface by a rope which makes an angle 45° with the horizontal. The
tension in the rope is 100 N when the box is dragged 10 m. The work done is
(a) 707.1 J
(b) 607.1 J
(c) 1414.2 J
(d) 900 J
Answer:
(a) 707.1 J
Solution:
The component of force acting along the surface is T cos θ

∴ Work done = T cos θ × x
= 10o cos 45° × 10
= 707.1 J

Question 61.
A position dependent force F = (7 – 2x + 3x²) N acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5 m. Work done is
(a) 35 J
(b) 70 J
(c) 135 J
(d) 270 J
Answer:
(c) 135 J
Solution:

Question 62.
In gravitational field, the work done in moving a body from one point into another depends on
(a) initial and final positions
(b) distance between them
(c) actual distance covered
(d) velocity of motion
Answer:
(c) initial and final positions

Question 63.
A particle of mass “m” moving with velocity v strikes a particle of mass “2m” at rest and sticks to it. The speed of the combined mass is

Answer:
(c) \(\frac{v}{3}\)
Solution:
According to conservation of linear momentum

Question 64.
A force of (\(10 \hat{i}-3 \hat{j}+6 \hat{k}\)) N acts on a body of 5 kg and displaces it from (\(6 \hat{i}+5 \hat{j}-3 \hat{k}\)) to (\(10 \hat{i}-2 \hat{j}+7 k\)) m. The work done is
(a) 100 J
(b) 0
(c) 121 J
(d) none of these
Answer:
(c) 121 J
Solution:

Question 65.
A 9 kg mass and 4 kg mass are moving with equal kinetic energies. The ratio of their momentum is
(a) 1 : 1
(b) 3 : 2
(c) 2 : 3
(d) 9 : 4.
Answer:
(b) 3 : 2
Solution:
Given that K.E are equal

Question 66.
If momentum of a body increases by 25% its kinetic energy will increase by
(a) 25%
(b) 50%
(c) 125%
(d) 56.25%
Answer:
(d) 56.25%
Solution:
Let momentum of p₁ = 100% momentum of p₂ = 125%.

Question 67.
A missile fired from a launcher explodes in mid air, its total
(a) kinetic energy increases
(b) momentum increases
(c) kinetic energy decreases
(d) momentum decreases
Answer:
(a) kinetic energy increases

Question 68.
A bullet hits and gets embedded in a wooden block resting on a horizontal friction less surface. Which of the following is conserved?
(a) momentum alone
(b) kinetic energy alone
(c) both momentum and kinetic energy
(d) no quantity is conserved
Answer:
(a) momentum alone

Question 69.
Two balls of equal masses moving with velocities 10 m/s and -7 m/s respectively collide elastically. Their velocities after collision will be
(a) 3 ms^-1 and 17 ms^-1
(b) -7 ms^-1 and 10 ms^-1
(c) 10 ms^-1 and -7 ms^-1
(d) 3 ms^-1 and -70 ms^-1
Answer:
(b) -7 ms^-1 and 10 ms^-1

Question 70.
A spring of negligible mass having a force constant of 10 Nm^-1 is compressed by a force to a distance of 4 cm. A block of mass 900 g is free to leave the top of the spring. If the spring is released, the speed of the block is
(a) 11.3 ms^-1
(b) 13.3 × 101 ms^-1
(c) 13.3 × 10^-2 ms^-1
(d) 13.3 × 10^-3 ms^-1
Answer:
(c) 13.3 × 10^-2 ms^-1
Solution:
We know that, the potential energy of the spring = \(\frac{1}{2}\)kx². Here the potential energy of the spring is converted into kinetic energy of the block.

Question 71.
A particle falls from a height ftona fixed horizontal plate and rebounds. If e is the coefficient ” of restitution, the total distance travelled by the particle on rebounding when it stops is

Answer:

S = h + 2e²h + 2e⁴h + 2e⁶h + …..
S = h + 2h (e² + e⁴ + e⁶ +…)
By using binomixal expansion we can write it as

Question 72.
If the force F acting on a body as a function of x then the work done in moving a body from x = 1 m to x = 3m is

(a) 6 J
(b) 4 J
(c) 2.5 J
(d) 1 J
Answer:
(b) 4 J

Question 73.
A boy “A” of mass 50 kg climbs up a staircase in 10 s. Another boy “B” of mass 60 kg climbs up a Same staircase in 15s. The ratio of the power developed by the boys “A” and “B” is

Answer:
(a) \(\frac{5}{4}\)
Solution:

Samacheer Kalvi 11th Physics Short Answer Questions

Question 1.
Define work, energy, power.
Answer:
Work: Work is said to be done by the force when the force applied on a body displaces it.
Energy: Energy is defined as the ability to do work.
Power: The rate of work done is called power.

Question 2.
Discuss the possibilities of work done to be zero.
Answer:
Work done is zero in the following cases.
(i) When the force is zero (F = 0). For example, ,a body moving on a horizontal smooth frictionless surface will continue to do so as no force (not even friction) is acting along the plane. (This is an ideal situation.)
(ii) When the displacement is zero (dr = 0). For example, when force is applied on a rigid wall it does not produce any displacement. Hence, the work done is zero as shown in figure.
(iii) When the force and displacement are perpendicular (0 = 90°) to each other, when a body moves on a horizontal direction, the gravitational force (mg) does not work on the body, since it acts at right angles to the displacement as shown in Figure (b). In circular motion the centripetal force does not do work on the object moving on a circle as it is always perpendicular to the displacement as shown in Figure (c).

Question 3.
Derive the relation between momentum and kinetic energy.
Answer:
Consider an object of mass m moving with a velocity v. Then its linear momentum is


Multiplying both the numerator and denominator of equation (i) by mass m

where | \(\vec{p}\) | is the magnitude of the momentum. The magnitude of the linear momentum can be obtained by

Note that if kinetic energy and mass are given, only the magnitude of the momentum can be calculated but not the direction of momentum. It is because the kinetic energy and mass are scalars.

Question 4.
How can an object move with zero acceleration (constant velocity) when the external force is acting on the object?
Answer:
It is possible when there is another force which acts exactly opposite to the external applied force. They both cancel each other and the resulting net force becomes zero, hence the object moves with zero acceleration.

Question 5.
Why should the object be moved at constant velocity when we define potential energy?
Answer:
If the object does not move at constant velocity, then it will have different velocities at the initial and final locations. According to work-kinetic energy theorem, the external force will impart some extra kinetic energy. But we associate potential energy to the forces like gravitational force, spring force and coulomb force. So the external agency should not impart any kinetic energy when the object is taken from initial to final location.

Question 6.
Derive an expression for potential energy near the surface of the earth.
Answer:
The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from ground to that height h with constant velocity. Let us consider a body of mass m being moved from ground to the height h against the gravitational force as shown.

The gravitational force \(\overrightarrow{\mathrm{F}}_{g}\) acting on the body is, \(\overrightarrow{\mathrm{F}}_{g}=-m g \hat{j}\) (as Gravitational potential energy the force is in y direction, unit vector \(\hat{j}\) is used). Here, negative sign implies that the force is acting vertically downwards. In order to move the body without acceleration (or with constant velocity), an external applied force \(\overrightarrow{\mathrm{F}}_{a}\), equal in magnitude but opposite to that of gravitational force \(\overrightarrow{\mathrm{F}}_{g}\) has to be applied on the body i.e., \(\overrightarrow{\mathrm{F}}_{a}=-\overrightarrow{\mathrm{F}}_{g}\).
This implies that \(\overrightarrow{\mathrm{F}}_{a}=+m g \hat{j}\). The positive sign implies that the applied force is in vertically upward direction. Hence, when the body is lifted up its velocity remains unchanged and thus its kinetic energy also remains constant.
The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from the ground to that height h.

Since the displacement and the applied force are in the same upward direction, the angle between them, θ = 0°. Hence, cos 0° = 1 and | \(\overrightarrow{\mathrm{F}}_{a}\) | = mg and | \(d \vec{r}\) | = dr.

Question 7.
Explain force displacement graph for a spring.
Answer:

Since the restoring spring force and displacement are linearly related as F = – kx, and are opposite in direction, the graph between F and x is a straight line with dwelling only in the second and fourth quadrant as shown in Figure. The elastic potential energy can be easily calculated by drawing a F – x graph. The shaded area (triangle) is the work done by the spring force.

Question 8.
Explain the potential energy – displacement graph for a spring.
Answer:
A compressed or extended spring will transfer its stored potential energy into kinetic energy of the mass attached to the spring. The potential energy-displacement graph is shown in Figure.

In a frictionless environment, the energy gets transferred from kinetic to potential and potential to kinetic repeatedly such that the total energy of the system remains constant. At the mean position,
∆KE = ∆U

Question 9.
Define unit of power.
Answer:
The unit of power is watt. One watt is defined as the power when one joule of work is done in one second.

Question 10.
Define average power and instantaneous power.
Answer:
The average power is defined as the ratio of the total work done to the total time taken.
P_av = total work done/total time taken The instantaneous power is defined as the power delivered at an instant
p_inst = dw/dt

Question 11.
Define elastic and inelastic collision.
Answer:
In any collision, if the total kinetic energy of the bodies before collision is equal to the total final kinetic energy of the bodies after collision then it is called as elastic collision.
In a collision the total initial kinetic energy of the bodies before collision is not equal to the . total final kinetic energy of the bodies after collision. Then it is called as inelastic collision.

Question 12.
What will happen to the potential energy of the system.
If (i) Two same charged particles are brought towards each other
(ii) Two oppositely charged particles are brought towards each other.
Answer:
(i) When the same charged particles are brought towards each other, the potential energy of the system will increase. Because work has to be done against the force of repulsion. This work done only stored as potential energy.
(ii) When two oppositely charged particles are brought towards each other, the potential energy of the system will decrease. Because work is done by the force of attraction between the charged particles.

Question 13.
Define the conservative and non-conservative forces. Give examples of each.
Answer:
Conservative force : e.g., Gravitational force, electrostatic force.
Non-Conservative force : e.g., forces of friction, viscosity.

Question 14.
A light body and a heavy body have same linear momentum. Which one has greater K.E ?
Answer:
Lighter body has more K.E. as K.E. = \(\frac{p^{2}}{2 m}\) and for constant p, K.E. \(\propto \frac{1}{m}\)

Question 15.
The momentum of the body is doubled, what % does its K.E change?
Answer:

Question 16.
A body is moving along a circular path. How much work is done by the centripetal force?
Answer:
W = FS cos 90° = 0.

Question 17.
Which spring has greater value of spring constant – a hard spring or a delicate spring?
Answer:
Hard spring.

Question 18.
Two bodies stick together after collision. What type of collision is in between these two bodies? .
Answer:
Inelastic collision.

Question 19.
State the two conditions under which a force does not work.
Answer:

1. Displacement is zero or it is perpendicular to force.
2. Conservative force moves a body over a closed path.

Question 20.
How will the momentum of a body changes if its K.E. is doubled?
Answer:
Momentum becomes \(\sqrt{2}\) times.

Question 21.
K.E. of a body is increased by 300 %. Find the % increase in its momentum.
Answer:

Question 22.
A light and a heavy body have same K.E., which of the two have more momentum and why?
Answer:
Heavier body.

Question 23.
Does the P.E. of a spring decreases or increases when it is compressed or stretched?
Answer:
Increases because W.D. on it when it increases is compressed or stretched.

Question 24.
Name a process in which momentum changes but K.E. does not.
Answer:
Uniform circular motion.

Question 25.
What happens to the P.E. of a bubble when it rises in water?
Answer:
Decreases.

Question 26.
A body is moving at constant speed over a frictionless surface. What is the work done by the weight of the body?
Answer:
W = 0.

Question 27.
Define spring constant of a spring.
Answer:
It is the restoring force set up in a string per unit extension.

Samacheer Kalvi 11th Physics Short Answer Questions 2 Marks

Question 28.
How much work is done by a coolie walking on a horizontal platform with a load on his head? Explain.
Answer:
W = 0 as his displacement is along the horizontal direction and in order to balance the load on his head, he applies a force on it in the upward direction equal to its weight. Thus angle between force and displacement is zero.

Question 29.
Mountain roads rarely go straight up the slope, but wind up gradually. Why?
Answer:
If roads go straight up then angle of slope 0 would be large so frictional force f = µ mg cos θ would be less and the vehicles may slip. Also greater power would be required.

Question 30.
A truck and a car moving with the same K.E. on a straight road. Their engines are simultaneously switched off which one will stop at a lesser distance?
Answer:
By Work – Energy Theorem,
Loss in K.E. = W.D. against the force × distance of friction

∴ Truck will stop in a lesser distance because of greater mass.

Question 31.
Is it necessary that work done in the motion of a body over a closed loop is zero for every force in nature? Why?
Answer:
No. W.D. is zero only in case of a conservative force.

Question 32.
How high must a body be lifted to gain an amount of P.E. equal to the K.E. it has when moving at speed 20 ms^-1. (The value of acceleration due to gravity at a place is 9.8 ms^-2).
Answer:

Question 33.
Give an example in which a force does work on a body but fails to change its K.E.
Answer:
When a body is pulled on a rough, horizontal surface with constant velocity. Work is done on the body but K.E. remains unchanged.

Question 34.
A bob is pulled sideway so that string becomes parallel to horizontal and released. Length of the pendulum is 2 m. If due to air resistance loss of energy is 10%, what is the speed with which the bob arrived at the lowest point.
Answer:

Question 35.
Two springs A and B are identical except that A is harder than B (K_A > K_B) if these are stretched by the equal force. In which spring will more work be done?
Answer:

Question 36.
Find the work done if a particle moves from position r₁ = to a position \((3 \hat{i}+2 \hat{j}-6 \hat{k})\) to a position \(\vec{r}_{2}=(14 \hat{i}+13 \hat{j}-9 \hat{k})\) under the effect of force \(\overrightarrow{\mathrm{F}}=(4 \hat{i}+\hat{j}+3 \hat{k}) \mathrm{N}\)
Answer:

Question 37.
Spring A and B are identical except that A is stiffer than B, i.e., force constant k_A > k_B. In which spring is more work expended if they are stretched by the same amount?
Answer:

Question 38.
A ball at rest is dropped from a height of 12 m. It loses 25% of its kinetic energy in striking the ground, find the height to which it bounces. How do you account for the loss in kinetic energy?
Answer:
If ball bounces to height h’, then
mgh’ = 75% of mgh
∴ h’ = 0.75 h = 9 m.

Question 39.
Which of the two kilowatt hour or electron volt is a bigger unit of energy and by what factor?
Answer:
kwh is a bigger unit of energy.

Question 40.
A spring of force constant K is cut into two equal pieces. Calculate force constant of each part.
Answer:
Force constant of each half becomes twice the force constant of the original spring.

Samacheer Kalvi 11th Physics Short Answer Questions 3 Marks

Question 41.
A car of mass 2000 kg is lifted up a distance of 30 m by a crane in 1 min. A second crane does the same job in 2 min. Do the cranes consume the same or different amounts of fuel? What is the power supplied by each crane? Neglect Power dissipation against friction.
Answer:
t₁ = 1 min = 60 s, t₂ = 2 min = 120 s
W = Fs = mgs = 5.88 × 105 J
As both cranes do same amount of work so both consume same amount of fuel.

Question 42.
20 J work is required to stretch a spring through 0.1m. Find the force constant of the spring. If the spring is further stretched through 0.1 m, calculate work done.
Answer:
P.E. of spring when stretched through a distance 01m,

when spring is further stretched through 01m, then P.E. will be :

Question 43.
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15 min. If the tank is 40 m above the ground, how much electric power is consumed by the pump. The efficiency of the pump is 30%.
Answer:

Question 44.
A ball bounces to 80% of its original height. Calculate the mechanical energy lost in each bounce.
Answer:
Let Initial P.E. = mgh
P.E. after first bounce = mg × 80% of h = 0.80 mgh
P.E. lost in each bounce = 0.20 mgh

Samacheer Kalvi 11th Physics Long Answer Questions

Question 1.
Obtain an expression for the critical vertical of a body revolving in a vertical circle
Answer:
Imagine that a body of mass (m) attached to one end of a massless and inextensible string executes circular motion in a vertical plane with the other end of the string fixed. The length of the string becomes the radius (r) of the circular path (See figure).

Let us discuss the motion of the body by taking the free body diagram (FBD) at a position where the position vector (\(\vec{r}\)) makes an angle θ with the vertically downward direction and the instantaneous velocity is as shown in Figure.
There are two forces acting on the mass.
1. Gravitational force which acts downward
2. Tension along the string.
Applying Newton’s second law on the mass, in the tangential direction,

The circle can be divided into four sections A, B, C, D for better understanding of the motion. The four important facts to be understood from the two equations are as follows:
(i) The mass is having tangential acceleration (g sin θ) for all values of θ (except θ = 0°), it is clear that this vertical circular motion is not a uniform circular motion.
(ii) From the equations (ii) and (i) it is understood that as the magnitude of velocity is not a constant in the course of motion, the tension in the string is also not constant.

Hence velocity cannot vanish, even when the tension vanishes.
These points are to be kept in mind while solving problems related to motion in vertical circle.
To start with let us consider only two positions, say the lowest point 1 and the highest point 2 as shown in Figure for further analysis. Let the velocity of the body at the lowest point 1 be \(\vec{v}_{1}\), at the highest point 2 be \(\vec{v}_{2}\) and \(\vec{v}\) at any other point. The direction of velocity is tangential to the circular path at all points. Let \(\overrightarrow{\mathrm{T}}_{1}\) be the tension in the string at the lowest point and \(\overrightarrow{\mathrm{T}}_{2}\) be , the tension at the highest point and \(\overrightarrow{\mathrm{T}}\) be the tension at any other point. Tension at each point acts towards the center. The tensions and velocities at these two points can be found by applying the law of conservation of energy.

For the lowest point (1)
When the body is at the lowest point 1, the gravitational force \(m \vec{g}\) which acts on the body (vertically downwards) and another one is the tension \(\overrightarrow{\mathrm{T}}_{1}\), acting vertically upwards, i.e. towards the center. From the equation (ii), we get

For the highest point (2)
At the highest point 2, both the gravitational force mg on the body and the tension T₂ act downwards, i.e. towards the center again.

From equations (iv) and (ii), it is understood that T₁ > T₂. The difference in tension T₁ – T₂ is obtained by subtracting equation (iv) from equation (ii).

The term can be found easily by applying law of conservation of energy at point 1 and also at point 2.
Note: The tension will not do any work on the mass as the tension and the direction of motion is always perpendicular.
The gravitational force is doing work on the mass, as it is a conservative force the total energy of the mass is conserved throughout the motion.
Total energy at point 1 (E₁) is same as the total energy at a point 2 (E₂)
E₁ = E₂
Potential energy at point 1, U₁ = 0 (by taking reference as point 1)

Similarly, Potential energy at point 2, U₂ = mg (2r) (h is 2r from point 1)

From the law of conservation of energy given in equation (vi), we get


Substituting equation (vii) in equation (iv) we get,

Therefore, the difference in tension is
T₁ – T₂ = 6 mg …(viii)
Minimum speed at the highest point (2)
The body must have a minimum speed at point 2 otherwise, the string will slack before reaching point 2 and the body will not loop the circle. To find this minimum speed let us take the tension T₂ = 0 in equation (iv).

The body must have a speed at point 2, \(v_{2} \geq \sqrt{g r}\) to stay in the circular path.
Maximum speed at the lowest point 1
To have this minimum speed (\(v_{2}=\sqrt{g r}\)) at point 2, the body must have minimum speed also at point 1.
By making use of equation (vii) we can find the minimum speed at point 1.

Substituting equation (ix) in (vii),

The body must have a speed at point 1, \(v_{1} \geq \sqrt{5 g r}\) to stay in the circular path.
From equations (ix) and (x), it is clear that the minimum speed at the lowest point 1 should be v 5 times more than the minimum speed at the highest point 2, so that the body loops without leaving the circle.

Question 2.
Obtain the expressions for the velocities of the two bodies after collision in the case of one dimensional elastic collision and discuss the special cases.
Answer:
Consider two elastic bodies of masses m₁ and m₂ moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure.

In order to have collision, we assume that the mass m₁ moves faster than mass m₂ i.e., u₁ > u₂.
For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same

From the law of conservation of linear momentum,
Total momentum before collision (p_i) = Total momentum after collision (p_f)

For elastic collision,
Total kinetic energy before collision KE_i = Total kinetic energy after collision KF_f

After simplifying and rearranging the terms,

Using the formula a² – b² = (a + b) (a – b), we can rewrite the above equation as
m₁(u₁ + v₁)(u₁ – v₁) = m₂ (v₂ + u₂) (v₂ – u₂) …(iv)
Dividing equation (iv) by (ii) gives,

This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for V₁ and v₂,
v₁ = v₂ + u₂ – u₁
Or v₂ = u₁ + v₁ – u₁
To find the final velocities v₁ and v₂ :
Substituting equation (vii) in equation (ii) gives the velocity of m₁ as
m₁ (u₁ – v₁ ) = m₂ (u₁ + v₁ – u₂ – u₂)
m₁u₁ – m₁v₁ = m₂(u₁ + v₁ – 2u₂)
m₁u₁ + 2m₂u₂ = m₁v₁ + m₂v₁

Similarly, by substituting (vi) in equation (ii) or substituting equation (viii) in equation (vii), we get the final velocity of m₂ as

Case 1: When bodies has the same mass i.e., m₁ = m₂,

The equations (x) and (xi) show that in one dimensional elastic collision, when two bodies of equal mass collide after the collision their velocities are exchanged.
Case 2: When bodies have the same mass i.e., m₁ = m₂ and second body (usually called target) is at rest (u₂ = 0),
By substituting m₁m₂ = and u₂ = 0 in equations (viii) and equations (ix) we get, from equation
(viii) ⇒ V₁ = 0 …(xii)
from equation (ix) ⇒ v₂ = u₁ … (xiii)
Equations (xii) and (xiii) show that when the first body comes to rest the second body moves with the initial velocity of the first body.
Case 3: The first body is very much lighter than the second body

Similarly, Dividing numerator and denominator of equation (ix) by m₂, we get

v₂ = 0
The equation (xiv) implies that the first body which is lighter returns back (rebounds) in the opposite direction with the same initial velocity as it has a negative sign. The equation (xv) implies that the second body which is heavier in mass continues to remain at rest even after collision. For example, if a ball is thrown at a fixed wall, the ball will bounce back from the wall with the same velocity with which it was thrown but in opposite direction.

Similarly,
Dividing numerator and denominator of equation (xiii) by m₁, we get

The equation (xvi) implies that the first body which is heavier continues to move with the same initial velocity. The equation (xvii) suggests that the second body which is lighter will move with twice the initial velocity of the first body. It means that the lighter body is thrown away from the point of collision.

Samacheer Kalvi 11th Physics Numerical Questions

Question 1.
A body is moving along z-axis of a coordinate system under the effect of a constant force F = Find the work done by the force in moving the body a distance of 2 m along z-axis.
Answer:

\(\mathrm{W}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{S}}=2 \mathrm{J}\)

Question 2.
Water is pumped out of a well 10 m deep by means of a pump rated 10 KW. Find the efficiency of the motor if 4200 kg of water is pumped out every minute. Take g = 10 m/s2.
Answer:

Question 3.
A railway carriage of mass 9000 kg moving with a speed of 36 kmph collides with a stationary carriage of same mass. After the collision, the carriages get coupled and move together. What is their common speed after collision? What type of collision is this?
Answer:
m₁ = 9000 kg, u₁ = 36 km/h = 10 m/s
m₂ = 9000 kg, u₂ = 0, v = v₁ = v₂ = ?
By conservation of momentum:
m₁u₁ + m₂u₂ = (m₁ + m₂)v
∴ v = 5 m/s

As total K.E. after collision < Total K.E. before collision
∴ collision is inelastic

Question 4.
In lifting a 10 kg weight to a height of 2m, 230 J energy is spent. Calculate the acceleration with which it was raised.
Answer:
W = mgh + mah = m(g + a)h
∴ a = 1.5 m/s².

Question 5.
A bullet of mass 0.02 kg is moving with a speed of 10 ms^-1. It can penetrate 10 cm of a wooden block, and comes to rest. If the thickness of the target would be 6 cm only, find the K.E. of the bullet when it comes out.

Question 6.
A man pulls a lawn roller through a distance of 20 m with a force of 20 kg weight. If he applies the force at an angle of 60° with the ground, calculate the power developed if he takes 1 min in doing so.
Answer:

Question 7.
A body of mass 0.3 kg is taken up an inclined plane to length 10 m and height 5 m and then allowed to slide down to the bottom again. The coefficient of friction between the body and the plane is 0.15. What is the
(i) work done by the gravitational force over the round trip?
(ii) work done by the applied force over the upward journey?
(iii) work done by frictional force over the round trip?
(iv) kinetic energy of the body at the end of the trip?
How is the answer to (iv) related to the first three answers?
Answer:


(i) W = FS = – mg sin θ × h = -14.7 J is the W.D. by gravitational force in moving plane.
W’ = FS = + mg sin θ × h = 14.7 J is the W.D. by gravitational force in moving the body down the inclined plane.
∴ Total W.D. round the trip, W₁ = W + W’ = 0
(ii) Force needed to move the body up the inclined plane,
F = mg sin θ + f_k = mg sin θ + µ_kR = mg sin θ + µ_k mg cos θ
∴ W.D. by force over the upward journey is
W₂ = F × l = mg (sin θ + µ_k cos θ)l = 18.5 J
(iii) W.D. by frictional force over the round trip,
W₃ = -fk(l + l) = -2fkl = -2µ_kcos θ l = -7.6 J
(iv) K.E. of the body at the end of round trip
= W.D. by net force in moving the body down the inclined plane
= (mg sin θ – µ_kcos θ) l
= 10.9 J
⇒ K.E. of body = net W.D. on the body.

Question 8.
Two identical 5 kg blocks are moving with same speed of 2 ms^-1 towards each other along a frictionless horizontal surface. The two blocks collide, stick together and come to rest. Consider the two blocks as a system. Calculate work done by
(i) external forces and
(ii) Internal forces.
Answer:
Here no external forces are acting on the system so :
\(\overrightarrow{\mathrm{F}}_{\mathrm{ext}}=0 \Rightarrow \mathrm{W}_{\mathrm{ext}}=0\)
According to work-energy theorem :
Total W.D. = Change in K.E.
or W_ext + = Final K.E. – Initial K.E.

Question 9.
A truck of mass 1000 kg accelerates uniformly from rest to a velocity of 15 ms^-1 in 5 seconds. Calculate
(i) its acceleration,
(ii) its gain in K.E.,
(iii) average power of the engine during this period, neglect friction.
Answer:

Question 10.
An elevator which can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms^-1. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power.
Answer:
Downward force on the elevator is :
F = mg + f = 22000 N
∴ Power supplied by motor to balance this force is :

Question 11.
To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed 18.0 kmh^-1 on a smooth road and colliding with a horizontally mounted spring of spring constant 6.25 × 10^-3 Nm^-1. What is the maximum compression of the spring?
Answer:
At maximum compression x_m, the K.E. of the car is converted entirely into the P.E. of the spring.

Students can Download Physics Chapter 3 Laws of Motion Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

Samacheer Kalvi 11th Physics Chapter 3 Laws of Motion Textual Questions Solved

Samacheer Kalvi 11th Physics Laws of Motion Multiple Choice Questions

Question 1.
When a car takes a sudden left turn in the curved road, passengers are pushed towards the right due to
(a) inertia of direction
(b) inertia of motion
(c) inertia of rest
(d) absence of inertia
Answer:
(a) inertia of direction

Question 2.
An object of mass m held against a vertical wall by applying horizontal force F as shown in the figure. The minimum value of the force F is [IIT JEE 1994]
(a) Less than mg
(b) Equal to mg
(c) Greater than mg
(d) Cannot determine
Answer:
(c) Greater than mg

Question 3.
A vehicle is moving along the positive x direction, if sudden brake is applied, then
(a) frictional force acting on the vehicle is along negative x direction
(b) frictional force acting on the vehicle is along positive x direction
(c) no frictional force acts on the vehicle
(d) frictional force acts in downward direction
Answer:
(a) frictional force acting on the vehicle is along negative x direction

The normal force formula is defined as the force that any surface exerts on any other object.

Question 4.
A book is at rest on the table which exerts a normal force on the book. If this force is considered as reaction force, what is the action force according to Newton’s third law?
(a) Gravitational force exerted by Earth on the book
(b) Gravitational force exerted by the book on Earth
(c) Normal force exerted by the book on the table
(d) None of the above
Answer:
(c) Normal force exerted by the book on the table

Question 5.
Two masses m₁ and m₂ are experiencing the same force where m₁ < m₂ The ratio of their acceleration \(\frac{a_{1}}{a_{2}}\) is –
(a) 1
(b) less than 1
(c) greater than 1
(d) all the three cases
Answer:
(c) greater than 1

Question 6.
Choose appropriate free body diagram for the particle experiencing net acceleration along negative y direction. (Each arrow mark represents the force acting on the system).

Answer:

Question 7.
A particle of mass m sliding on the smooth double inclined plane (shown in figure) will experience –
(a) greater acceleration along the path AB
(b) greater acceleration along the path AC
(c) same acceleration in both the paths
(d) no acceleration in both the paths

Answer:
(a) greater acceleration along the path AC

Question 8.
Two blocks of masses m and 2m are placed on a smooth horizontal surface as shown. In the first case only a force F₁  is applied from the left. Later only a force F₂  is applied from the right. If the force acting at the interface of the two blocks in the two cases is same, then F₁ : F₂ is [Physics Olympiad 2016]

(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer:
(c) 2 : 1

Question 9.
Force acting on the particle moving with constant speed is –
(a) always zero
(b) need not be zero
(c) always non zero
(d) cannot be concluded
Answer:
(b) need not be zero

Question 10.
An object of mass m begins to move on the plane inclined at an angle 0. The coefficient of static friction of inclined surface is lay. The maximum static friction experienced by the mass is –
(a) mg
(b) µ_s mg
(c) µ_s mg sin θ
(d) µ_s mg cos θ
Answer:
(d) µ_s mg cos θ

Question 11.
When the object is moving at constant velocity on the rough surface –
(a) net force on the object is zero
(b) no force acts on the object
(c) only external force acts on the object
(d) only kinetic friction acts on the object
Answer:
(a) net force on the object is zero

Question 12.
When an object is at rest on the inclined rough surface –
(a) static and kinetic frictions acting on the object is zero
(b) static friction is zero but kinetic friction is not zero
(c) static friction is not zero and kinetic friction is zero
(d) static and kinetic frictions are not zero
Answer:
(c) static friction is not zero and kinetic friction is zero

Question 13.
The centrifugal force appears to exist –
(a) only in inertial frames
(b) only in rotating frames
(c) in any accelerated frame
(d) both in inertial and non-inertial frames
Answer:
(b) only in rotating frames

Question 14.
Choose the correct statement from the following –
(a) Centrifugal and centripetal forces are action reaction pairs
(b) Centripetal forces is a natural force
(c) Centrifugal force arises from gravitational force
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion
Answer:
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion

Question 15.
If a person moving from pole to equator, the centrifugal force acting on him –
(a) increases
(b) decreases
(c) remains the same
(d) increases and then decreases
Answer:
(a) increases

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions

Question 1.
Explain the concept of inertia. Write two examples each for inertia of motion, inertia of rest and inertia of direction.
Answer:
The inability of objects to move on its own or change its state of motion is called inertia. Inertia means resistance to change its state. There are three types of inertia:
1. Inertia of rest:
The inability of an object to change its state of rest is called inertia of rest.
Example:

• When a stationary bus starts to move, the passengers experience a sudden backward push.
• A book lying on the table will remain at rest until it is moved by some external agencies.

2. Inertia of motion:
The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.
Example:

• When the bus is in motion, and if the brake is applied suddenly, passengers move forward and hit against the front seat.
• An athlete running is a race will continue to run even after reaching the finishing point.

3. Inertia of direction:
The inability of an object to change its direction of motion on its own is called inertia of direction.
Example:

• When a stone attached to a string is in whirling motion, and if the string is cut suddenly, the stone will not continue to move in circular motion but moves tangential to the circle.
• When a bus moving along a straight line takes a turn to the right. The passengers are thrown towards left.

Question 2.
State Newton’s second law.
Answer:
The force acting on an object is equal to the rate of change of its momentum –
\(\overline{\mathrm{F}}\) = \(\frac{d \bar{p}}{d t}\)

Question 3.
Define one newton.
Answer:
One newton is defined as the force which acts on 1 kg of mass to give an acceleration 1 ms^-2 in the direction of the force.

Question 4.
Show that impulse is the change of momentum.
Answer:
According to Newton’s Second Law
F = \(\frac {dp}{dt}\) i.e. dp = Fdt
Integrate it over a time interval from t_i  to t_f

P_i → initial momentum of the object at t_i
P_f → Final momentum of the object at t_f
P_f – P_i = ∆p = change in momentum during the time interval ∆t.
\(\int_{t_{i}}^{t_{f}} \mathrm{F} \cdot d t=\mathrm{J}\) is called the impulse.
If the force is constant over the time interval ∆t, then

Hence the proof.

Question 5.
Using free body diagram, show that it is easy to pull an object than to push it.
Answer:
When a body is pushed at an arbitrary angle θ [0 to \(\frac {π}{2}\)], the applied force F can be resolved into two components as F sin 0 parallel to the surface and F cos 0 perpendicular to the surface as shown in figure. The total downward force acting on the body is mg + F cos θ. It implies that the normal force acting on the body increases. Since there is no acceleration along the vertical direction the normal force N is equal to
N_push = mg + F cos θ …………(1)
As a result the maximal static friction also increases and is equal to
\(f_{S}^{\max }\) = \(\mu_{r} \mathrm{N}_{\mathrm{push}}\) = µ_s(mg + F cos θ) ……(2)
Equation (2) shows that a greater force needs to be applied to push the object into motion.

When an object is pulled at an angle θ, the applied force is resolved into two components as shown in figure. The total downward force acting on the object is –
N_pull = mg – F cos θ ………….(3)

Equation (3) shows that the normal force is less than – N_push. From equations (1) and (3), it is easier to pull an object than to push to make it move.

Question 6.
Explain various types of friction. Suggest a few methods to reduce friction.
Answer:
There are two types of Friction:
(1) Static Friction:
Static friction is the force which opposes the initiation of motion of an object on the surface. The magnitude of static frictional force f_s lies between
\(0 \leq f_{s} \leq \mu_{s} \mathrm{N}\)
where, µ_s – coefficient of static friction
N – Normal force

(2) Kinetic friction:
The frictional force exerted by the surface when an object slides is called as kinetic friction. Also called as sliding friction or dynamic friction,
f_k – µ_kN
where µ_k – the coefficient of kinetic friction
N – Normal force exerted by the surface on the object

Methods to reduce friction:
Friction can be reduced

• By using lubricants
• By using Ball bearings
• By polishing
• By streamlining

Question 7.
What is the meaning by ‘pseudo force’?
Answer:
Pseudo force is an apparent force which has no origin. It arises due to the non-inertial nature of the frame considered.

Question 8.
State the empirical laws of static and kinetic friction.
Answer:
The empirical laws of friction are:

• Friction is independent of surface of contact.
• Coefficient of kinetic friction is less than coefficient of static friction.
• The direction of frictional force is always opposite to the motion of one body over the other.
• Frictional force always acts on the object parallel to the surface on which the objet is placed,
• The magnitude of frictional force between any two bodies in contact is directly proportional to the normal reaction between them.

Question 9.
State Newton’s third law.
Answer:
Newton’s third law states that for every action there is an equal and opposite reaction.

Question 10.
What are inertial frames?
Answer:
Inertial frame is the one in which if there is no force on the object, the object moves at constant velocity.

Question 11.
Under what condition will a car skid on a leveled circular road?
Answer:
On a leveled circular road, if the static friction is not able to provide enough centripetal ’force to turn, the vehicle will start to skid
\(\mu_{s}<\frac{v^{2}}{r g}\)

Samacheer Kalvi 11th Physics Laws of Motion Long Answer Questions

Question 1.
Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it.
Answer:
In nature, conservation laws play a very important role. The dynamics of motion of bodies can be analysed very effectively using conservation laws. There are three conservation laws in mechanics. Conservation of total energy, conservation of total linear momentum, and conservation of angular momentum. By combining Newton’s second and third laws, we can derive the law of conservation of total linear momentum. When two particles interact with each other, they exert equal and opposite forces on each other.

The particle 1 exerts force \(\overrightarrow{\mathrm{F}}_{12}\) on particle 2 and particle 2 exerts an exactly equal and opposite force \(\overrightarrow{\mathrm{F}}_{12}\) on particle 1 according to Newton’s third law.
\(\overrightarrow{\mathrm{F}}_{12}\) = –\(\overrightarrow{\mathrm{F}}_{12}\) ……..(1)
In terms of momentum of particles, the force on each particle (Newton’s second law) can be written as –
\(\overrightarrow{\mathrm{F}}_{12}\) = \(\frac{d \vec{p}_{1}}{d t}\) and \(\overrightarrow{\mathrm{F}}_{21}\) = \(\frac{d \vec{p}_{2}}{d t}\) ………(2)
Here \(\vec{p}_{1}\) is the momentum of particle 1 which changes due to the force \(\overrightarrow{\mathrm{F}}_{12}\) exerted by particle 2. Further Po is the momentum of particle \(\vec{p}_{2}\) This changes due to \(\overrightarrow{\mathrm{F}}_{21}\) exerted by particle 1.
Substitute equation (2) in equation (1)
\(\frac{d \vec{p}_{1}}{d t}\) = – \(\frac{d \vec{p}_{2}}{d t}\) …………(3)
\(\frac{d \vec{p}_{1}}{d t}\) + \(\frac{d \vec{p}_{2}}{d t}\) = 0 ………(4)
\(\frac {d}{dt}\)(\(\vec{p}_{1}\) + \(\vec{p}_{2}\)) = 0
It implies that \(\vec{p}_{1}\) + \(\vec{p}_{2}\) = constant vector (always).
\(\vec{p}_{1}\) + \(\vec{p}_{2}\) is the total linear momentum of the two particles (\(\vec{p}_{tot}\) = \(\vec{p}_{1}\) + \(\vec{p}_{2}\)).It is also called as total linear momentum of the system. Here, the two particles constitute the system. From this result, the law of conservation of linear momentum can be stated as follows.

If there are no external forces acting on the system, then the total linear momentum of the system (\(\vec{p}_{tot}\)) is always a constant vector. In other words, the total linear momentum of the system is conserved in time. Here the word ‘conserve’ means that \(\vec{p}_{1}\) and \(\vec{p}_{2}\) can vary, in such a way that \(\vec{p}_{1}\) + \(\vec{p}_{2}\) is a constant vector.

The forces \(\overrightarrow{\mathrm{F}}_{12}\) and \(\overrightarrow{\mathrm{F}}_{21}\) are called the internal forces of the system, because they act only between the two particles. There is no external force acting on the two particles from outside. In such a case the total linear momentum of the system is a constant vector or is conserved.

Meaning of law of conservation of momentum:
1. The Law of conservation of linear momentum is a vector law. It implies that both the magnitude and direction of total linear momentum are constant. In some cases, this total momentum can also be zero.

2. To analyse the motion of a particle, we can either use Newton’s second law or the law of conservation of linear momentum. Newton’s second law requires us to specify the forces involved in the process. This is difficult to specify in real situations. But conservation of linear momentum does not require any force involved in the process. It is convenient and hence important.

For example, when two particles collide, the forces exerted by these two particles on each other is difficult to specify. But it is easier to apply conservation of linear momentum during the collision process.

Examples:
Consider the firing of a gun. Here the system is Gun+bullet. Initially the gun and bullet are at rest, hence the total linear momentum of the system is zero. Let \(\vec{p}_{1}\) be the momentum of the bullet and \(\vec{p}_{2}\) the momentum of the gun before firing. Since initially both are at rest,

Total momentum before firing the gun is zero, \(\vec{p}_{1}\) + \(\vec{p}_{2}\) = 0.
According to the law of conservation of linear momentum, total linear momentum has to be zero after the firing also.

When the gun is fired, a force is exerted by the gun on the bullet in forward direction. Now the momentum of the bullet changes from \(\vec{p}_{1}\) to \(\vec{p}_{1}\) To conserve the total linear momentum of the system, the momentum of the gun must also change from \(\vec{p}_{2}\) to \(\vec{p}_{2}^{\prime}\). Due to the conservation of linear momentum, \(\vec{p}_{1}\) + \(\vec{p}_{2}^{\prime}\) = 0.

It implies that \(\vec{p}_{1}^{\prime}\)= \(\vec{p}_{2}^{\prime}\), the momentum of the gun is exactly equal, but in the opposite direction to the momentum of the bullet. This is the reason after firing, the gun suddenly moves backward with the momentum (-\(\vec{p}_{2}^{\prime}\)). It is called ‘recoil momentum’. Th is is an example of conservation of total linear momentum.

Question 2.
What are concurrent forces? State Lami’s theorem.
Answer:
Concurrent force:
A collection of forces is said to be concurrent, if the lines of forces act at a common point.

Lami’s Theorem:
If a system of three concurrent and coplanar forces is in equilibrium, then Lami’s theorem states that the magnitude of each force of the system is proportional to sine of the angle between the other two forces.
i.e. |\(\overrightarrow{\mathrm{F}}_{1}\)|∝ sin α, |\(\overrightarrow{\mathrm{F}}_{2}\)| ∝ sin β, |\(\overrightarrow{\mathrm{F}}_{2}\)| ∝ sin γ,

Question 3.
Explain the motion of blocks connected by a string in

1. Vertical motion
2. Horizontal motion.

Answer:
When objects are connected by strings and When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.

Case 1:
Vertical motion:
Consider two blocks of masses m₁ and m₂ (m₁> m₂) connected by a light and in-extensible string that passes over a pulley as shown in Figure.

Let the tension in the string be T and acceleration a. When the system is released, both the blocks start moving, m₂ vertically upward and m_k, downward with same acceleration a. The gravitational force m₁g on mass m₁ is used in lifting the mass m₂. The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.

Applying Newton’s second law for mass m₂ T \(\hat{j}\) – m₂ g \(\hat{j}\) = m₂ a \(\hat{j}\) The left hand side of the above equation is the total force that acts on m₂ and the right hand side is the product of mass and acceleration of m₂ in y direction.
By comparing the components on both sides, we get
T = m₂ g = m₂ a ……….(1)
Similarly, applying Newton’s second law for mass m₂
T \(\hat{j}\) – m₁ g\(\hat{j}\) = m₁a\(\hat{j}\)
As mass m₁ moves downward (-\(\hat{j}\)), its acceleration is along (-\(\hat{j}\))
By comparing the components on both sides, we get
T = m₁g = -m₁a
m₁g – T = m₁a ………..(2)
Adding equations (1) and (2), we get
m₁g – m₂g = m₁a + m₂a
(m₁ – m₂)g = (m₁ + m₂)a …………(3)
From equation (3), the acceleration of both the masses is –
a = (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ………..(4)
If both the masses are equal (m₁ = m₂), from equation (4)
a = 0
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
T = m₂g = m₂(\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))
T = m₂g + m₂ (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ……….(5)
By taking m₂g common in the RHS of equation (5)

Equation (4) gives only magnitude of acceleration.
For mass m₁, the acceleration vector is given by \(\vec{a}\) = –\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\)\(\hat{j}\)
For mass m₂, the acceleration vector is given by \(\vec{a}\) = \(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\) \(\hat{j}\)

Case 2:
Horizontal motion:
In this case, mass m₂ is kept on a horizontal table and mass m₁, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface

As both the blocks are connected to the un stretchable string, if m₁ moves with an acceleration a downward then m₂ also moves with the same acceleration a horizontally.
The forces acting on mass m₂ are

• Downward gravitational force (m₂g)
• Upward normal force (N) exerted by the surface
• Horizontal tension (T) exerted by the string

The forces acting on mass m₁ are

• Downward gravitational force (m₁g)
• Tension (T) acting upwards

The free body diagrams for both the masses is shown in figure.

Applying Newton’s second law for m₁
T\(\hat{i}\) – m₁g\(\hat{j}\) = -m₁a\(\hat{j}\) (alongy direction)
By comparing the components on both sides of the above equation,
T – m₁g = -m₁a …………(1)
Applying Newton’s second law for m₂
Ti = m₁ai (along x direction)
By comparing the components on both sides of above equation,
T = m₂a ………….(2)
There is no acceleration along y direction for m₂.
N\(\hat{j}\) – m₂g\(\hat{j}\) = 0
By comparing the components on both sides of the above equation
N – m₂g = 0
N = m₂g ……….(3)
By substituting equation (2) in equation (1), we can find the tension T
m₂a – m₁g = -m₁a
m₂a + m₁a = m₁g
a = \(\frac{m_{1}}{m_{1}+m_{2}}\)g …………(4)
Tension in the string can be obtained by substituting equation (4) in equation (2)
T = \(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\)g ………..(5)

Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

Question 4.
Briefly explain the origin of friction. Show that in an inclined plane, angle of friction is pqual to angle of repose.
Answer:
If a very gentle force in the horizontal direction is given to an object at rest on the table it does not move. It is because of the opposing force exerted by the surface on the object which resists its motion. This force is called the frictional force. During the time of Newton and Galileo, frictional force was considered as one of the natural forces like gravitational force. But in the twentieth century, the understanding on atoms, electron and protons has changed the perspective.

The frictional force is actually the electromagnetic force between the atoms on the two surfaces. Even well polished surfaces have irregularities on the surface at the microscopic level. The component of force parallel to the inclined plane (mg sin θ) tries to move the object down. The component of force perpendicular to the inclined plane (mg cos θ) is balanced by the Normal force (N).
N = mg cos θ ………(1)

When the object just begins to move, the static friction attains its maximum value
f_s = \(f_{s}^{\max }\) = µ_s N
This friction also satisfies the relation
\(f_{s}^{\max }\) = µ_s mg sin θ ……….(2)
Equating the right hand side of equations (1) and (2),
(\(f_{s}^{\max }\))/N = sin θ / cos θ
From the definition of angle of friction, we also know that
tan θ = µ_s ………..(3)
in which θ is the angle of friction.
Thus the angle of repose is the same as angle of friction. But the difference is that the angle of repose refers to inclined surfaces and the angle of friction is applicable to any type of surface.

Question 5.
State Newton’s three laws and discuss their significance.
Answer:
First Law:
Every object continues to be in the state of rest or of uniform motion (constant velocity) unless there is external force acting on it.

Second Law:
The force acting on an object is equal to the rate of change of its momentum

Third Law:
For every action there is an equal and opposite reaction.

Significance of Newton’s Laws:
1. Newton’s laws are vector laws. The equation \(\overline{\mathrm{F}}\) = m\(\overline{\mathrm{a}}\) is a vector equation and essentially it is equal to three scalar equations. In Cartesian coordinates, this equation can be written as F_x\(\hat{i}\) + F_y\(\hat{j}\) + F_z\(\hat{j}\) = ma_x\(\hat{i}\) + ma_y\(\hat{j}\) + ma_z\(\hat{j}\)
By comparing both sides, the three scalar equations are

F_x = ma_x The acceleration along the x-direction depends only on the component of force acting along the x – direction.
F_y = ma_y The acceleration along the y direction depends only on the component of force acting along the y – direction.
F_z = ma_z The acceleration along the z direction depends only on the component of force acting along the z – direction.
From the above equations, we can infer that the force acting along y direction cannot alter the acceleration along x direction. In the same way, F_z cannot affect a_y and a_x. This understanding is essential for solving problems.

2. The acceleration experienced by the body at time t depends on the force which acts on the body at that instant of time. It does not depend on the force which acted on the body before the time t. This can be expressed as
\(\overline{\mathrm{F}}\)(t) = m\(\overline{\mathrm{a}}\)(t)
Acceleration of the object does not depend on the previous history of the force. For example, when a spin bowler or a fast bowler throws the ball to the batsman, once the ball leaves the hand of the bowler, it experiences only gravitational force and air frictional force. The acceleration of the ball is independent of how the ball was bowled (with a lower or a higher speed).

3. In general, the direction of a force may be different from the direction of motion. Though in some cases, the object may move in the same direction as the direction of the force, it is not always true. A few examples are given below.

Case 1:
Force and motion in the same direction:
When an apple falls towards the Earth, the direction of motion (direction of velocity) of the apple and that of force are in the same downward direction as shown in the Figure.

Case 2:
Force and motion not in the same direction:
The Moon experiences a force towards the Earth. But it actually moves in elliptical orbit. In this case, the direction of the force is different from the direction of motion as shown in Figure.

Case 3:
Force and motion in opposite direction:
If an object is thrown vertically upward, the direction of motion is upward, but gravitational force is downward as shown in the Figure.

Case 4:
Zero net force, but there is motion:
When a raindrop gets detached from the cloud it experiences both downward gravitational force and upward air drag force. As it descends towards the Earth, the upward air drag force increases and after a certain time, the upward air drag force cancels the downward gravity. From then on the raindrop moves at constant velocity till it touches the surface of the Earth. Hence the raindrop comes with zero net force, therefore with zero acceleration but with non-zero terminal velocity. It is shown in the Figure

4. If multiple forces \(\overrightarrow{\mathrm{F}}_{1}\), \(\overrightarrow{\mathrm{F}}_{2}\), \(\overrightarrow{\mathrm{F}}_{3}\),……. \(\overrightarrow{\mathrm{F}}_{n}\) act on the same body, then the total force (\(\overrightarrow{\mathrm{F}}_{net}\)) is equivalent to the vectorial sum of the individual forces. Their net force provides the acceleration.
\(\overrightarrow{\mathrm{F}}_{net}\) = \(\overrightarrow{\mathrm{F}}_{1}\) + \(\overrightarrow{\mathrm{F}}_{2}\) + \(\overrightarrow{\mathrm{F}}_{3}\) + ……… + \(\overrightarrow{\mathrm{F}}_{n}\)

Newton’s second law for this case is –
\(\overrightarrow{\mathrm{F}}_{net}\) = m\(\overline{\mathrm{a}}\)
In this case the direction of acceleration is in the direction of net force.
Example:
Bow and arrow

5. Newton’s second law can also be written in the following form.
Since the acceleration is the second derivative of position vector of the body \(\left(\vec{a}=\frac{d^{2} \vec{r}}{d t^{2}}\right)\)
the force on the body is –
\(\overline{\mathrm{F}}\) = m\(\frac{d^{2} \vec{r}}{d t^{2}}\)
From this expression, we can infer that Newton’s second law is basically a second order ordinary differential equation and whenever the second derivative of position vector is not zero, there must be a force acting on the body.

6. If no force acts on the body then Newton’s second law, m = \(\frac{d \vec{v}}{d t}\) = 0
It implies that \(\overline{\mathrm{v}}\) = constant. It is essentially Newton’s first law. It implies that the second law is consistent with the first law. However, it should not be thought of as the reduction of second law to the first when no force acts on the object. Newton’s first and second laws are independent laws. They can internally be consistent with each other but cannot be derived from each other.

7. Newton’s second law is cause and effect relation. Force is the cause and acceleration is the effect. Conventionally, the effect should be written on the left and cause on the right hand side of the equation. So the correct way of writing Newton’s second law is –
m\(\overline{\mathrm{a}}\) = \(\overline{\mathrm{F}}\) or \(\frac{d \vec{p}}{d t}\) = \(\overline{\mathrm{F}}\)

Question 6.
Explain the similarities and differences of centripetal and centrifugal forces.
Answer:
Salient features of centripetal and centrifugal forces.
Centripetal Force:

• It is a real force which is exerted on the body by the external agencies like gravitational force, tension in the string, normal force etc.
• Acts in both inertial and non-inertial frames
• It acts towards the axis of rotation or center of the circle in circular motion
  \(\left|\overrightarrow{\mathrm{F}}_{\mathrm{C}_{\mathrm{P}}}\right|\) = mω²r = \(\frac{m v^{2}}{r}\)
• Real force and has real effects
• Origin of centripetal force is interaction between two objects.
• In inertial frames centripetal force has to be included when free body diagrams are drawn.

Centrifugal Force:

• It is a pseudo force or fictitious force which cannot arise from gravitational force, tension force, normal force etc.
• Acts only in rotating frames (non-inertial frame)
• It acts outwards from the axis of rotation or radially outwards from the center of the circular motion
  \(\left|\overrightarrow{\mathrm{I}}_{\mathrm{C}_{\mathrm{f}}}\right|\) = mω²r = \(\frac{m v^{2}}{r}\)
• Pseudo force but has real effects
• Origin of centrifugal force is inertia. It does not arise from interaction. In an inertial frame the object’s inertial motion appears as centrifugal force in the rotating frame.
• In inertial frames there is no centrifugal force. In rotating frames, both centripetal and centrifugal force have to be included when free body diagrams are drawn.

Question 7.
Briefly explain ‘centrifugal force’ with suitable examples.
Answer:
To use Newton’s first and second laws in the rotational frame of reference, we need to include a Pseudo force called centrifugal force. This centrifugal force appears to act on the object with respect to rotating frames.

Circular motion can be analysed from two different frames of reference. One is the inertial frame (which is either at rest or in uniform motion) where Newton’s laws are obeyed. The other is the rotating frame of reference which is a non – inertial frame of reference as it is accelerating.

When we examine the circular motion from these frames of reference the situations are entirely different. To use Newton’s first and second laws in the rotational frame of reference, we need to include a pseudo force called ‘centrifugal force’. This ‘centrifugal force’ appears to act on the object with respect to rotating frames. To understand the concept of centrifugal force, we can take a specific case and discuss as done below.

Free body diagram of a particle including the centrifugal force Consider the case of a whirling motion of a stone tied to a string. Assume that the stone has angular velocity ω in the inertial frame (at rest). If the motion of the stone is observed from a frame which is also rotating along with the stone with same angular velocity ω then, the stone appears to be at rest.

This implies that in addition to the inward centripetal force – mω²r there must be an equal and opposite force that acts on the stone outward with value +mω²r . So the total force acting on the stone in a rotating frame is equal to zero (-mω²r + mω² r = 0). This outward force +mω²r is called the centrifugal force. The word ‘centrifugal’ means ‘flee from center’.

Note that the ‘centrifugal force’ appears to act on the particle, only when we analyse the motion from a rotating frame. With respect to an inertial frame there is only centripetal force which is given by the tension in the rstring. For this reason centrifugal force is called as a ‘pseudo force’. A pseudo force has no origin. It arises due to the non inertial nature of the frame considered. When circular motion problems are solved from a rotating frame of reference, while drawing free body diagram of a particle, the centrifugal force should necessarily be included as shown in the figure.

Question 8.
Briefly explain ‘rolling friction’.
Answer:
The invention of the wheel plays a crucial role in human civilization. One of the important applications is suitcases with rolling on coasters. Rolling wheels makes it easier than carrying luggage. When an object moves on a surface, essentially it is sliding on it. But wheels move on the surface through rolling motion. In rolling motion when a wheel moves on a surface, the point of contact with surface is always at rest. Since Rolling and kinetic friction the point of contact is at rest, there is no relative motion between the wheel and surface.

Hence the frictional fore is very less. At the same time if an object moves without a wheel, there is a relative motion between the object and the surface. As a result frictional force is larger. This makes it difficult to move the object. The figure shows the difference between rolling and kinetic friction. Ideally in pure rolling, motion of the point of contact with the surface should be at rest, but in practice it is not so.

Due to the elastic nature of the surface at the point of contact there will be some deformation on the object at this point on the wheel or surface as shown in figure. Due to this deformation, there will be minimal friction between wheel and surface. It is called ‘rolling friction. In fact, rolling friction’ is much smaller than kinetic friction.

Question 9.
Describe the method of measuring angle of repose.
Answer:
When objects are connected by strings and When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.

Case 1:
Vertical motion:
Consider two blocks of masses m₁ and m₂ (m₁> m₂) connected by a light and in extensible string that passes over a pulley as shown in Figure.

Let the tension in the string be T and acceleration a. When the system is released, both the blocks start moving, m₂ vertically upward and m_k, downward with same acceleration a. The gravitational force m₁g on mass m₁ is used in lifting the mass m₂. The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.

Applying Newton’s second law for mass m₂
T \(\hat{j}\) – m₂g\(\hat{j}\) = m₂a\(\hat{j}\) The left hand side of the above equation is the total force that acts on m2 and the right hand side is the product of mass and acceleration of m₂ in y direction.
By comparing the components on both sides, we get
T = m₂g = m₂a ……….(1)

Similarly, applying Newton’s second law for mass m₂
T \(\hat{j}\) – m₁g\(\hat{j}\) = m₁a\(\hat{j}\)
As mass mj moves downward (-\(\hat{j}\)), its acceleration is along (-\(\hat{j}\))
By comparing the components on both sides, we get
T = m₁g = -m₁a
m₁g – T = m₁a ………..(2)

Adding equations (1) and (2), we get
m₁g – m₂g = m₁a + m₂a
(m₁ – m₂)g = (m₁ + m₂)a …………(3)
From equation (3), the acceleration of both the masses is –
a = (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ………..(4)
If both the masses are equal (m₁ = m₂), from equation (4)
a = 0
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
T = m₂g = m₂(\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))
T = m₂g + m₂ (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ……….(5)
By taking m2g common in the RHS of equation (5)

Equation (4) gives only magnitude of acceleration
For mass m₁, the acceleration vector is given by \(\vec{a}\) = –\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\)\(\hat{j}\)
For mass m₂, the acceleration vector is given by \(\vec{a}\) = \(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\) \(\hat{j}\)

Case 2:
Horizontal motion:
In this case, mass m₂ is kept on a horizontal table and mass m₁, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface

As both the blocks are connected to the un stretchable string, if m₁ moves with an acceleration a downward then m₂ also moves with the same acceleration a horizontally.
The forces acting on mass m₂ are

1. Downward gravitational force (m₂g)
2. Upward normal force (N) exerted by the surface
3. Horizontal tension (T) exerted by the string

The forces acting on mass m₁ are

1. Downward gravitational force (m₁g)
2. Tension (T) acting upwards

The free body diagrams for both the masses is shown in figure 2.

Applying Newton’s second law for m₁
T\(\hat{i}\) – m₁g\(\hat{j}\) = -m₁a\(\hat{j}\) (alongy direction)
By comparing the components on both sides of the above equation,
T – m₁g = -m₁a …………(1)
Applying Newton’s second law for m₂
T\(\hat{i}\) = m₁a\(\hat{i}\) (along x direction)
By comparing the components on both sides of above equation,
T = m₂a ………….(2)
There is no acceleration along y direction for m₂.
N\(\hat{j}\) – m₂g\(\hat{j}\) = 0
By comparing the components on both sides of the above equation
N – m₂g = 0
N = m₂g ……….(3)
By substituting equation (2) in equation (1), we can find the tension T
m₂a – m₁g = -m₁a
m₂a + m₁a = m₁g
a = \(\frac{m_{1}}{m_{1}+m_{2}}\)g …………(4)
Tension in the string can be obtained by substituting equation (4) in equation (2)
T = \(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\)g ………..(5)

Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

Question 10.
Explain the need for banking of tracks.
Answer:
In a leveled circular road, skidding mainly depends on the coefficient of static friction py. The coefficient of static friction depends on the nature of the surface which has a maximum limiting value. To avoid this problem, usually the outer edge of the road is slightly raised compared to inner edge. This is called banking of roads or tracks. This introduces an inclination, and the angle is called banking angle. “Let the surface of the road make angle 9 with horizontal surface. Then the normal force makes the same angle 9 with the vertical. When the car takes a turn.

there are two forces acting on the car:
(a) Gravitational force mg (downwards)
(b) Normal force N (perpendicular to surface)
We can resolve the normal force into two components N cos θ and N sin θ. The component balances the downward gravitational force ‘mg’ and component will provide the necessary centripetal acceleration. By using Newton second law.

The banking angle 0 and radius of curvature of the road or track determines the safe speed of the car at the turning. If the speed of car exceeds this safe speed, then it starts to skid outward but frictional force comes into effect and provides an additional centripetal force to prevent the outward skidding.

At the same time, if the speed of the car is little lesser than safe speed, it starts to skid inward and frictional force comes into effect, which reduces centripetal force to prevent inward skidding. However if the speed of the vehicle is sufficiently greater than the correct speed, then frictional force cannot stop the car from skidding.

Question 11.
Calculate the centripetal acceleration of Moon towards the Earth.
Answer:
The centripetal acceleration is given by a = \(\frac{v^{2}}{r}\) This expression explicitly depends on Moon’s speed which is nontrivial. We can work with the formula
ω²R_m = a_m
a_m is centripetal acceleration of the Moon due to Earth’s gravity, ω is angular velocity
R_m is the distance between Earth and the Moon, which is 60 times the radius of the Earth.
R_m = 60R = 60 x 6.4 x 10⁶ = 384 x 10⁶ m
As we know the angular velocity ω = \(\frac { 2π}{ T }\) and T = 27.3 days = 27.3 x 24 x 60 x 60 second = 2.358 x 10⁶ sec.
By substituting these values in the formula for acceleration
a₆ = \(\frac{\left(4 \pi^{2}\right)\left(384 \times 10^{6}\right)}{\left(2.358 \times 10^{8}\right)^{2}}\) = 0.00272 ms^-2

Samacheer Kalvi 11th Physics Laws of Motion Conceptual Questions

Question 1.
Why it is not possible to push a car from inside?
Answer:
While trying to push a car from outside, he pushes the ground backwards at an angle. The ground offers an equal reaction in the opposite direction, so car can be moved. But the person sits inside means car and the person becomes a single system, and the force given will be a internal force. According to Newton’s third law, total internal force acting on the system is zero and it cannot accelerate the system.

Question 2.
There is a limit beyond which the polishing of a surface increases frictional resistance rather than decreasing it why?
Answer:
Friction is a contact force. Friction is directly proportional to area of contact. In the normal surfaces there are bumps to interlock the surfaces in contact. But the surfaces are polished beyond certain limit. The area of contact will be increased and the molecules come closer to each other. It increases electrostatic force between the molecules. As a result it increases friction.

Question 3.
Can a single isolated force exist in nature? Explain your answer.
Answer:
No. According to Newton’s third law, for every action, there is an equal and opposite reaction. So, whatever case we consider, if there is an action there is always a reaction. So it is impossible.

Question 4.
Why does a parachute descend slowly?
Answer:
A parachute descends slowly because the surface area of parachute is large so that air gives more resistance when it descends down.

Question 5.
When walking on ice one should take short steps. Why?
Answer:
Let R represent the reaction offered by the ground. The vertical component R cos θ will balance the weight of the person and the horizontal component R sin θ will help the person to walk forward.
Now, normal reaction = R cos θ
Friction force = R sin θ
Coefficient of friction, µ = \(\frac {R sin θ}{R cos θ }\) = tan θ
In a long step, θ is more. So tan θ is more. But p has a fixed value. So, there is danger of slipping in a long step.

Question 6.
When a person walks on a surface, the frictional force exerted by the surface on the person is opposite to the direction of motion. True or false?
Answer:
False. In frictional force exerted by the surface on the person is in the direction of his motion. Frictional force acts as an external force to move the person. When the person trying to move, he gives a push to ground on the backward direction and by Newton’s third law he is pushed by the ground in the forward direction. Hence frictional force acts along the direction of motion.

Question 7.
Can the coefficient of friction be more than one?
Answer:
Yes. The coefficient of friction can be more than one in some cases such as silicone rubber. Coefficient of friction is the ratio of frictional force to normal force, i.e. F = μR. If p is greater than one means frictional force is greater than normal force. But in general case the value is usually between 0 and 1.

Question 8.
Can we predict the direction of motion of a body from the direction of force on it?
Answer:
Yes. The direction of motion is always opposite to the force of kinetic friction. By using the principle of equilibrium, the direction of force of static friction can be determined. When the object is in equilibrium, the frictional force must point in the direction which results as a net force is zero.

Question 9.
The momentum of a system of particles is always conserved. True or false?
Answer:
True. The total momentum of a system of particles is always constant i.e. conserved. When no external force acts on it.

Samacheer Kalvi 11th Physics Laws of Motion Numerical Problems

Question 1.
A force of 50 N act on the object of mass 20 kg. shown in the figure. Calculate the acceleration of the object in x and y directions.
Answer:
Given F = 50 N and m = 20 kg
(1) component of force along x – direction
F_x = F cos θ
= 50 x cos 30° = 43.30 N
a_x = \(\frac{F_{x}}{m}\) = \(\frac {43.30}{20}\) =2.165 ms^-2

(2) Component of force along y – direction
F_y = F sin θ = 50 sin 30° = 25 N
a_y = \(\frac{F_{y}}{m}\) = \(\frac {25}{20}\) = 1.25 ms^-2

Question 2.
A spider of mass 50 g is hanging on a string of a cob web as shown in the figure. What is the tension in the string?
Answer:
Given m = 50 g, T = ?
Tension in the string T = mg
= 50 x 10^-2 x 9.8 = 0.49 N

Question 3.
What is the reading shown in spring balance?

Answer:
When a spring balance hung on a rigid support and load is attached at its other end, the weight of the load exerts a force on the rigid support in turn support exerts equal and opposite force on that load, so that balance will be stretched. This is the principle of spring balance. Flence the answer is 4 kg.
Given: m = 2 kg, 0 = 30°.
Resolve the weight into its component as mg sin θ and mg cos θ.
Here mg sin θ acts parallel to the surface
∴ W = mg sin θ
= 2 x 9.8 x sin 30° = 2 x 9.8 x \(\frac {1}{2}\) = 9.8 N

Question 4.
The physics books are stacked on each other in the sequence: +1 volumes 1 and 2; +2 volumes 1 and 2 on a table
(a) Identify the forces acting on each book and draw the free body diagram.
(b) Identify the forces exerted by each book on the other.
Answer:
Let m₁, m₂, m₃, m_4, are the masses of +1 volume I and II and +2 volumes I & II

(a) Force on book m₄

• Downward gravitational force acting downward (m₃g)
• Upward normal force (N₃) exerted by book of mass m₃
  |

(b) Force on book m₃

• Downward gravitational force (m₃g)
• Downward force exerted by m₄ (N₄)
• Upward force exerted by m₂ (N₂)
  

(c) Force on book m₂

• Downward gravitational force (m₂g)
• Downward force exerted by m₃ (N₃)
• Upward force exerted by m₁ (CN₁)
  

(d) Force on book m₁

• Downward gravitational force exerted by earth (m₁g)
• Downward force exerted by m₂ (N₂)
• Upward force exerted by the table (N_table)
  

Question 5.
A bob attached to the string oscillates back and forth. Resolve the forces acting on the bob in to components. What is the acceleration experienced by the bob at an angle θ.
Answer:
The gravitational force (mg) acting downward can be resolved into two components as mg cos θ and mg sin θ
T – tension exerted by the string.
Tangential force F_T = ma_T = mg sin θ
∴ Tangential acceleration a_T = g sin θ
Centripetal force Fc = ma_c = T – mg cos θ
a_c = \(\frac { T – mg cos θ }{ m }\)

Question 6.
Two masses m₁ and m₂ are connected with a string passing over a friction-less pulley fixed at the comer of the table as shown in the figure. The coefficient of static friction of mass m₁ with the table is µ_s Calculate the minimum mass m₃ that may be placed on m₁to prevent it from sliding. Check if m₁ = 15 kg, m₂ = 10 kg, m₃ = 25 and µ_s = 0.2.

Solution:
Let m₃ is the mass added on m₁
Maximal static friction
\(f_{s}^{\max }\) = µ_sN = µ_s (m₁ + m₃ )g
Here
N = (m₁ + m₃ )g
Tension acting on string = T = m₂ g
Equate (1) and (2)
µ_s(m₁ + m₃) = m₂g
µ_sm₁ + µ_sm₃ = m₂
m₃ = \(f_{s}^{\max }\) – m₁

(ii) Given,
m₁ = 15 kg, m₂ = 10 kg : m₃ = 25 kg and µ_s = 0.2
m₃ = \(f_{s}^{\max }\) – m₁
m₃ = \(\frac {10}{ 0.2 }\) – 15 = 50 – 15 = 35 kg
The minimum mass m₃ = 35 kg has to be placed on ml to prevent it from sliding. But here m₃ = 25 kg only.
The combined masses (m₁ + m₃) will slide.

Question 7.
Calculate the acceleration of the bicycle of mass 25 kg as shown in Figures 1 and 2.

Answer:
Given:
Mass of bicycle m = 25 kg
Fig. I:
Net force acting in the forward direction, F = 500 – 400 = 100 N
acceleration a = \(\frac { F}{ m }\) = \(\frac { 100 }{25}\) = 4 ms^-2

Fig. II:
Net force acting on bicycle F = 400 – 400 = 0
∴ acceleration a = \(\frac { F}{ m }\) = \(\frac { 0}{25}\) = 0

Question 8.
Apply Lami’s theorem on sling shot and calculate the tension in each string?

Answer:
Given F = 50 N, θ = 30°
Here T is resolved into its components as T sin θ and T cos θ as shown.
According to Lami’s theorem,
\(\frac { F}{sin θ}\) = \(\frac { T}{sin (180 – θ)}\) = \(\frac { T}{sin (180 – θ)}\)
\(\frac { F}{sin θ}\) = \(\frac { T}{sin θ}\)
\(\frac { F}{2 sin θ cos θ}\) = \(\frac { T}{sin θ}\) [T = \(\frac { T}{2 cos θ}\) ]
T = \(\frac {T}{2 cos θ}\) = \(\frac { 50}{ 2 cos 30}\) = 28.868 N

Question 9.
A football player kicks a 0.8 kg ball and imparts it a velocity 12 ms^-1. The contact between the foot and ball is only for one – sixtieth of a second. Find the average kicking force.
Answer:
Given,
Mass of the ball = 0.8 kg
Final velocity (V) =12 ms^-1 and time t =\(\frac {1}{60}\) s
Initial velocity = 0
We know the average kicking force
F = ma = \(\frac {m(v – u)}{t}\) = \(\frac{0.8(12-0)}{\left(\frac{1}{60}\right)}\)
F = 576 N

Question 10.
A stone of mass 2 kg is attached to a string of length 1 meter. The string can withstand maximum tension 200 N. What is the maximum speed that stone can have during the whirling motion?
Solution:
Given,
Mass of a stone = 2 kg,
length of a string = 1 m
Maximum tension = 200 N
The force acting on a stone in the whirling motion is centripetal force. Which is provided by tension of the string.
T_max = F_max = \(\frac{m \mathrm{V}_{\mathrm{max}}^{2}}{r}\)
200 = \(v_{\max }^{2}\) = 100
\(v_{\max }^{2}\) = 10 ms^-1

Question 11.
Imagine that the gravitational force between Earth and Moon is provided by an invisible string that exists between the Moon and Earth. What is the tension that exists in this invisible string due’ to Earth’s centripetal force? (Mass of the Moon = 7.34 x 1022 kg, Distance between Moon and Earth = 3.84 x 108 m).

Solution:
Given,
Mass of the moon = 7.34 x 1022 kg
Distance between moon and earth = 3.84 x 108 m
Centripetal force = F = \(\frac{m \mathrm{V}^{2}}{r}\) = \(\frac{7.34 \times 10^{22} \times\left(1.023 \times 10^{3}\right)^{2}}{3.84 \times 10^{8}}\) = 2 x 10²⁰

Question 12.
Two bodies of masses 15 kg and 10 kg are connected with light string kept on a smooth surface. A horizontal force F = 500 N is applied to a 15 kg as shown in the figure. Calculate the tension acting in the string.

Answer:
Given,
m₁  = 15 kg, m₂  = 10 kg, F = 500 N
Tension acting in the string T = \(\frac{m_{2}}{m_{1}+m_{2}}\) F
T = \(\frac {10}{25}\) x 500 = 200 N

Question 13.
People often say “For every action there is an equivalent opposite reaction”. Here they meant ‘action of a human’. Is it correct to apply Newton’s third law to human actions? What is meant by ‘action’ in Newton third law? Give your arguments based on Newton’s laws.
Answer:
Newton’s third law is applicable to only human’s physical actions which involves physical force. Third law is not applicable to human’s psychological actions or thoughts.

Question 14.
A car takes a turn with velocity 50 ms^-1 on the circular road of radius of curvature To m. Calculate the centrifugal force experienced by a person of mass 60 kg inside the car?
Answer:
Given,
Mass of a person = 60 kg
Velocity of the car = 50 ms^-1
Radius of curvature = 10 m
Centrifugal force F = \(\frac{m \mathrm{V}^{2}}{r}\) = \(\frac{60 \times(50)^{2}}{10}\) = 15,000 N

Question 15.
A long stick rests on the surface. A person standing 10 m away from the stick. With what minimum speed an object of mass 0.5 kg should he thrown so that it hits the stick. (Assume the coefficient of kinetic friction is 0.7).
Answer:
Given,
Distance (s) = 10 m
Mass of the object (m) = 0.5 kg
Coefficient of kinetic friction (µ) = 0.7
Work done in moving a body in horizontal surface ω = µ_R x s = µmg x s
This work done is equal to initial kinetic energy of the object
\(\frac{1}{2} m v^{2}\) = µ mg s
\(\left|v^{2}\right|\) = 2 µgs = 2 x 0.7 x 9.8 x 10
v² = 14 x 9.8 = 137. 2
v = 11. 71 ms^-1

Samacheer Kalvi 11th Physics Laws of Motion Additional Questions Solved

Samacheer Kalvi 11th Physics Laws of Motion Multiple Choice Questions

Question 1.
The concept “force causes motion” was given by –
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(b) Aristotle

Question 2.
Who decoupled the motion and force?
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(a) Galileo

Question 3.
The inability of objects to move on its own or change its state of motion is called as –
(a) force
(b) momentum
(c) inertia
(d) impulse
Answer:
(c) inertia

Question 4.
Inertia means –
(a) inability
(b) resistance to change its state
(c) movement
(d) inertial frame
Answer:
(b) resistance to change its state

Question 5.
When a bus starts to move from rest, the passengers experience a sudden backward push is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(c) Inertia of rest

Question 6.
If the brake is applied in the moving bus suddenly, passengers move forward is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(a) Inertia of motion

Question 7.
In whirling motion, if the string is cut suddenly, the stone moves tangential to circle is an –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(b) Inertia of direction

Question 8.
Newtons laws are applicable in –
(a) Inertial frame
(b) non inertial frame
(c) in any frame
(d) none
Answer:
(a) Inertial frame

Question 9.
The accelerated train is an example for –
(a) inertial frame
(b) non-inertial frame
(c) both (a) and (b)
(d) none of the above
Answer:
(b) non-inertial frame

Question 10.
Rate of change of momentum of an object is equal to –
(a) acceleration
(b) work done
(c) force
(d) impulse
Answer:
(c) force

Question 11.
The product of mass and velocity is –
(a) force
(b) impulse
(c) momentum
(d) acceleration
Answer:
(c) momentum

Question 12.
Unit of momentum –
(a) kg ms^-2
(b) kg ms^-1
(c) MLT^-2
(d) MLT^-1
Answer:
(b) kg ms^-1

Question 13.
According to Newton’s third law –
(a) F₁₂ = F₂₁
(*) F₁₂ = -F₂₁
(c) F₁₂ + F₂₁ = 0
(d) F₁₂ x F₂₁ = 0
Answer:
(a) F₁₂ = F₂₁

Question 14.
According to Newton’s third law –
(a) \(\overrightarrow{\mathrm{F}_{12}}=\overrightarrow{\mathrm{F}_{21}}\)
(b) \(\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}\)
(c) \(\mathrm{F}_{12}+\mathrm{F}_{21}\) = 0
(d) \(\mathrm{F}_{12}x\mathrm{F}_{21}\) = 0
Answer:
(b) \(\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}\)

Question 15.
The law which is valid in both inertial and non-inertial frame is –
(a) Newton’s first law
(b) Newton’s second law
(c) Newton’s third law
(d) none
Answer:
(c) Newton’s third law

Question 16.
When a force is applied on a body, it can change –
(a) velocity
(b) momentum
(c) direction of motion
(d) all the above
Answer:
(d) all the above

Question 17.
The rate of change of velocity is 1 ms^-2 when a force is applied on the body of mass 75 gm the force is –
(a) 75 N
(b) 0.75 N
(c) 0.075 N
(d) 0.75 x 10^-3 N
Answer:
(c) Force is given by
F = m a
= 75 gm x 1 cm s^-2 = 75 x 10^-3 x 1 = 75 x 10^-3 = 0.075 N

Question 18.
The action and reaction forces acting on –
(a) same body
(b) different bodies
(c) either same or different bodies
(d) none of the above
Answer:
(b) different bodies

Question 19.
Newton’s first law of motion gives the concept of –
(a) velocity
(b) energy
(c) momentum
(d) Inertia
Answer:
(d) Inertia

Question 20.
Inertia of a body has direct dependence on –
(a) velocity
(b) area
(c) mass
(d) volume
Answer:
(c) mass

Question 21.
If a car and a scooter have the same momentum, then which one is having greater speed?
(a) scooter
(b) car
(c) both have same velocity
(d) data insufficient
Answer:
(a) scooter

Question 22.
Newton’s second law gives –
(a) \(\overrightarrow{\mathrm{F}} \propto \frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(b) \(\overrightarrow{\mathrm{F}}=\frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(c) \(\overrightarrow{\mathrm{F}}=m \vec{a}\)
(d) all the above
Answer:
(d) all the above

Question 23.
1 dyne is –
(a) 10⁵N
(b) 10^-5N
(c) 1N
(d) 10^-3N
Answer:
(b) 10^-5N

Question 24.
If same force is acting on two masses m₁ and m₂, and the accelerations of two bodies are a₁ and a₂ respectively, then –
(a) \(\frac{a_{2}}{a_{1}}=\frac{m_{2}}{m_{1}}\)
(b) \(\frac{a_{1}}{a_{2}}=\frac{m_{1}}{m_{2}}\)
(c) \(\frac{a_{1}}{a_{2}}=\frac{m_{2}}{m_{1}}\)
(d) m₁ a₁ + m₂a₂ = 0
Answer:
(c) \(\frac{a_{1}}{a_{2}}=\frac{m_{2}}{m_{1}}\)

Question 25.
If a force \(\overline{\mathrm{F}}\) = 3\(\hat{i}\) – 4\(\hat{j}\) N produces an acceleration of 10 ms^-2 on a body, then the mass of a body is –
(a) 10 kg
(b) 9 kg
(c) 0.9 kg
(d) 0.5 kg
Answer:
\(\overline{\mathrm{F}}\) = 3\(\hat{i}\) – 4\(\hat{j}\)
Magnitude:
|\(\overline{\mathrm{F}}\)| = \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5N
F = ma
⇒ m = \(\frac{|\mathrm{F}|}{a}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\) = 0.5 kg

Question 26.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop?
(a) 0.75 s
(b) 1.33 s
(c) 6 s
(d) 35 s
Answer:
Acceleration a = \(\frac{-F}{m}\) = \(\frac{50}{20}\) = – 2.5 ms^-2
u = l5 ms^-1
v = 0
t = ?
v = u + at
0 = 15 – 2.5t
t = \(\frac{15}{2.5}\) = 6s

Question 27.
Rain drops come down with –
(a) zero acceleration and non zero velocity
(b) zero velocity with non zero acceleration
(c) zero acceleration and non zero net force
(d) none
Answer:
(a) zero acceleration and non zero velocity

Question 28.
If force is the cause then the effect is –
(a) mass
(b) potential energy
(c) acceleration
(d) Inertia
Answer:
(c) acceleration

Question 29.
In free body diagram, the object is represented by a –
(a) line
(b) arrow
(c) circle
(d) point
Answer:
(d) point

Question 30.
When an object of mass m slides on a friction less surface inclined at an angle 0, then normal force exerted by the surface is –
(a) g cos θ
(b) mg cos θ
(c) g sin θ
(d) mg tan θ
Answer:
(b) mg cos θ

Question 31.
The acceleration of the sliding object in an inclined plane –
(a) g cos θ
(b) mg cos θ
(c) g sin θ
(d) mg sin θ
Answer:
(c) g sin θ

Question 32.
The speed of an object sliding in an inclined plane at the bottom is –
(a) mg cos θ
(b) \(\sqrt{2 s g sin θ}\)
(c) \(\sqrt{2 s g cos θ}\)
(d) \(\sqrt{2 s g tan θ}\)
Answer:
(b) \(\sqrt{2 s g sin θ}\)

Question 33.
The acceleration of two bodies of mass m₁ and m₂ in contact on a horizontal surface is –
(a) \(a=\frac{\mathbf{F}}{m_{1}}\)
(b) \(a=\frac{F}{m_{2}}\)
(c)  \(a=\frac{\mathrm{F}}{m_{1}+m_{2}}\)
(d) \(a=\frac{\mathrm{F}}{m_{1} m_{2}}\)
Answer:
(c)  \(a=\frac{\mathrm{F}}{m_{1}+m_{2}}\)

Question 34.
Two blocks of masses m₁ and m₂ (m₁ > m₂) in contact with each other on frictionless, horizontal surface. If a horizontal force F is given on m₁, set into motion with acceleration a, then reaction force on mass m₁ by m₂, is –
(a) \(\frac{\mathrm{F} m_{1}}{m_{1}+m_{2}}\)
(b) \(\frac{m_{1} m_{2}}{\mathrm{F} m_{1}}\)
(c) \(\frac{m_{1} m_{2}}{\mathrm{F} m_{2}}\)
(d) \(\frac{\mathrm{F} m_{2}}{m_{1}+m_{2}}\)
Answer:
(d) \(\frac{\mathrm{F} m_{2}}{m_{1}+m_{2}}\)

Question 35.
If two masses m₁ and m₂ (m₁ > m₂) tied to string moving over a frictionless pulley, then acceleration of masses –
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g
(b) \(\frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)}\) g
(c) \(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\) g
(d) \(\frac{m_{1} m_{2}}{2 m_{1} m_{2}}\) g
Answer:
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g

Question 36.
if two masses m₁ and m₂ (m₁ > m₂)tied to string moving over a frictionless pulley, then acceleration of masses –
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g
(b) \(\frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)}\) g
(c) \(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\) g
(d) \(\frac{m_{1} m_{2}}{2 m_{1} m_{2}}\) g
Answer:
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g

Question 37.
Three massses is in contact as shown. If force F is applied to mass m₁, the acceleration of three masses is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Answer:
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Question 38.
Three masses in contact is as shown above. If force F is applied to mass m₁ then the contact force acting on mass m₂ is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)

Question 39.
Three masses is contact as shown. It force F is applied to mass m₁, then the contact force acting on mass m₃ is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Question 40.
Two masses connected with a string. When a force F is applied on mass m₂. The acceleration produced is –

(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}}\)
(b) \(\frac{\mathbf{F}}{m_{1}-m_{2}}\)
(c) \(\frac{m_{1}+m_{2}}{\mathrm{F}}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}}\)

Question 41.
Two masses connected with a string. When a force F is applied on mass m₂. The force acting on m₁ is –
(a) \(\frac{m_{1} \mathrm{F}}{m_{1}+m_{2}}\)
(b) \(\frac{m_{2} \mathrm{F}}{m_{1}+m_{2}}\)
(c) \(\frac{m_{1}+m_{2}}{m_{1}} \mathbf{F}\)
(d) \(\frac{m_{1}+m_{2}}{m_{2}} \mathbf{F}\)
Answer:
(b) \(\frac{m_{2} \mathrm{F}}{m_{1}+m_{2}}\)

Question 42.
If a block of mass m lying on a frictionless inclined plane of length L height h and angle of inclination θ, then the velocity at its bottom is –
(a) g sin θ
(b) g cos θ
(c) \(\sqrt{2 g h}\)
(d) \(\sqrt{2 a sin θ}\)
Answer:
(c) \(\sqrt{2 g h}\)

Question 43.
If a block of mass m lying on a frictionless inclined plane of length L, height h and angle of inclination θ, then the time take taken to reach the bottom is –
(a) g sing θ
(b) sin θ \(\sqrt{\frac{2 h}{g}}\)
(c) sin θ \(\sqrt{\frac{g}{h}}\)
(d) \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)
Answer:
(d) \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)

Question 44.
A rocket works on the principle of conservation of –
(a) energy
(b) mass
(c) angular momentum
(d) linear momentum
Answer:
(b) mass

Question 45.
A bomb at rest explodes. The total momentum of all its fragments is –
(a) zero
(b) infinity
(c) always 1
(d) always greater then 1
Answer:
(a) zero

Question 46.
A block of mass m₁ is pulled along a horizontal friction-less surface by a rope of mass m₂ If a force F is given at its free end. The net force acting on the block is –
(a) \(\frac{m_{1} \mathrm{F}}{m_{1}-m_{2}}\)
(b) F
(c) \(\frac{m_{2} \mathrm{F}}{\left(m_{1}+m_{2}\right)}\)
(d) \(\frac{m_{1} \mathrm{F}}{\left(m_{1}+m_{2}\right)}\)
Answer:
(b) F

Question 47.
A block of mass m is pulled along a horizontal surface by a rope. The tension in the rope will be same at all the points –
(a) if the rope is accelerated
(b) if the rope is mass less
(c) always
(d) none of the above
Answer:
(b) if the rope is mass less

Question 48.
The lines of forces act at a common point is called as –
(a) concurrent forces
(b) co-planar forces
(c) equilibrium
(d) resultant
Answer:
(a) concurrent forces

Question 49.
If the lines of forces act in the same plane, they can be –
(a) concurrent forces
(b) coplanar forces
(c) either concurrent force or coplanar forces
(d) Lami’s force
Answer:
(d) concurrent forces

Question 50.
Lami’s theorem is applicable only when the system of forces are is –
(a) same plane
(b) different plane
(c) equilibrium
(d) none of the above
Answer:
(c) equilibrium

Question 51.
Due to the action of internal forces of the system, the total linear momentum of the system is –
(a) a variable
(b) a constant
(c) always zero
(d) always infinity
Answer:
(c) always zero

Question 52.
The velocity with which a gun suddenly moves backward after firing is –
(a) linear velocity
(b) positive velocity
(c) recoil velocity
(d) v₁ + v₂
Answer:
(c) recoil velocity

Question 53.
If a very large force acts on an object for a very short duration, then the force is called as –
(a) Newtonian force
(b) impulsive force
(c) concurrent force
(d) coplanar force
Answer:
(A) impulsive force

Question 54.
The unit of impulse is –
(a) Nm
(b) Ns
(c) Nm²
(d) Ns^-2
Answer:
(b) Ns

Question 55.
The force which always opposes the relative motion between an object and the surface where it is placed is –
(a) concurrent force
(b) frictional force
(c) impulsive force
(d) coplanar force
Answer:
(b) frictional force

Question 56.
The force which opposes the initiation of motion of an object on the surface is –
(a) static friction
(b) kinetic friction
(c) friction
(d) zero
Answer:
(d) static friction

Question 57.
When the object is at rest, the resultant of gravitational force and upward normal force is –
(a) Static force
(b) zero
(c) one
(d) infinity
Answer:
(b) zero

Question 58.
The magnitude of static frictional force d lies between –
(a) 0 ≤ f ≤ µ_sN
(b) 0 ≥f ≥ µ_sN
(c) 0 and 1
(d) 0 and minimal static frictional force.
Answer:
(a) 0 ≤ f ≤ µ_sN

Question 59.
The unit of co-efficient of static friction is –
(a) N
(b) N m
(c) N s
(d) no unit
Answer:
(d) no unit

Question 60.
If the object is at rest and no external force is applied on the object, the static friction acting on the object is –
(a) µ_sN
(b) zero
(c) one
(d) infinity
Answer:
(d) no unit

Question 61.
When object begins to slide, the static friction acting on the object attains –
(a) zero
(b) minimum
(c) maximum
(d) infinity
Answer:
(c) maximum

Question 62.
The static friction does not depend upon –
(a) the area of contact
(b) normal force
(c) the magnitude of applied force
(d) none of the above
Answer:
(a) the area of contact

Question 63.
Which of the following pairs of materials has minimum amount of coefficient of static friction is –
(a) Glass and glass
(b) wood and wood
(c) ice and ice
(d) steel and steel
Answer:
(c) ice and ice

Question 64.
Kinetic friction is also called as –
(a) sliding friction
(b) dynamic friction
(c) both (a) and (b)
(d) static friction
Answer:
(c) both (a) and (b)

Question 65.
The unit of coefficient of kinetic friction is/has –
(a) Nm
(b) Ns
(c) Nm²
(d) no unit
Answer:
(d) no unit

Question 66.
The nature of materials in mutual contact decides –
(a) µs
(b) µk
(c) µs or µk
(d) none
Answer:
(c) µs or µk

Question 67.
Coefficient of kinetic friction is less than –
(a) O
(b) one
(c) µs
(d) µsN
Answer:
(c) µs

Question 68.
The static friction –
(a) increases linearly
(b) is constant
(c) zero
(d) varies parabolically
Answer:
(a) increases linearly

Question 69.
The kinetic friction –
(a) increases linearly
(b) is constant
(c) zero
(d) varies parabolically
Answer:
(b) is constant

Question 70.
Kinetic friction is independent of –
(a) nature of materials
(b) temperature of the surface
(c) applied force
(d) none of the above
Answer:
(c) applied force

Question 71.
The angle between the normal force and the resultant force of normal force and maximum frictional force is –
(a) angle of friction
(b) angle of repose
(c) angle of inclination
(d) none of the above
Answer:
(a) angle of friction

Question 72.
The angle friction θ is given by –
(a) tan µ_s
(b) tan^-1 µ_s
(c) \(\frac{f S^{\mathrm{max}}}{N}\)
(d) sin^-1 µ_s
Answer:
(b) tan^-1 µ_s

Question 73.
The angle of inclined plane with the horizontal such that an object placed on it begins to slide is –
(a) angle of friction
(b) angle of repose
(c) angle of response
(d) angle of retardation
Answer:
(b) angle of repose

Question 74.
Comparatively, which of the following has lesser value than others?
(a) static friction
(b) kinetic friction
(c) Rolling friction
(d) skiping friction
Answer:
(c) Rolling friction

Question 75.
The origin of friction is –
(a) electrostatic interaction
(b) electromagnetic interaction magnetic
(c) photon interaction
(d) interaction
Answer:
(b) electromagnetic interaction

Question 76.
Friction can be reduced by –
(a) polishing
(b) lubricating
(c) using ball bearings
(d) all the above
Answer:
(c) using ball bearings

Question 77.
For a particle revolving in a circular path, the acceleration of the particle is –
(a) along the tangent
(b) along the radius
(c) along the circumference of the circle
(d) zero
Answer:
(b) along the radius

Question 78.
A particle moves along a circular path under the action of a force. The work done by the force is –
(a) Positive and non zero
(b) zero
(c) Negative and non zero
(d) none of the above
Answer:
(b) zero

Question 79.
A bullet hits and gets embedded in a solid block resting on a horizontal frictionless table. Which of the following is conserved?
(a) Momentum and kinetic energy
(b) kinetic energy alone
(c) Momentum alone
(d) potential energy alone
Answer:
(c) Momentum alone

Question 80.
The origin of the centripetal force can be –
(a) gravitational force
(b) frictional force
(c) coulomb force
(d) all the above
Answer:
(d) all the above

Question 81.
Centripetal acceleration is –
(a) \(\frac{m v^{2}}{r}\)
(b) \(\frac{v^{2}}{r}\)
(c) r v²
(d) rω
Answer:
(b) \(\frac{v^{2}}{r}\)

Question 82.
Centripetal acceleration is –
(a) \(\frac{m v^{2}}{r}\)
(b) r ω²
(c) rv²
(d) rω
Answer:
(c) rω²

Question 83.
The centripetal force is –
(a) \(\frac{m v^{2}}{r}\)
(b) rω²
(c) both (a) and (b)
(d) none
Answer:
(c) both (a) and (b)

Question 84.
When a car is moving on a circular track the centripetal force is due to –
(a) gravitational force
(b) frictional force
(c) magnetic force
(d) elastic force
Answer:
(b) frictional force

Question 85.
If the road is horizontal then the normal force and gravitational force are –
(a) equal and along the same direction
(b) equal and opposite
(c) unequal and along the same direction
(d) unequal and opposite
Answer:
(b) equal and opposite

Question 86.
The velocity of a car for safe turn on leveled circular road –
(a) \(v \leq \sqrt{\mu_{s} r g}\)
(b) \(v \geq \sqrt{\mu_{s} r g}\)
(c) \(v=\sqrt{\mu_{s} rg}\)
(d) \(v \leq \mu_{s} rg\)
Answer:
(a) \(v \leq \sqrt{\mu_{s} r g}\)

Question 87.
In a leveled circular road, skidding mainly depends on –
(a) µ_s
(b) µ_k
(c) acceleration
(d) none
Answer:
(a) µ_s

Question 88.
The speed of a car to move on the banked road so that it will have safe turn is –
(a) µ_srg
(b) \(\sqrt{r g \tan \theta}\)
(c) rg tan θ
(d) r²g tan θ
Answer:
(b) \(\sqrt{r g \tan \theta}\)

Question 89.
Centrifugal force is a –
(a) pseudo force
(b) real force
(c) forced acting towards center
(d) none of the above
Answer:
(a) pseudo force

Question 90.
Origin of centrifugal force is due to –
(a) interaction between two
(b) inertia
(c) electromagnetic interaction
(d) inertial frame
Answer:
(b) inertia

Question 91.
Centripetal force acts in –
(a) inertial frame
(b) non inertial frame
(c) both (a) and (h)
(d) linear motion
Answer:
(c) both (a) and (b)

Question 92.
Centrifugal force acts in –
(a) inertial frame
(b) non inertial frame
(c) both (a) and (b)
(d) linear motion
Answer:
(b) non inertial frame

Question 93.
A cricket ball of mass loo g moving with a velocity of 20 ms-¹ is brought to rest by a player in 0.05s the impulse of the ball is –
(a) 5 Ns
(b) – 2 Ns
(c) – 2.5 Ns
(d) zero
Answer:
(b) – 2 Ns
mass = 0.1 kg
Initial velocity t = 20 ms^-1
Final velocity y = 0
Change in momentum in impulse = m(v – u) = 0.1(0 – 20) = – 2 Ns

Question 94.
If a stone tied at the one end of a string of length 0.5 m is whirled in a horizontal circle with a constant speed 6 ms^-1  then the acceleration of the shone is –
(a) 12 ms^-2
(b) 36 ms^-2
(c) 2π² ms^-2
(d) 72 ms^-2
Answer:
(d) Centripetal acceleration = \(\frac{v^{2}}{r}\) = \(\frac{6^{2}}{0.5}\) = \(\frac{36}{0.5}\) = 72 ms^-2

Question 95.
A block of mass 3 kg is at rest on a rough inclined plane with angle of inclination 30° with horizontal. If .is 0.7, then the frictional force is –
(a) 17.82 N
(b) 1.81 N
(c) 3.63 N
(d) 2.1 N
Answer:
(a) Frictional force = µmg cos θ = 0.7 x 3 x 9.8 cos 30° = 17.82 N

Question 96.
Two masses 2 kg and 4 kg are tied at the ends of a mass less string and which is passing over a friction-less pulley. The tension in the string is –
(a) 3.68 N
(b) 78.4 N
(c) 26 N
(d) 13.26 N
Answer:
(c) Tension in the string T = \(\frac{2 m_{1} m_{2}}{\left(m_{1}+m_{2}\right)}\)g
T = \(\frac{2 x 2 x 4}{2 + 4}\) x 9.8 = \(\frac{16}{6}\) x 9.8 = 26.13 N

Question 97.
A bomb of 10 kg at rest explodes into two pieces of mass 4 kg and 6 kg. if the velocity of 4 kg mass is 6 ms-¹ then the velocity of 6 kg is –
(a) – 4 ms^-1
(b) – 6 ms^-1
(c) – 24 ms^-1
(d) – 2.2 ms^-1
Answer:
(a) According to law of conservation of momentum
m₁v₁ + m₂v₂ = 0
v₂ = –\(\frac{m_{1} v_{1}}{m_{2}}\) = \(\frac{4 x 6}{6}\) = -4ms^-1

Question 98.
A body is subjected under three concurrent forces and it is in equilibrium. The resultant of any two forces is –
(a) coplanar with the third force
(b) is equal and opposite to third force
(c) both (a) and (b)
(d) none of the above
Answer:
(c) both (a) and (b)

Question 99.
An impulse is applied to a moving object with the force at an angle of 20° with respect to velocity vector. The angle between the impulse vector and the change in momentum vector is –
(a) 0°
(b) 30°
(c) 60°
(d) 120°
Answer:
(a) Impulse and change in momentum are in same direction. So the angle is zero.

Question 100.
A bullet of mass m and velocity v₁ is fired into a large block of wood of mass M. The final velocity of the system is-
(a) \(\frac{v_{1}}{m+\mathrm{M}}\)
(b) \(\frac{m v_{1}}{m+\mathrm{M}}\)
(c) \(\frac{m+m}{m} v_{1}\)
(d) \(\frac{m+m}{m-M} v_{1}\)
Answer:
(b) \(\frac{m v_{1}}{m+\mathrm{M}}\)

Question 101.
A block of mass 2 kg is placed on the floor. The co – efficient of static friction is 0.4. The force of friction between the block and floor is –
(a) 2.8 N
(b) 7.8 N
(c) 2 N
(d) zero
Answer:
(b) The force required to move = = µR = µmg = 0.4 x 2 x 9.8 = 7.84 N

Question 102.
A truck weighing 1000 kg is moving with velocity of 50 km/h on smooth horizontal roads. A mass of 250 kg is dropped into it. The velocity with which it moves now is –
(a) 12.5 km/h
(b) 20 km/h
(c) 40 km/h
(d) 50 km/h
Answer:
(c) According to law of conservation of linear momentum
m₂ v₂ = (m₁ + m₂)v₂
v₂ = \(\frac{m_{1} v_{1}}{m_{1}+m_{2}}\) = \(\frac{1000 \times 50}{1250}\) = 40 km/h

Question 103.
A body of mass loo g is sliding from an inclined plane of inclination 30°. if u = 1.7, then the frictional force experienced is –
(a) \(\frac{3.4}{\sqrt{3}}\)N
(b) 1.47 N
(c) \(\frac{\sqrt{3}}{3.4}\)N
(d) 1.38 N
Answer:
(b) Frictional force F = µ mg cos θ = 1.7 x 0.1 x 10 cos 30°= \(\frac{1.7}{2}\) x \(\sqrt{3}\) = 1.47 N

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (1 Mark)

Question 1.
A passenger sitting in a car at rest, pushes the car from within. The car doesn’t move, why?
Answer:
For motion, there should be external force.

Question 2.
Give the magnitude and directions of the net force acting on a rain drop falling with a constant speed.
Answer:
as \(\overline{\mathrm{a}}\) = 0 so \(\overline{\mathrm{F}}\) = 0.

Question 3.
Why the passengers in a moving car are thrown outwards when it suddenly takes a turn?
Answer:
Due to inertia of direction.

Question 4.
You accelerate your car forward. What is the direction of the frictional force on a package resting on the floor of the car?
Answer:
The package in the accelerated car (a non inertial frame) experiences a Pseudo force in a direction opposite to that of the motion of the car. The frictional force on the package which acts opposite to this pseudo force is thus in the same direction (forward) as that of the car.

Question 5.
What is the purpose of using shockers in a car?
Answer:
To decrease the impact of force by increasing the time for which force acts.

Question 6.
Why are types made of rubber not of steel?
Answer:
Since coefficient of friction between rubber and road is less than the coefficient of friction between steel and road.

Question 7.
Wheels are made circular. Why?
Answer:
Rolling friction is less than sliding friction.

Question 8.
If a ball is thrown up in a moving train, it comes back to the thrower’s hands. Why?
Answer:
Both during its upward and downward motion, the ball continues to move inertia of motion with the same horizontal velocity as the train. In this period, the ball covers the same horizontal distance as the train and so it comes back to the thrower’s hand.

Question 9.
Calculate the force acting on a body which changes the momentum of the body at the rate of 1 kg-m/s² .
Answer:
As F = rate change of momentum
F = 1 kg-m/s² = 1N

Question 10.
On a rainy day skidding takes place along a curved path. Why?
Answer:
As the friction between the types and road reduces on a rainy day.

Question 11.
Why does a gun recoils when a bullet is being fired?
Answer:
To conserve momentum.

Question 12.
Why is it difficult to catch a cricket ball than a tennis ball even when both are moving with the same velocity?
Answer:
Being heavier, cricket ball has higher rate of change of momentum during motion so more force sumed.

Question 13.
The distance travelled by a moving body is directly proportional to time. Is any external force acting on it?
Answer:
As s ∝ t, so acceleration a = 0, therefore, no external force is acting on the body.

Question 14.
Calculate the impulse necessary to stop a 1500 kg car moving at a speed of 25 ms^-1.
Answer:
Use formula I = change in momentum = m(v – u) (Impulse – 37500 Ns)

Question 15.
Lubricants are used between the two parts of a machine. Why?
Answer:
To reduce friction and so to reduce wear and tear.

Question 16.
What provides the centripetal force to a car taking a turn on a level road?
Answer:
Force of friction between the type and road provides centripetal force.

Question 17.
A body is acted upon by a number of external forces. Can it remain at rest?
Answer:
Yes, if the external forces acting on the body can be represented in magnitude and direction by the sides of a closed polygon taken in the same order.

Question 18.
Bodies of larger mass need greater initial effort to put them in motion. Why?
Answer:
As F = ma so for given a, more force will be required to put a large mass in motion.

Question 19.
An athlete runs a certain distance before taking a long jump Why?
Answer:
So that inertia of motion may help him in his muscular efforts to take a longer jump.

Question 20.
Action and reaction forces do not balance each other. Why?
Answer:
As they acts on different bodies.

Question 21.
The wheels of vehicles are provided with mudguards. Why?
Answer:
When the wheel rotates at a high speed, the mud sticking to the wheel flies off tangentially, this is due to inertia of direction. If order that the flying mud does not spoil the clothes of passer by the wheels are provided with mudguards.

Question 22.
China wares are wrapped in straw paper before packing. Why?
Answer:
The straw paper between the China ware increases the Time of experiencing the jerk during transportation. Hence impact of force reduces on China wares.

Question 23.
Why is it difficult to walk on a sand?
Answer:
Less reaction force.

Question 24.
The outer edge of a curved road is generally raised over the inner edge Why?
Answer:
In addition to the frictional force, a component of reaction force also provides centripetal force.

Question 25.
Explain why the water doesn’t fall even at the top of the circle when the bucket full of water is upside down rotating in a vertical circle?
Answer:
Weight of the water and bucket is used up in providing the necessary centripetal force at the top of the circle.

Question 26.
Why does a speedy motor cyclist bends towards the center of a circular path while taking a turn on it?
Answer:
So that in addition of the frictional force, the horizontal component of the normal reaction also provides the necessary centripetal forces.

Question 27.
An impulse is applied to a moving object with a force at an angle of 20° wr.t. velocity vector, what is the angle between the impulse vector and change in momentum vector ?
Answer:
Impulse and change in momentum are along the same direction. Therefore angle between these two vectors is zero.

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (2 Marks)

Question 28.
A man getting out of a moving bus runs in the same direction for a certain distance. Comment.
Answer:
Due to inertia of motion.

Question 29.
If the net force acting upon the particle is zero, show that its linear momentum remains constant.
Answer:
As F x \(\frac {dp}{dt}\)
when F = 0, \(\frac {dp}{dt}\) = 0 so P = constant

Question 30.
A force of 36 dynes is inclined to the horizontal at an angle of 60°. Find the acceleration in a mass of 18 g that moves in a horizontal direction.
Answer:
F = 36 dyne at an angle of 60°
F_x = F cos 60° = 18 dyne
F_x = ma_x
So a_x = \(\frac{F_{x}}{m}\) = 1 cm /s²

Question 31.
The motion of a particle of mass m is described by h = ut + \(\frac {1}{2}\) gt². Find the force acting on particle.
Answer:
a = ut + \(\frac {1}{2}\) gt²
find a by differentiating h twice w.r.t.
a = g
As F = ma so F = mg (answer)

Question 32.
A particle of mass 0.3 kg is subjected to a force of F = -kx with k= 15 Nmr^-1. What will be its initial acceleration if it is released from a point 20 cm away from the origin?
Answer:
As F = ma so F = -kx = ma
a = \(\frac {-kx}{m}\)
for x = 20 cm, ⇒ a = -10 m/s².

Question 33.
A 50 g bullet is fired from a 10 kg gun with a speed of 500 ms-¹. What is the speed of the recoil of the gun?
Answer:
Initial momentum = 0
Using conservation of linear momentum mv + MV = 0
V = \(\frac {-mv}{M}\) ⇒ V = 2.5 m/s

Question 34.
Smooth block is released at rest on a 45° incline and then slides a distance d. If the time taken of slide on rough incline is n times as large as that to slide than on a smooth incline. Show that coefficient of friction, µ = \(\left(1-\frac{1}{n^{2}}\right)\)
Answer:
When there is no friction, the block slides down the inclined plane with acceleration. a = g sin θ
when there is friction, the downward acceleration of the block is a’ = g (sin θ – µ cos θ)
As the block Slides a distance d in each case so
d = \(\frac {1}{2}\) at² = \(\frac {1}{2}\) a’t’²
\(\frac{a}{a^{\prime}}=\frac{t^{\prime 2}}{t^{2}}=\frac{(n t)^{2}}{t^{2}}\) = n²
or \(\frac {g sin θ}{g(sin θ – µ cos θ)}\) = n²
Solving, we get (Using θ = 45°)
µ = 1 – \(\frac{1}{n^{2}}\)

Question 35.
A spring balance is attached to the ceiling of a lift. When the lift is at rest spring balance reads 49 N of a body hang on it. If the lift moves:

1. Downward
2. upward, with an acceleration of 5 ms²
3. with a constant velocity.

What will be the reading of the balance in each case?
Answer:
1.  R = m(g – a) = 49 N
so = m = \(\frac {49}{9.8}\) = 5 kg
R = 5 (9.8 – 5)
R = 24 N

2. R = m(g + a)
R = 5 (9.8 + 5)
R = 74 N

3.  as a = 0 so R = mg = 49 N

Question 36.
A bob of mass 0.1 kg hung from the ceiling of room by a string 2 m long is oscillating. At its mean position the speed of a bob is 1 ms^-1. What is the trajectory of the ‘oscillating bob if the string is cut when the bob is –

1. At the mean position
2. At its extreme position.

Answer:

1. Parabolic
2. vertically downwards

Question 37.
A block placed on a rough horizontal surface is pulled by a horizontal force F. Let f be the force applied by the rough surface on the block. Plot a graph of f versus F.
Answer:

Unto point A, f = F (50 Long as block is stationary) beyond A, when F increases, block starts moving f remains constant.

Question 38.
A mass of 2 kg is suspended with thread AB. Thread CD of the same type is attached to the other end of 2 kg mass.

• Lower end of the lower thread is pulled gradually, hander and hander is the downward direction so as to apply force on AB Which of the thread will break & why?
• If the lower thread is pulled with a jerk, what happens?

Answer:

• Thread AB breaks down
• CD will break.

Question 39.
A block of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is p and the acceleration due to gravity is g, calculate the minimum force required to be applied by the finger to held the block against the wall?
Answer:

For the block not to fall f = Mg
But f = µR = µF so
µF = Mg
F = \(\frac {Mg}{µ}\)

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (3 Marks) & Numericals

Question 40.
A block of mass 500 g is at rest on a horizontal table. What steady force is required to give the block a velocity of 200 cm s^-2 in 4 s?
Answer:
Use F – ma
a = \(\frac {v – u}{ t }\) = \(\frac {200- 0}{ 4 }\) = 50 cm/s²
F = 500 x 50 = 25,000 dyne.

Question 41.
A force of 98 N is just required to move a mass of 45 kg on a rough horizontal surface. Find the coefficient of friction and angle of friction?
Answer:
F = 48 N,R = 45 x 9.8 = 441 N
µ = \(\frac {F’}{ R}\) = 0.22
Angle of friction θ = tan^-1 0.22 = 12°24′

Question 42.
Calculate the force required to move a train of 2000 quintal up on an incline plane of 1 in 50 with an acceleration of 2 ms^-2. The force of friction per quintal is 0.5 N.
Answer:
Force of friction = 0.5 N per quintal
f = 0.5 x 2000 = 1000 N
m = 2000 quintals = 2000 x 100 kg
sin θ = \(\frac {1}{50}\), a – 2 m/s²
In moving up an inclined plane, force required against gravity
mg sin θ = 39200 N
And force required to produce acceleration = ma
= 2000 x 100 x 2 = 40,0000 N
Total force required = 1000 + 39,200 + 40,0000 = 440200 N.

Question 43.
A force of 100 N gives a mass m₁, an acceleration of 10 ms^-2 and of 20 ms^-2 to a mass m₂.
What acceleration must be given to it if both the masses are tied together?
Answer:
Suppose, a = acceleration produced if m₁ and m₂ are tied together,
F = 100 N
Let a₁ and a₂ be the acceleration produced in m₁ and m₂ respectively.
∴ a₁ and a₂ = 20ms^-2 (given)
Again m₁ = \(\frac{\mathrm{F}}{a_{1}}\) and m₂ = \(\frac{\mathrm{F}}{a_{2}}\)
⇒ m₁ = \(\frac {100}{10}\) = 10kg
and m₂ = \(\frac {100}{20}\) = 5kg
∴ m₁ + m₂ = 10 + 5 = 15
so, a = \(\frac{\mathbf{F}}{m_{1}+m_{2}}\) = \(\frac {100}{15}\) = \(\frac {20}{3}\) = 6.67 ms²

Question 44.
The pulley arrangement of figure are identical. The mass of the rope is negligible. In (a) mass m is lifted up by attaching a mass (2m) to the other end of the rope. In (b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2 mg. In which case, the acceleration of m is more?

Answer:
Case (a):
a = \(\frac {2m – m}{2m + m}\) g = a = \(\frac {g}{3}\)
Case (b):
FBD of mass m
ma’ = T – mg
ma’ = 2 mg – mg
⇒ ma’ = mg
a’ = g
So in case (b) acceleration of m is more.

Question 45.
Figure shows the position-time graph of a particle of mass 4 kg. What is the

(a) Force on the particle for t < 0, t > 4s, 0 < t < 4s?
(b) Impulse at t = 0 and t = 4s?
(Consider one dimensional motion only)
Answer:
(a) For t < 0. No force as Particles is at rest. For t > 4s, No force again particle comes at rest.
For 0 < t < 4s, as slope of OA is constant so velocity constant i.e., a = 0, so force must be zero.

(b) Impulse at t = 0
Impulse = change in momentum
I = m(v – w) = 4(0 – 0.75) = 3 kg ms^-1
Impulse at t = 4s
1 = m(v – u) = 4 (0 – 0.75) = -3 kg ms^-1

Question 46.
What is the acceleration of the block and trolley system as the figure, if the coefficient of kinetic friction between the trolley and the surface is 0.04? Also Calculate friction in the string: Take g = 10 m/s², mass of the string is negligible.

Answer:
Free body diagram of the block
30 – T = 3a
Free body diagram of the trolley

T – f_k = 20 a ………….(2)
where f_k = µ_k= 0.04 x 20 x 10 = 8 N
Solving (i) & (ii), a = 0.96 m/s² and T = 27.2 N

Question 47.
Three blocks of masses ml = 10 kg, m² = 20 kg are connected by strings on smooth horizontal surface and pulled by a force of 60 N. Find the acceleration of the system and frictions in the string.

Solution:
All the blocks more with common acceleration a under the force F = 60 N.
F = (m₁ + m₂ + m₃)a
a = \(\frac{\mathrm{F}}{\left(m_{1}+m_{2}+m_{3}\right)}\) = 1 m/s²
to determine, T₁ →Free body diagram of m₁.

T₁ = m₁a = 10 x 1 = 10 N
to determine, T₂ →Free body diagram of m₃

F – T₂ = m₃a
Solving, we get T₂ = 30 N

Question 48.
The rear side of a truck is open and a box of 40 kg mass is placed 5m away from the open end. The coefficient of friction between the box and the surface below it is 0.15 on a straight road, the truck starts from rest and accelerates with 2 m/s². At what distance from the starting point does the box fall off the truck ? (ignore the size of the box)
Answer:
Force on the box due to accelerated motion of the truck
F = ma = 40 x 2 = 80 N (in forward direction)
Reaction on the box, F’ = F = 80 N (in backward direction)
Force of limiting friction, f = µR = 0 .15 x 40 x 10 = 60 N
Net force on the box in backward direction is P = F’ f = 80 – 60 = 20 N
Backward acceleration in the box = a= \(\frac {p}{m}\) = \(\frac {20}{40}\) = 0.5 ms^-2
t = time taken by the box to travel s = 5 m and falls off the truck, then from
s = ut + \(\frac {1}{2}\) at²
5 = 0 x t + \(\frac {1}{2}\) x 0.5 x t²
t = 4.47
If the truck travels a distance x during this time
then x = 0 x 4.34 +\(\frac {1}{2}\) x 2 x (4.471)²
x = 19.98 m

Question 49.
A block slides down as incline of 30° with the horizontal. Starting from rest, it covers 8 m in the first 2 seconds. Find the coefficient of static friction.
Use s = ut + \(\frac {1}{2}\) at²
a = \(\frac{2 s}{t^{2}}\) at² as u = 0
µ = \(\frac{g sin θ – a}{g Cos θ }\)
Putting the value and solving, µ = 0.11

Question 50.
A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m/s² . The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the:
(a) Force on the floor of the helicopter by the crew and passenger.
(b) Action of the rotor of the helicopter on the surrounding air
(c) Force on the helicopter due to the surrounding air (g = 10 m/s² )
Answer:
(a) Force on the floor of the helicopter by the crew and passengers
= apparent weight of crew and passengers
= 500(10+ 15)
=12500 N

(b) Action of rotor of helicopter on surrounding air is Obviously vertically downwards, because helicopter rises on account of reaction of this force. Thus force of action
= (2000 + 500) (10 + 15)
= 2500 x 25
= 62,500 N

(c) Force on the helicopter due to surrounding air is obviously a reaction. As action and reaction are equal and opposite, therefore
Force of reaction F’ = 62,500 vertically upwards.

Question 51.
A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is (µ). Let the mass of the box be m.

1.  At what angle of inclination θ of the plane to the horizontal will the box just start to slide down the plane ?
2. What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to a > θ.
3. What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed ?
4. What is the force needed to be applied upwards along the plane to pk kg f make the box move up the plane with acceleration a ?

Answer:
1. When the box just starts sliding
µ = tanθ
or 0 = tan^-1 µ

2. Force acting on the box down the plane
= mg (sin a – µ cos a)

3. Force needed mg (sin a + µ cos a)

4. Force needed = mg (sin a + µ cos a) + ma.

Question 52.
Two masses of 5 kg and 3 kg are suspended with help of mass less in extensible string as shown. Calculate T₁ and T₂ when system is going upwards with acceleration m/s². (Use g 9.8 m/s²)
Answer:
According Newton’s second law of motion
(1) T₁ – (m₁ + m₂)g = (m₁ + m₂)a
T₁ = (m₁ + m₂)(a + g) = (5 + 3) (2 + 9.8)
T₁ = 94.4 N

(2) T₂ – m₂g = m₂a
T₂ = m₂ (a + g)
T₂ = 3(2 + 9.8)
T₂ = 35.4 N

Question 53.
There are few forces acting at a Point P produced by strings as shown, which is at rest. Find the forces F₁ & F₁

Answer:
Using Resolution of forces IN and 2N and then applying laws of vector addition. Calculate for F₁ & F₁.
F₁ = \(\frac{1}{\sqrt{2}}\) N, F₂ = \(\frac{3}{\sqrt{2}}\)N

Question 54.
A hunter has a machine gun that can fire 50g bullets with a velocity of 150 ms A 60 kg tiger springs at him with a velocity of 10 ms^-1. How many bullets must the hunter fire into the target so as to stop him in his track?
Answer:
Given m = mass of bullet = 50 gm = 0.50 kg
M = mass of tiger = 60 kg
v = Velocity of bullet – 150 m/s
V = Velocity of tiger = – 10 m/s
(v It is coming from opposite direction n = no. of bullets fired per second at the tiger so as to stop it.)
P_i = 0, before firing ……..(i)
P_f = n (mv) + MV …………(ii)
∴ From the law of conservation of momentum,
P_i = P_f
⇒ 0 = n (mv) + MV
n = \(\frac{MV}{mv}\) = \(\frac{-60 \times(-10)}{0.05 \times 150}\) = 80

Question 55.
Two blocks of mass 2 kg and 5 kg are connected by an ideal string passing over a pulley. The block of mass 2 kg is free to slide on a surface inclined at an angle of 30° with the horizontal whereas 5 kg block hangs freely. Find the acceleration of the system and the tension in the string.

Let a be the acceleration of the system and T be the Tension in the string. Equations of motions for 5 kg and 2 kg blocks are
5g – T = 5a
T – 2g sin θ – f = 2a
where f = force of limiting friction
= µR = µ mg cos θ = 0.3 x 2 g x cos 30°
Solving (1) & (2)
a = 4.87 m/s²

You can Download Samacheer Kalvi 11th 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 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 1.
Construct a quadratic equation with roots 7 and -3.
Solution:
Let the given roots be α = 7 and β = -3
Sum of the roots α + β = 7 + (-3)
α + β = 7 – 3 = 4
Product of the roots αβ = (7)(-3)
αβ = -21
The required quadratic equation is
x² – (sum of two roots) x + Product of the roots = 0
x² – 4x – 21 = 0

Question 2.
A quadratic polynomial has one of its zeros 1 + \(\sqrt{5}\) and it satisfies p(1) = 2. Find the quadratic polynomial.
Solution:
Given α = 1 + \(\sqrt{5}\) So, β = 1 – \(\sqrt{5}\)

The quadratic polynomial is
p(x) = x² – (α + β)x + αβ
p(x) = k (x² – 2x – 4)
p( 1) = k(1 – 2 – 4) = -5 k
Given p (1) = 2

The Polynomial Roots Calculator will find the roots of any polynomial with just one click.

Question 3.
If α and β are the roots of the quadratic equation x² + \(\sqrt{2}\)x + 3 = 0, form a quadratic polynomial with zeroes 1/α, 1/β.
Solution:
α and β are the roots of the equation x² + \(\sqrt{2}\)x + 3 = 0

Question 4.
If one root of k(x – 1)² = 5x – 7 is double the other root, show that k = 2 or – 25.
Solution:
k(x – 1)² = 5x – 7
(i.e.,) k(x² – 2x + 1) – 5x + 7 = 0
x² (k) + x(-2k – 5) + k + 1 = 0
kx² – x(2k + 5) + (k + 7) = 0
Here it is given that one root is double the other.
So let the roots to α and 2α

2(4k² + 25 + 20k) = 9k (k + 7)
2(4k² + 25 + 20k) = 9k² + 63k
8k² + 50 + 40k – 9k² – 63k = 0
-k² – 23k + 50 = 0
k² + 23k – 5o = 0
(k + 25)(k – 2) = 0
k = -25 or 2

Question 5.
If the difference of the roots of the equation 2x² – (a + 1)x + a – 1 = 0 is equal to their product then prove that a = 2.
Solution:

Question 6.
Find the condition that one of the roots of ax² + bx + c may be
(i) negative of the other
(ii) thrice the other
(iii) reciprocal of the other.
Solution:
(i) Let the roots be α and -β
Sum of the roots = – b/a = 0 ⇒ b = 0

(ii) Let the roots be α, 3α

Question 7.
If the equations x² – ax + b = 0 and x² – ex + f = 0 have one root in common and if the second equation has equal roots that ae = 2(b + f).
Solution:

Question 8.
Discuss the nature of roots of
(i) -x² + 3x + 1 = 0
(ii) 4x² – x – 2 = 0
(iii) 9x² + 5x = 0
Solution:
(i) -x² + 3x + 1 = 0
x² – 3x – 1 = 0 ———- (1)
Compare this equation with the equation
ax² + bx + c = 0 ——– (2)
we have a = 1, b = -3, c = -1
Discriminant = b² – 4ac
b² – 4ac = (-3)² – 4 × 1 × – 1
= 9 + 4 =13
b² – 4ac = 13 > 0
∴ The two roots are real and distinct.

(ii) 4x² – x – 2 = 0
4x² – x – 2 = 0 ——(3)
Compare this equation with the equation
ax² + bx + c = 0 (4)
we have a = 4 , b = – 1, c = – 2
Discriminant = b² – 4ac
b² – 4ac = (-1)² – 4 (4) (-2)
= 1 + 32
= 33
b² – 4ac = 33 >0
∴ The two roots are real and distinct.

(iii) 9x² + 5x = 0
9x² + 5x = 0 ——- (5)
Compare this equation with the equation
ax² + bx + c = 0 ——– (6)
we have a = 9, b = 5 , c = 0
Discriminant = b² – 4ac
b² – 4ac = 5² – 4 × 9 × 0
b² – 4ac = 25 > 0
∴ The two roots are real and distinct.

Question 9.
Without sketching the graphs find whether the graphs of the following functions will intersect the x- axis and if so in how many points.
(i) y = x² + x + 2
(ii) y = x² – 3x – 1
(iii) y = x² + 6x + 9
Solution:

Completing the Square Calculator is a free online tool that displays the variable value for the quadratic equation using completing the square method.

Question 10.
Write f(x) = x² + 5x + 4 in completed square form.
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 Additional Questions

Question 1.
Find the values of k so that the equation x² = 2x (1 + 3k) + 7(3 + 2k) = 0 has real and equal roots.
Solution:
The equation is x² – x(2) (1 + 3k) + 7 (3 + 2k) = 0
The roots are real and equal
⇒ ∆ = 0 (i.e.,) b² – 4ac = 0
Here a = 1, b = -2 (1 + 3k), c = 7(3 + 2k)
So b² – 4ac = 0
⇒ [-2 (1 + 3k)]² – 4(1) (7) (3 + 2k) = 0
(i.e.,) 4 (1 + 3k)² – 28 (3 + 2k) = 0
(÷ by 4) (1 + 3k)² – 7(3 + 2k) = 0
1 + 9k² + 6k – 21 – 14k = 0
9k² – 8k – 20 = 0
(k – 2)(9k + 10) = 0

To solve the quadratic inequalities ax² + bx + c < 0 (or) ax² + bx + c > 0

Question 2.
If the sum and product of the roots of the quadratic equation ax² – 5x + c = 0 are both equal to 10 then find the values of a and c.
Solution:
The given equation is ax² – 5x + c = 0
Let the roots be α and β Given α + β = 10 and αβ = 10

Question 3.
If α and β are the roots of the equation 3x² – 4x + 1 = 0, form the equation whose roots are \(\frac{\alpha^{2}}{\beta}\) and \(\frac{\beta^{2}}{\alpha}\)
Solution:

Question 4.
If one root of the equation 3x² + kx – 81 = 0 is the square of the other then find k.
Solution:

Question 5.
If one root of the equation 2x² – ax + 64 = 0 is twice that of the other then find the value of a.
Solution:

You can Download Samacheer Kalvi 11th 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 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Get the free “Partial Fraction Calculator” widget for your website, blog, WordPress, Blogger, or iGoogle.

Resolve the following rational expressions into partial fractions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Equating nuemarator on bothsides we get

Question 6.

Solution:

Equating numerator on both sides
(x – 2)² = A(x² + 1) + (Bx + c)(x)
Put x = 0
1 = A
Equating co-eff of x²
1 = A + B
(i.e.,) 1 + B = 1 ⇒ B = 0
put x = 1
A(2) + B + C = 0 (i.e.,) 2A + B + C = 0
2 + 0 + C = 0 ⇒ C = -2

Question 7.

Solution:
Since numerator and denominator are of same degree
we have divide the numerator by the denominator

Substituting the value in ….(1)

Question 8.

Solution:
Numerator is of greater degree than the denominator
So dividing Numerator by the denominator

⇒ 21x + 31 = A(x + 3) + B(x + 2)
Put x = -3
-63 + 31 = B(-1)
B = 32
Put x = -2
-42 + 31 = A(1) + B(0)
A = -11

Question 9.

Solution:

Question 10.

Solution:

Equating Numerator on both sides we get
6x² – x + 1 = A(x² + 1) + (Bx + c)(x + 1)
6 + 1 + 1 = A(2) + 0 ⇒ 2A = 8 ⇒ A = 4
Equating co-eff of x²
6 = A + B
(i.e.,) 4 + B = 6 ⇒ B = 6 – 4 = 2
put x = 0
1 = A+ C
4 + C = 1 ⇒ C = 1 – 4 = -3

Question 11.

Solution:
Since Numerator and are of same degree divide Numerator by the denominator

equating Numerator on both sides we get
x – 5 = A(x + 3) + B(x – 1)
Put x = -3
-3 -5 = A(0) + B(-4)
-4B = -8 ⇒ B = 2
Put x = 1
1 – 5 = A(4) + B(0)
4A = -4 ⇒ A = -1

Question 12.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 Additional Questions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Equating nemerator on b/s
9 = A(x+2)² + B(x – 1)(x + 2) + C(x – 1)
Put x = -2
9 = A(0) + B(0) + C(-3)
-3C = 9 ⇒ C = -3
Put x = 1
9 = A (1 + 2)² + B (0) + C(0)
9A = 9
A = 1
Put x = 0
9 = 4A – 2B – C
9 = 4(1) – 2B + 3
9 – 7 = -2B
2 = -2B
B = -1

Question 4.

Solution:

Question 5.

Solution:

0 = 0 + B(1 + 2)
3B = 0 ⇒ B = 0
Put x = -2
(-2)³ – 1 = A(-2 – 1) + B(0)
-8 – 1 = -3A
-9 = -3A
A = 9/3 ⇒ A = 3

Students can Download Physics Chapter 5 Motion of System of Particles and Rigid Bodies Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 5 Motion of System of Particles and Rigid Bodies

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Textual Questions Solved

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Multiple Choice Questions

Question 1.
The center of mass of a system of particles does not depend upon,
(a) position of particles
(b) relative distance between particles
(c) masses of particles
(d) force acting on particle
Answer:
(d) force acting on particle

Question 2.
A couple produces, [AIPMT 1997, AIEEE 2004]
(a) pure rotation
(b) pure translation
(c) rotation and translation
(d) no motion [AIPMT 1997]
Answer:
(a) pure rotation

Question 3.
A particle is moving with a constant velocity along a line parallel to positive X – axis. The magnitude of its angular momentum with respect to the origin is –
(a) zero
(b) increasing with x
(c) decreasing with x
(d) remaining constant [IIT 2002]
Answer:
(d) remaining constant

Question 4.
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force 30 N?
(a) 0.25 rad s^-2
(b) 25 rad s^-2
(c) 5 m s^-2
(d) 25 m s^-2
[NEET 2017]
Answer:
(b) 25 rad s^-2

Question 5.
A closed cylindrical container is partially filled with water. As the container rotates in a horizontal plane about a perpendicular bisector, its moment of inertia,
(a) increases
(b) decreases
(c) remains constant
(d) depends on direction of rotation. [IIT 1998]
Answer:
(a) increases

Use this kinetic energy calculator to help you find out the energy of an object in motion. This KE calculator makes use of the kinetic energy formula

Question 6.
A rigid body rotates with an angular momentum L. If its kinetic energy is halved, the angular momentum becomes,
(a) L
(b) L / 2
(c) 2 L
(d) L / 2 [AFMC 1998, AIPMT 2015]
Answer:
(d) L / 2

Question 7.
A particle undergoes uniform circular motion. The angular momentum of the particle remain conserved about –
(a) the center point of the circle.
(b) the point on the circumference of the circle
(c) any point inside the circle.
(d) any point outside the circle. [IIT 2003]
Answer:
(a) the center point of the circle.

Question 8.
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along –
(a) a line perpendicular to the plane of rotation
(b) the line making an angle of 45° to the plane of rotation
(c) the radius
(d) tangent to the path [AIPMT 2012]
Answer:
(a) a line perpendicular to the plane of rotation

Question 9.
Two discs of same moment of inertia rotating about their regular axis passing through center and perpendicular to the plane of disc with angular velocities ω₁ and ω₁. They are brought in to contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is-
(a) \(\frac {1}{4}\) I(ω₁ – ω₂)ω²
(b) I(ω₁ – ω₂)ω₂
(c) \(\frac {1}{8}\) I(ω₁ – ω₂)ω²
(d) \(\frac {1}{2}\) I(ω₁ – ω₂)ω²
Answer:
(a) \(\frac {1}{4}\) I(ω₁ – ω₂)ω²

Question 10.
A disc of moment of inertia I_a is rotating in a horizontal plane about its symmetry axis with a constant angular speed to. Another disc initially at rest of moment of inertia I_b is dropped coaxially on to the rotating disc. Then, both the discs rotate with same constant angular speed. The loss of kinetic energy due to friction in this process is-

Answer:

Question 11.
The ratio of the acceleration for a solid sphere (mass m and radius R) rolling down an incline of angle 0 without slipping and slipping down the incline without rolling is –
(a) 5 : 7
(b) 2 : 3
(c) 2 : 5
(d) 7 : 5
[AIPMT 2014]
Answer:
(a) 5 : 7

Question 12.
From a disc of radius R a mass M, a circular hole of diameter R, whose rim passes through the center is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis passing through it?
(a) 15MR²/32
(b) 13MR²/32
(c) 11MR²/32
(d) 9MR²/32 [NEET 2016]
Answer:
(b) 13MR²/32

Question 13.
The speed of a solid sphere after rolling down from rest without sliding on an inclined plane of vertical height h is,
(a) \(\sqrt{\frac{4}{3} g h}\)
(b) \(\sqrt{\frac{10}{7} g h}\)
(c) \(\sqrt{2gh}\)
(d) \(\sqrt{\frac{1}{2} g h}\)
Answer:
(a) \(\sqrt{\frac{4}{3} g h}\)

Question 14.
The speed of the center of a wheel rolling on a horizontal surface is vQ. A point on the rim in level with the center will be moving at a speed of speed of,
(a) zero
(b) v₀
(c) \(\sqrt{2}\)v₀
(d) 2 v₀
[PMT 1992, PMT 2003, IIT 2004]
Answer:
(c) \(\sqrt{2}\)v₀

Question 15.
A round object of mass m and radius r rolls down without slipping along an inclined plane. The fractional force,
(a) dissipates kinetic energy as heat.
(b) decreases the rotational motion.
(c) decreases the rotational and transnational motion ,
(d) converts transnational energy into rotational energy [PMT 2005]
Answer:
(d) converts transnational energy into rotational energy

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions

Question 1.
Define center of mass.
Answer:
The center of mass of a body is defined as a point where the entire mass of the body appears to be concentrated.

Question 2.
Find out the center of mass for the given geometrical structures.
(a) Equilateral triangle
(b) Cylinder
(c) Square
Answer:

(a) For equilateral triangle, center of mass lies at its centro-id.
(b) For cylinder, center of mass lies at its geometrical center.
(c) For square, center of mass lies at the point where the diagonals meet.

Question 3.
Define torque and mention its unit.
Answer:
Torque is defined as the moment of the external applied force about a point or axis of rotation. The expression for torque is,
\(\vec{\tau}\) = \(\vec{r}\) x \(\vec{F}\)

Question 4.
What are the conditions in which force cannot produce torque?
Answer:
The forces intersect (or) passing through the axis of rotation cannot produce torque as the perpendicular distance between the forces is 0 i.e. r = 0.
∴ \(\vec{\tau}\) = \(\vec{r}\) x \(\vec{F}\) = 0

Question 5.
Give any two examples of torque in day – to – day life.
Answer:

• The opening and closing of a door about the hinges.
• Turning of a nut using a wrench.

Question 6.
What is the relation between torque and angular momentum?
Answer:
We have the expression for magnitude of angular momentum of a rigid body as, L = I ω. The expression for magnitude of torque on a rigid body is, τ = I α.
We can further write the expression for torque as,
τ = I\(\frac {dω}{dt}\) (∴ α = \(\frac {dω}{dt}\))
Where, ω is angular velocity and α is angular acceleration. We can also write equation,
τ = \(\frac {d(Iω)}{dt}\)
τ = \(\frac {dL}{dt}\)

Question 7.
What is equilibrium?
Answer:
A rigid body is said to be in mechanical equilibrium where both its linear momentum and angular momentum remain constant.

Question 8.
How do you distinguish between stable and unstable equilibrium?
Answer:
Stable Kquilibrium:

• The body tries to come back to equilibrium if slightly disturbed and released.
• The center of mass of the body shifts slightly higher if disturbed from equilibrium.
• Potential energy of the body is minimum and it increases if disturbed.

Unstable Equilibrium:

• The body cannot come back to equilibrium if slightly disturbed and released.
• The center of mass of the body shifts slightly lower if disturbed from equilibrium.
• Potential energy of the body is not minimum and it decreases if disturbed.

Question 9.
Define couple.
Answer:
A pair of forces which are equal in magnitude but opposite in direction and separated by a perpendicular distance so that their lines of action do not coincide that causes a turning effect is called a couple.

Question 10.
State principle of moments.
Answer:
Principle of moment states that when an object is in equilibrium the sum of the anticlockwise moments about a point is equal to the sum of the clockwise moments.

Question 11.
Define center of gravity.
Answer:
The center of gravity of a body is the point at which the entire weight of the body acts, irrespective of the position and orientation of the body.

Question 12.
Mention any two physical significance of moment of inertia
Answer:
Moment of inertia for point mass,
I = \(m_{i} r_{i}^{2}\)
Moment of inertia for bulk object,
I = ∑\(m_{i} r_{i}^{2}\)

Question 13.
What is radius of gyration?
Answer:
The radius of gyration of an object is the perpendicular distance from the axis of rotation to an equivalent point mass, which would have the same mass as well as the same moment of inertia of the object.

Question 14.
State conservation of angular momentum.
Answer:
The law of conservation of angular momentum states that when no external torque acts on the body the net angular momentum of a rotating rigid body remains constant.

Question 15.
What are the rotational equivalents for the physical quantities, (i) mass and (ii) force?
Answer:
The rotational equivalents for (i) mass and (ii) force are moment of inertia and torque respectively.

Question 16.
What is the condition for pure rolling?
Answer:
In pure rolling, there is no relative motion of the point of contact with the surface when the rolling object speeds up or shows down. It must accelerate or decelerate respectively.

Question 17.
What is the difference between sliding and slipping?
Sliding:

• Velocity of center of mass is greater than Rω i.e. V_CM > Rω.
• Velocity of transnational motion is greater than velocity of rotational motion.
• Resultant velocity acts in the forward direction.

Slipping:

• Velocity of center of mass is lesser than Rω. i.e. V_CM < Rω
• Velocity of translation motion is lesser than velocity of rotational motion.
• Resultant velocity acts in the backward direction.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Long Answer Questions

Question 1.
Explain the types of equilibrium with suitable examples.
Answer:

• Transnational motion – A book resting on a table.
• Rotational equilibrium – A body moves in a circular path with constant velocity.
• Static equilibrium – A wall – hanging, hanging on the wall.
• Dynamic equilibrium – A ball decends down in a fluid with its terminal velocity.
• Stable equilibrium – A table on the floor
• Unstable equilibrium – A pencil standing on its tip.
• Neutral equilibrium – A dice rolling on a game board.

Question 2.
Explain the method to find the center of gravity of a irregularly shaped lamina.
Answer:
There is also another way to determine the center of gravity of an irregular lamina. If we suspend the lamina from different points like P, Q, R as shown in figure, the vertical lines I PP’, QQ’, RR’ all pass through the center of gravity. Here, reaction force acting at the point of suspension and the gravitational force acting at the center of gravity cancel each other and the torques caused by them also cancel each other.

Determination of center of gravity of plane lamina by suspending

Question 3.
Explain why a cyclist bends while negotiating a curve road? Arrive at the expression for angle of bending for a given velocity.
Answer:
Let us consider a cyclist negotiating a circular level road (not banked) of radius r with a speed v. The cycle and the cyclist are considered as one system with mass m. The center gravity of the system is C and it goes in a circle of radius r with center at O. Let us choose the line OC as X – axis and the vertical line through O as Z – axis as shown in Figure.

The system as a frame is rotating about Z – axis. The system is at rest in this rotating frame. To solve problems in rotating frame of reference, we have to apply a centrifugal force (pseudo force) on the system which will be \(\frac{m v^{2}}{r}\) This force will act through the center of gravity. The forces acting on the system are,

• gravitational force (mg)
• normal force (N)
• frictional force (f)
• centrifugal force (\(\frac{m v^{2}}{r}\)).

As the system is in equilibrium in the rotational frame of reference, the net external force and net external torque must be zero. Let us consider all torques about the point A in Figure.
For rotational equilibrium,
τ_net = 0
The torque due to the gravitational force about point A is (mg AB) which causes a clockwise turn that is taken as negative. The torque due to the centripetal force is I BC which causes an (\(\frac{m v^{2}}{r}\) BC) Which causes an anticlockwise turn that is taken as positive.


While negotiating a circular level road of radius r at velocity v, a cyclist has to bend by an angle 0 from vertical given by the above expression to stay in equilibrium (i.e. to avoid a fall).

Question 4.
Derive the expression for moment of inertia of a rod about its center and perpendicular to the rod.
Answer:
Let us consider a uniform rod of mass (M) and length (l) as shown in figure. Let us find an expression for moment of inertia of this rod about an axis that passes through the center of mass and perpendicular to the rod. First an origin is to be fixed for the coordinate system so that it coincides with the center of mass, which is also the geometric center of the rod. The rod is now along the x axis. We take an infinitesimally small mass (dm) at a distance (x) from the origin. The moment of inertia (dI) of this mass (dm) about the axis is, dI = (dm) x²

As the mass is uniformly distributed, the mass per unit length (λ) of the rod is, λ = \(\frac {M}{l}\)
The (dm) mass of the infinitesimally small length as, dm = λ dx = \(\frac {M}{l}\) dx
The moment of inertia (I) of the entire rod can be found by integrating dl,

As the mass is distributed on either side of the origin, the limits for integration are taken from to – l/2 to l/2.

Question 5.
Derive the expression for moment of inertia of a uniform ring about an axis passing through the center and perpendicular to the plane.
Answer:
Let us consider a uniform ring of mass M and radius R. To find the moment of inertia of the ring about an axis passing through its center and perpendicular to the plane, let us take an infinitesimally small mass (dm) of length (dx) of the ring. This (dm) is located at a distance R, which is the radius of the ring from the axis as shown in figure.
The moment of inertia (dl) of this small mass (dm) is,
dI = (dm)R²
The length of the ring is its circumference (2πR). As the mass is uniformly distributed, the mass per unit length (λ) is,
λ = \(\frac {mass}{lengh}\) = \(\frac {M}{2πR}\)
The mass (dm) of the infinitesimally small length is,
dm = λ dx = \(\frac {M}{2πR}\) dx
Now, the moment of inertia (I) of the entire ring is,
To cover the entire length of the ring, the limits of integration are taken from 0 to 2πR.

Question 6.
Derive the expression for moment of inertia of a uniform disc about an axis passing through the center and perpendicular to the plane.
Answer:
Consider a disc of mass M and radius R. This disc is made up of many infinitesimally small rings as shown in figure. Consider one such ring of mass (dm) and thickness (dr) and radius (r). The moment of inertia (dl) of this small ring is,
dI = (dm)R²
As the mass is uniformly distributed, the mass per unit area (σ) is σ = \(\frac {mass}{area}\) = \(\frac{M}{\pi R^{2}}\)
The mass of the infinitesimally small ring is,
dm = σ 2πr dr = \(\frac{\mathrm{M}}{\pi \mathrm{R}^{2}}\) 2πr dr
where, the term (2πr dr) is the area of this elemental ring (2πr is the length and dr is the thickness), dm = \(\frac{2 \mathrm{M}}{\mathrm{R}^{2}}\) r dr
dI = \(\frac{2 \mathrm{M}}{\mathrm{R}^{2}}\) r³ dr
The moment of inertia (I) of the entire disc is,

Question 7.
Discuss conservation of angular momentum with example.
Answer:
When no external torque acts on the body, the net angular momentum of a rotating rigid body remains constant. This is known as law of conservation of angular momentum.
τ = \(\frac {dL}{dt}\)
If τ = 0 then, L = constant.
As the angular momentum is L = Iω, the conservation of angular momentum could further be written for initial and final situations as,
I_iω_i = I_iω_i (or) Iω = constant
The above equations say that if I increases ω will decrease and vice – versa to keep the angular momentum constant.

There are several situations where the principle of conservation of angular momentum is applicable. One striking example is an ice dancer as shown in Figure A. The dancer spins slowly when the hands are stretched out and spins faster when the hands are brought close to the body.

Stretching of hands away from body increases moment of inertia, thus the angular velocity decreases resulting in slower spin. When the hands are brought close to the body, the moment of inertia decreases, and thus the angular velocity increases resulting in faster spin. A diver while in air as in Figure B curls the body close to decrease the moment of inertia, which in turn helps to increase the number of somersaults in air.

Question 8.
State and prove parallel axis theorem.
Answer:
Parallel axis theorem:
Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.

If I_C is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is – given by the relation,
I = I_C + M d²
Let us consider a rigid body as shown in figure. Its moment of inertia about an axis AB passing through the center of mass is I_C. DE is another axis parallel to AB at a perpendicular distance d from AB. The moment of inertia of the body about DE is I. We attempt to get an expression for I in terms of I_C. For this, let us consider a point mass m on the body at position x from its center of mass.

The moment of inertia of the point mass about the axis DE is, m (x + d)². The moment of inertia I of the whole body about DE is the summation of the above expression.
I = ∑ m (x + d)²
This equation could further be written as,
I = ∑ m(x² + d² + 2xd)
1= ∑ (mx² + md² + 2 dmx)
l = ∑ mx² + md² + 2d ∑ mx
Here, ∑ mx² is the moment of inertia of the body about the center of mass. Hence,I_C = ∑ mx²
The term, ∑ mx = 0 because, x can take positive and negative values with respect to the axis AB. The summation (∑ mx) will be zero.
Thus, I = I_C + ∑ m d² = I_C + (∑m) d²
Here, ∑ m is the entire mass M of the object (∑ m = M).
I = I_C + Md²

Question 9.
State and prove perpendicular axis theorem.
Answer:
Perpendicular axis theorem:
This perpendicular axis theorem holds good only for plane laminar objects. The theorem states that the moment of inertia of a plane laminar body about an axis perpendicular to its plane is equal to the sum of moments of inertia about two perpendicular axes lying in the plane of the body such that all the three axes are mutually perpendicular and have a common point.

Let the X and Y – axes lie in the plane and Z – axis perpendicular to the plane of the laminar object. If the moments of inertia of the body about X and Y-axes are I_X and I_Y respectively – and I_Z is the moment of inertia about Z-axis, then the perpendicular axis theorem could be expressed as,
I_Z = I_X + I_Y

To prove this theorem, let us consider a plane laminar object of negligible thickness on which lies the origin (O). The X and Y – axes lie on the plane and Z – axis is perpendicular to it as shown in figure. The lamina is considered to be made up of a large number of particles of mass m. Let us choose one such particle at a point P which has coordinates (x, y) at a distance r from O.

The moment of inertia of the particle about Z – axis is, mr².
The summation of the above expression gives the moment of inertia of the entire lamina about Z – axis as, I_Z = ∑ mr²
Here, r² = x² + y²
Then, I_Z = ∑ m (x² + y²)
I_Z = ∑ m x² + ∑ m y²
In the above expression, the term ∑ m x² is the moment of inertia of the body about the Y-axis and similarly the term ∑ m y²is the moment of inertia about X- axis. Thus,
I_X = ∑ m y² and I_Y = ∑ m x²
Substituting in the equation for Iz gives,
I_Z = I_X + I_Y
Thus, the perpendicular axis theorem is proved.

Question 10.
Discuss rolling on inclined plane and arrive at the expression for the acceleration.
Answer:
Let us assume a round object of mass m and radius R is rolling down an inclined plane without slipping as shown in figure. There are two forces acting on the object along the inclined plane. One is the component of gravitational force (mg sin θ) and the other is the static frictional force (f). The other component of gravitation force (mg cos θ) is cancelled by the normal force (N) exerted by the plane. As the motion is happening along the incline, we shall write the equation for motion from the free body diagram (FBP) of the object.

For transnational motion, mg sin θ is the supporting force and f is the opposing force, mg sin θ f = ma
For rotational motion, let us take the torque with respect to the center of the object. Then mg sin 0 cannot cause torque as it passes through it but the frictional force f can set torque of Rf = Iα
By using the relation, a = rα, and moment of inertia I = mK² we get,
Rf = mK² \(\frac {a}{R}\); f = ma \(\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\)
Now equation becomes,
mg sin θ – ma \(\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\) = ma
mg sin θ = ma + ma \(\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\)
a \(\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\) = g sin θ
After rewriting it for acceleration, we get,
a = \(\frac{g \sin \theta}{\left(1+\frac{K^{2}}{R^{2}}\right)}\)
We can also find the expression for final velocity of the rolling object by using third equation of motion for the inclined plane.
v² = u² + 2as. If the body starts rolling from rest, u = 0. When h is the vertical height of the incline, the length of the incline s is, s = \(\frac {h}{sin θ}\)
By taking square root,

The time taken for rolling down the incline could also be written from first equation of motion as, v = u + at. For the object which starts rolling from rest, u = 0. Then,
t = \(\frac {v}{a}\)

The equation suggests that for a given incline, the object with the least value of radius of gyration K will reach the bottom of the incline first.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Conceptual Questions

Question 1.
When a tree is cut, the cut is made on the side facing the direction in which the tree is required to fall. Why?
Answer:
A cut on the tree is made on the side facing the direction in which the tree is required to fall because that side will no longer be supported by the normal force from the bottom, therefore the gravitational force tries to rotate it. So the torque given by the gravity to the tree makes the tree fall on the side as anticipated.

Question 2.
Why does a porter bend forward while carrying a sack of rice on his back?
Answer:
When a porter carries a sack of rice, the line of action of his center of gravity will go away from the body. It affects the balance, to avoid this he bends. By which center of gravity will realign within the body again. So balance is maintained.

Question 3.
Why is it much easier to balance a meter scale on your finger tip than balancing on a match stick?
Answer:
A meter scale is larger then a match stick. So the center of gravity for meter scale is higher than a matchstick when we keep it vertically. It is easier to balance the object whose center of gravity is higher than the object whose centro of gravity is lower. So, it is hard to balance a match stick than a meter scale.

Question 4.
Two identical water bottles one empty and the other filled with water are allowed to roll down an inclined plane. Which one of them reaches the bottom first? Explain your answer.
Answer:
Mass of the empty water bottle mostly concentrated on its surface. So moment of inertia of empty water bottle is more than the bottle filled with water. As we know, moment of inertia is inversely proportional to angular velocity. Therefore, the bottle filled with water whirls with greater speed and reaches the ground first.

Question 5.
Write the relation between angular momentum and rotational kinetic energy. Draw a graph for the same. For two objects of same angular momentum, compare the moment of inertia using the graph.
Answer:
Let a rigid body of moment of inertia I rotate with angular velocity ω.
The angular momentum of a rigid body is, L = Iω
The rotational kinetic energy of the rigid body is, KE = \(\frac { 1 }{ 2 }\) Iω².
By multiplying the numerator and denominator of the above equation with I, we get a relation between L and KE as,

It resembles with y = Kx². If angular momentum is same for two objects, kinetic energy is inversely proportional to moment of inertia.
Moment of inertia of the object whose kinetic energy is lesser will have greater magnitude.

Question 6.
Three identical solid spheres move down through three inclined planes A, B and C all same dimensions. A is without friction, B is undergoing pure rolling and C is rolling with slipping. Compare the kinetic energies E_A, E_B and E_C at the bottom.
Answer:
Even though, the three identical solid spheres of same dimensions move down through three different inclined plane, according to the law of conservation of energy, the potential energy possessed by these three solid spheres will be converted into kinetic energies. So the kinetic energies E_A, E_B and E_C are equal at the bottom.

Question 7.
Give an example to show that the following statement is false. Any two forces acting on a body can be combined into single force that would have same effect.
Answer:
A single force i.e. resultant of two forces acting on a body depends upon the angle between them also. The simple example for this is if two forces 5 N and 5 N acting on the object in the opposite direction, the single resultant force acting on the body is zero. But, if two forces acting on the object along the same direction, then the resultant i.e. the single force is 5 + 5 = 10 N. Hence the given statement “any two forces acting on a body can be combined into single force that would leave same effect” is wrong.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Numerical Problems

Question 1.
A uniform disc of mass 100 g has a diameter of 10 cm. Calculate the total energy of the disc when rolling along a horizontal table with a velocity of 20 cm s^-2.
Answer:
Given,
Mass of the disc = 100 g = 100 x 10^-3 kg = \(\frac { 1 }{ 10 }\)kg
Velocity of disc = 20 cm s^-1 = 20 x 10^-2 ms^-1 = 0.2 ms^-1

Question 2.
A particle of mass 5 units is moving with a uniform speed of v = \(3 \sqrt{2}\) units in the XOY plane along the line y = x + 4. Find the magnitude of angular momentum.
Answer:
Given,
Mass = 5 units
Speed = v = \(3 \sqrt{2}\) units
Y = X + 4
Angular momentum = L = m(\(\bar{r} \times \bar{v}\))
= m(x\(\hat{i}\) +y\(\hat{j}\))x(v\(\hat{i}\) + v\(\hat{j}\)) = m[xv\(\hat{k}\)-vy\(\hat{k}\)] = m[xv\(\hat{k}\)– v(x + 4)\(\hat{k}\)]
L = -mv\(\hat{k}\) = -4 x 5 x \(3 \sqrt{2}\)\(\hat{k}\) = – 60\(\sqrt{2}\)\(\hat{k}\)
L = 60\(\sqrt{2}\) units.

Question 3.
A fly wheel rotates with a uniform angular acceleration. If its angular velocity increases from 20π rad/s to 40π rad/s in 10 seconds, find the number of rotations in that period.
Answer:
Given,
Initial angular velocity ω₀ = 20 π rad/s
Final angular velocity ω = 40 π rad/s
Time t = 10 s
Solution:
Angular acceleration α = \(\frac{\omega-\omega_{0}}{t}\) = \(\frac {40π – 20π }{ 10 }\)
α = 2π rad/s²
According to equation of motion for rotational motion

The number of rotations = n = \(\frac {θ}{ 2π }\)
n = \(\frac {300π}{ 2π }\) = 150 rotations.

Question 4.
A uniform rod of mass m and length / makes a constant angle 0 with an axis of rotation which passes through one end of the rod. Find the moment of inertia about this gravity is.
Answer:
Moment of inertia of the rod about the axis which is passing through its center of gravity is


Moment of inertia of a uniform rod of mass m and length l about one axis which passes through one end of the rod

Question 5.
Two particles P and Q of mass 1 kg and 3 kg respectively start moving towards each other from rest under mutual attraction. What is the velocity of their center of mass?
Answer:
Given,
Mass of particle P = 1 kg Mass of particle Q = 3 kg
Solution:
Particles P and Q forms a system. Here no external force is acting on the system,

We know that M = \(\frac {d}{dt}\) (V_CM ) = f
It means that, C.M. of an isolated system remains at rest when no external force is acting and internal forces do not change its center of mass.

Question 6.
Find the moment of inertia of a hydrogen molecule about an axis passing through its center of mass and perpendicular to the inter-atomic axis.
Given: mass of hydrogen atom 1.7 x 10²⁷kg and inter atomic distance is equal to 4 x 10^-10m.
Answer:
Given,
Inter-atomic distance : 4 x 10^-10 m
Mass of H₂ atom : 1.7 x 10^-27 kg
Moment of inertia of H₂ =

Question 7.
On the edge of a wall, we build a brick tower that only holds because of the bricks’ own weight. Our goal is to build a stable tower whose overhang d is greater than the length l of a single brick. What is the minimum number of bricks you need?
(Hint: Find the center of mass for each brick and add.)
Answer:
Given:

Length of the brick = l
Length of the overhang = d
The mono of bricks can be decided only by using the concept of position of center of gravity. The first brick is in contact with the ground and it will not fall over.
Let one end of brick 2 is coinciding with the center of brick 1 i.e. x = 0.
∴ The position of n brick is
x_n = (n – 1) \(\frac {L}{4}\)
The center of gravity is in the midway between the center of brick 2 and the center of brick n.
position of G =

brick tower will fall when G >\(\frac {L}{4}\) it shows that n > 4.

Question 8.
The 747 boing plane is landing at a speed of 70 m s^-1. Before touching the ground, the wheels are not rotating. How long a skid mark do the wing wheels leave (assume their mass is 100 kg which is distributed uniformly, radius is 0.7 m, and the coefficient of friction with the ground is 0.5)?
Answer:
The types of the plane will leave a skid mark if the speed of the types in contact with ground is lesser than the velocity of the plane. The condition for this is –
v > ω
(When the type attained an angular velocity of V/R)
The types will stop the skidding and starts the rolling.
The forces acting on the wheel after the plane touches down are,
N – P Normal force W – weight
The wheel is not accelerating means
N = ω
The torque about the center of the wheel is
τ = RF = µωR
The angular acceleration is

According to equation of motion, time taken to stop the skidding by the wheel is,

The six mark will have a length of
l = vt = 70 x 0.03 = 2.1 m
Note:
The 747 is resting on the runway, supported by 16 wheels under the wing, and 2 under the nose total length is 68.63 m. The normal force experienced by plane through its 16 wheels is ω = 232 KN.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Additional Questions Solved

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Multiple Choice Questions

Question 1.
The changes produced by the deforming forces in a rigid body are –
(a) very large
(b) infinity
(c) negligibly small
(d) small
Answer:
(c) negligibly small

Question 2.
When a rigid body moves all particles that constitute the body follows-
(a) same path
(b) different paths
(c) either same or different path
(d) circular path
Answer:
(b) different path

Question 3.
For bodies of regular shape and uniform mass distribution, the center of mass is at –
(a) the comers
(b) inside the objects
(c) the point where the diagonals meet
(d) the geometric center
Answer:
(d) the geometric center

Question 4.
For square and rectangular objects center of mass lies at –
(a) the point where the diagonals meet
(b) at the comers
(c) on the center surface
(d) any point
Answer:
(a) the point where the diagonals meet

Question 5.
Center of mass may lie –
(a) within the body
(b) outside the body
(c) both (a) and (b)
(d) only at the center
Answer:
(c) both (a) and (b)

Question 6.
The dimension of point mass is –
(a) positive
(b) negative
(c) zero
(d) infinity
Answer:
(c) zero

Question 7.
The motion of center of mass of a system of two particles is unaffected by their internal forces –
(a) irrespective of the actual directions of the internal forces
(b) only if they are along the line joining the particles
(c) only if acts perpendicular to each other
(d) only if acting opposite
Answer:
(a) irrespective of the actual directions of the internal forces

Question 8.
A circular plate of diameter 10 cm is kept in contact with a square plate of side 10 cm. The density of the material and the thickness are same everywhere. The center of mass of the system will be
(a) inside the circular plate
(b) inside the square plate
(c) At the point of contact
(d) outside the system
Answer:
(6) inside the square plate

Question 9.
The center of mass of a system of particles does not depend on
(a) masses of particles
(b) position of the particles
(c) distribution of masses
(d) forces acting on the particles
Answer:
(d) forces acting on the particles

Question 10.
The center of mass of a solid cone along the line from the center of the base to the vertex is at –
(a) \(\frac { 1 }{ 2 }\) th of its height
(b) \(\frac { 1 }{ 3 }\) of its height
(c) \(\frac { 1 }{ 4 }\) th of its height
(d) \(\frac { 1 }{ 5 }\) th of its height
Answer:
(d) \(\frac { 1 }{ 5 }\) th of its height

Question 11.
All the particles of a body are situated at a distance of X from origin. The distance of the center of mass from the origin is –
(a) ≥ r
(b) ≤ r
(c) = r
(d) > r

Question
A free falling body breaks into three parts of unequal masses. The center of mass of the three parts taken together shifts horizontally towards –
(a) heavier piece
(b) lighter piece
(c) does not shift horizontally
(d) depends on vertical velocity
Answer:
(c) does not shift horizontally

Question 13.
The distance between the center of carbon and oxygen atoms in the gas molecule is 1.13 A. The center of mass of the molecule relative to oxygen atom is –
(a) 0.602 Å
(b) 0.527 Å
(c) 1.13 Å
(d) 0.565 Å
Answer:
(b) 0.527 Å
Given,
Inter atomic distance = 1.13 Å
Mass of carbon atom = 14
Mass of oxygen atom = 16
Let C.M. of molecule lies at a distance of X from oxygen atom-
i.e. m₁r₁ = m₂r₂
16 X = 14(1.13 – X)
30 X = 15.82
X = 0.527 Å

Question 14.
The unit of position vector of center of mass is-
(a) kg
(b) kg m²
(c) m
(d) m²
Answer:
(c) m

Question 15.
The sum of moments of masses of all the particles in a system about the center of mass is-
(a) minimum
(b) maximum
(c) zero
(d) infinity
Answer:
(c) zero

Question 16.
The motion of center of mass depends on-
(a) external forces acting on it
(b) internal forces acting within it
(c) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(a) external forces acting on it

Question 17.
Two particles P and Q move towards with each other from rest with the velocities of 10 ms^-1 and 20 ms^-1 under the mutual force of attraction. The velocity of center of mass is-
(a) 15 ms^-1
(b) 20 ms^-1
(c) 30 ms^-1
(d) zero
Answer:
(d) zero

Question 18.
The reduced mass of the system of two particles of masses 2 m and 4 m will be –
(a) 2 m
(b) \(\frac {2 }{ 3 }\)y m
(c) \(\frac {3}{ 2 }\)y m
(d) \(\frac { 4 }{ 3 }\)m
Answer:
(d) \(\frac { 4 }{ 3 }\)m

Question 19.
The motion of the center of mass of a system consists of many particles describes its –
(a) rotational motion
(b) vibratory motion
(c) oscillatory motion
(d) translator y motion
Answer:
(c) oscillatory motion

Question 20.
The position of center of mass can be written in the vector form as –

Answer:

Question 21.
The positions of two masses m₁ and m₂ are x₁ and x₂. The position of center of mass is –

Answer:

Question 22.
In a two particle system, one particle lies at origin another one lies at a distance of X. Then the position of center of mass of these particles of equal mass is –

Answer:
(a) \(\frac {X}{2}\)

Question 23.
Principle of moments is –

Answer:

Question 24.
Infinitesimal quantity means –
(a) collective particles
(b) extremely small
(c) nothing
(d) extremely larger
Answer:
(b) extremely small

Question 25.
In the absence of external forces the center of mass will be in a state of –
(a) rest
(b) uniform motion
(c) may be at rest or in uniform motion
(d) vibration
Answer:
(c) may be at rest or in uniform motion

Question 26.
The activity of the force to produce rotational motion in a body is called as –
(a) angular momentum
(b) torque
(c) spinning
(d) drive force
Answer:
(b) torque

Question 27.
The moment of the external applied force about a point or axis of rotation is known as –
(a) angular momentum
(b) torque
(c) spinning
(d) drive force
Answer:
(b) torque

Question 28.
Torque is given as –
(a) \(\vec{r}\) . \(\vec{F}\)
(b) \(\vec{r}\) x \(\vec{F}\)
(c) \(\vec{F}\) x \(\vec{r}\)
(d) r F cos θ
Answer:
(b) \(\vec{r}\) x \(\vec{F}\)

Question 29.
The magnitude of torque is –
(a) rF sin θ
(b) rF cos θ
(c) rF tan θ
(d) rF
Answer:
(a) rF sin θ

Question 30.
The direction of torque ácts –
(a) along \(\vec{F}\)
(b) along \(\vec{r}\) & \(\vec{F}\)
(c) Perpendicular to \(\vec{r}\)
(d) Perpendicular to both \(\vec{r}\) & \(\vec{F}\)
Answer:
(d) Perpendicular to both \(\vec{r}\) & \(\vec{F}\)

Question 31.
The unit of torque is –
(a) is
(b) Nm^-2
(c) Nm
(d) Js^-1
Answer:
(c) Nm

Question 32.
The direction of torque is found using –
(a) left hand rule
(b) right hand rule
(c) palm rule
(d) screw rule
Answer:
(b) right hand rule

Question 33.
if the direction of torque is out of the paper then the rotation produced by the torque is –
(a) clockwise
(b) anticlockwise
(c) straight line
(d) random direction
Answer:
(a) clockwise

Question 34.
If the direction of the torque is inward the paper then the rotation is –
(a) clockwise
(b) anticlockwise
(c) straight line
(d) random direction
Answer:
(a) clockwise

Question 35.
if \(\vec{r}\) and \(\vec{F}\) are parallel or anti parallel, then the torque is –
(a) zero
(b) minimum
(c) maximum
(d) infinity
Answer:
(a) zero

Question 36.
The maximum possible value of torque is –
(a) zero
(b) infinity
(c) \(\vec{r}\) + \(\vec{F}\)
(d) rF
Answer:
(d) rF

Question 37.
The relation between torque and angular acceleration is –
(a) \(\vec{τ}\) = \(\frac{1}{\alpha}\)
(b) \(\vec{α}\) = \(\frac{\vec{\tau}}{\mathrm{I}}\)
(c) \(\vec{α}\) = I \(\vec{τ}\)
(d) \(\vec{τ}\) = \(\frac{\vec{\alpha}}{\mathrm{I}}\)
Answer:
(b) \(\vec{α}\) = \(\frac{\vec{\tau}}{\mathrm{I}}\)

Question 38.
Angular momentum is –
(a) \(\vec{P}\) x \(\vec{r}\)
(b) \(\vec{r}\) x \(\vec{P}\)
(c) \(\overrightarrow{\frac{r}{\vec{p}}}\)
(d) \(\vec{r}\) . \(\vec{P}\)
Answer:
(b) \(\vec{r}\) x \(\vec{P}\)

Question 39.
The magnitude of angular momentum is given by –
(a) rp
(b) rp sin θ
(c) rp cos θ
(d) rp tan θ
Answer:
(b) rp sin θ

Question 40.
Angular momentum is associated with –
(a) rotational motion
(b) linear motion
(c) both (a) and (b)
(d) circular motion only
Answer:
(c) both (a) and (b)

Question 41.
Angular momentum acts perpendicular to –
(a) \(\vec{r}\)
(b) \(\vec{P}\)
(c) both \(\vec{r}\) and \(\vec{P}\)
(d) plane of the paper
Answer:
(c) both \(\vec{r}\) and \(\vec{P}\)

Question 42.
Angular momentum is given by –
(a) \(\frac {I}{ω}\)
(b) τω
(c) Iω
(d) \(\frac {ωI}{2}\)
Answer:
(c) Iω

Question 43.
The rate of change of angular momentum is –
(a) Torque
(b) angular velocity
(c) centripetal force
(d) centrifugal force
Answer:
(a) Torque

Question 44.
The forces acting on a body when it is at rest –
(a) is gravitational force
(b) Normal force
(c) both gravitational as well as normal force
(d) No force is acting
Answer:
(c) both gravitational as well as normal force

Question 45.
The net force acting on a body when it is at rest is –
(a) gravitational force
(b) Normal force
(c) Sum of gravitational and normal force
(d) zero
Answer:
(d) zero

Question 46.
If net force acting on a body is zero, then the body is in –
(a) transnational equilibrium
(b) rotational equilibrium
(c) both (a) and (b)
(d) none
Answer:
(a) transnational equilibrium

Question 47.
If the net torque acting on the body is zero, then the body is in –
(a) transnational equilibrium
(b) rotational equilibrium
(c) mechanical equilibrium
(d) none
Answer:
(b) rotational equilibrium

Question 48.
when the net force and net torque acts on the body is zero then the body is in –
(a) transnational equilibrium
(b) rotational equilibrium
(c) mechanical equilibrium
(d) none
Answer:
(d) none

Question 49.
When the net force and net torque acts on the body is zero then the body is in –
(a) static equilibrium
(b) Dynamic equilibrium
(c) both (a) and (b)
(d) transnational equilibrium
Answer:
(c) both (a) and (b)

Question 50.
When two equal and opposite forces acting on the body at two different points, it may give –
(a) net force
(b) torque
(c) stable equilibrium
(d) none
Answer:
(b) torque

Question 51.
The torque in rotational motion is analogous to in transnational motion –
(a) linear momentum
(b) mass
(c) couple
(d) force
Answer:
(d) force

Question 52.
Which of the following example does not constitute a couple?
(a) steering a car
(b) turning a pen cap
(c) ball rolls on the floor
(d) closing the door
Answer:
(c) ball rolls on the floor

Questioner 53.
If the linear momentum and angular momentum are zero, then the object is said to be in –
(a) stable equilibrium
(b) unstable equilibrium
(c) neutral equilibrium
(d) all the above
Answer:
(d) all the above

Question 54.
When the body is disturbed, the potential energy remains same, then the body is in –
(a) stable equilibrium
(b) unstable equilibrium
(c) neutral equilibrium
(d) all the above
Answer:
(c) neutral equilibrium

Question 55
The point where the entire weight of the body acts is called as –
(a) center of mass
(b) center of gravity
(c) both (a) and (b)
(d) pivot
Answer:
(b) center of gravity

Question 56.
The forces acting on a cyclist negotiating a circular Level road is /are –
(a) gravitational force
(b) centrifugal force
(c) frictional force
(d) all the above
Answer:
(d) all the above

Question 57.
While negotiating a circular level road a cyclist has to bend by an angle θ from vertical to stay in an equilibrium is-
(a) \(\tan \theta=\frac{r g}{r^{2}}\)
(b) θ = \(\tan ^{-1}\left(\frac{v^{2}}{r g}\right)\)
(c) θ = \(\sin ^{-1}\left(\frac{r g}{r^{2}}\right)\)
(d) zero
Answer:
(b) θ = \(\tan ^{-1}\left(\frac{v^{2}}{r g}\right)\)

Question 58.
Moment of inertia for point masses –
(a) m²r
(b) rw²
(c) mr²
(d) zero
Answer:
(c) mr²

Question 59.
Moment of inertia for bulk object –
(a) rm²
(b) rw²
(c) \(m_{i} r_{i}^{2}\)
(d) \(\Sigma m_{i} r_{i}^{2}\)
Answer:
(d) \(\Sigma m_{i} r_{i}^{2}\)

Question 60.
For rotational motion, moment of inertia is a measure of –
(a) transnational inertia
(b) mass
(c) rotational inertia
(d) invariable quantity
Answer:
(c) rotational inertia

Question 61.
Unit of moment of inertia –
(a) kgm
(b) mkg^-2
(c) kgm²
(d) kgm^-1
Answer:
(c) kgm²

Question 62.
Dimensional formula for moment of inertia is –
(a) [ML^-2]
(b) [M²L^-1]
(c) [M^-2]
(d) [ML²]
Answer:
(d) [ML²]

Question 63.
Moment of inertia of a body is a –
(a) variable quantity
(b) invariable quantity
(c) constant quantity
(d) measure of torque
Answer:
(a) variable quantity

Question 64.
Moment of inertia of a thin uniform rod about an axis passing through the center of mass and perpendicular to the length is –
(a) \(\frac { 1 }{ 3 }\)Ml²
(b) \(\frac { 1 }{ 12 }\)Ml²
(c) \(\frac { 1 }{ 2 }\)M(l² + b² )
(d) Ml²
Answer:
(b) \(\frac { 1 }{ 12 }\)Ml²

Question 65.
Moment of inertia ofa thin uniform rod about an axis passing through one end and perpendicular to the length is-
(a) \(\frac { 1 }{ 3 }\)Ml²
(b) \(\frac { 1 }{ 12 }\)Ml²
(c) \(\frac { 1 }{ 2 }\)M(l² + b² )
(d) Ml²
Answer:
(a) \(\frac { 1 }{ 3 }\)Ml²

Question 66.
Moment of inertia of a thin uniform rectangular sheet about an axis passing through the center of mass and perpendicular to the plane of the sheet is-
(a) \(\frac { 1 }{ 3 }\)Ml²
(b) \(\frac { 1 }{ 12 }\)Ml²
(c) \(\frac { 1 }{ 2 }\)M(l² + b² )
(d) Ml²
Answer:
(c) \(\frac { 1 }{ 2 }\)M(l² + b² )

Question 67.
Moment of inertia of a thin uniform ring about an axis passing through the center of gravity and perpendicular to the plane is –
(a) MR²
(b) 2 MR²
(c) \(\frac { 1 }{ 2 }\)MR²
(d) \(\frac { 3 }{ 2 }\)MR²
Answer:
(a) MR²

Question 68.
Moment of inertia of a thin uniform ring about an axis passing through the center and lying on the plane (along diameter) is –
(a) MR²
(b) 2 MR²
(c) \(\frac { 1 }{ 2 }\) MR²
(d) \(\frac { 2 }{ 3 }\)MR²
Answer:
(c) \(\frac { 1 }{ 2 }\) MR²

Question 69.
Moment of inertia of a thin uniform disc about an axis passing through the center and perpendicular to the plane is –
(a) MR²
(b) 2 MR²
(c) \(\frac { 1 }{ 2 }\) MR²
(d) \(\frac { 2 }{ 3 }\)MR²
Answer:
(c) \(\frac { 1 }{ 2 }\) MR²

Question 70.
Moment of inertia of a thin uniform disc about an axis passing through the center lying on the plane (along diameter is)
(a) MR²
(b) \(\frac { 1 }{ 2 }\) MR²
(c) \(\frac { 3 }{ 2 }\) MR²
(d) \(\frac { 1 }{ 4 }\) MR²
Answer:
(d) \(\frac { 1 }{ 4 }\) MR²

Question 71.
Moment of inertia of a thin uniform hollow cylinder about an axis of the cylinder is –
(a) MR²
(b) \(\frac { 1 }{ 2 }\) MR²
(c) \(\frac { 3 }{ 2 }\) MR²
(d) \(\frac { 1 }{ 4 }\) MR²
Answer:
(a) MR²

Question 72.
Moment of inertia of a thin uniform hollow cylinder about an axis of the cylinder is –
(a) MR²
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)
(c) \(\frac { 1 }{ 2 }\) MR²
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)
Answer:
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)

Question 73.
Moment of inertia of a uniform solid cylinder about an axis passing through the center and along the axis of the cylinder is –
(a) MR²
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)
(c) \(\frac { 1 }{ 2 }\) MR²
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)
Answer:
(c) \(\frac { 1 }{ 2 }\) MR²

Question 74.
Moment of inertia of a uniform solid cylinder about as axis passing perpendicular to the length and passing through the center is –
(a) MR²
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)
(c) \(\frac { 1 }{ 2 }\) MR²
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)
Answer:
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)

Question 75.
Moment of inertia of a thin hollow sphere about an axis passing through the center along its diameter is
(a) \(\frac { 2 }{ 3 }\)MR²
(b) \(\frac { 5 }{ 3 }\)MR²
(c) \(\frac { 7 }{ 5 }\)MR²
(d) \(\frac { 2 }{ 5 }\)MR²
Answer:
(a) \(\frac { 2 }{ 3 }\)MR²

Question 76.
Moment of inertia of a thin hollow sphere about an axis passing through the edge along its tangent is –
(a) \(\frac { 2 }{ 3 }\)MR²
(b) \(\frac { 5 }{ 3 }\)MR²
(c) \(\frac { 7 }{ 5 }\)MR²
(d) \(\frac { 2 }{ 5 }\)MR²
Answer:
(b) \(\frac { 5 }{ 3 }\)MR²

Question 77.
torment of inertia of a uniform solid sphere about an axis passing through the center along its diameter is –
(a) \(\frac { 2 }{ 3 }\)MR²
(b) \(\frac { 5 }{ 3 }\)MR²
(c) \(\frac { 7 }{ 5 }\)MR²
(d) \(\frac { 2 }{ 5 }\)MR²
Answer:
(d) \(\frac { 2 }{ 5 }\)MR²

Question 78.
Moment of inertia of a uniform solid sphere about an axis passing through the edge along its tangent is –
(a) \(\frac { 2 }{ 3 }\)MR²
(b) \(\frac { 5 }{ 3 }\)MR²
(c) \(\frac { 7 }{ 5 }\)MR²
(d) \(\frac { 2 }{ 5 }\)MR²
Answer:
(c) \(\frac { 7 }{ 5 }\)MR²

Question 79.
The ratio of K²/R² of a thin uniform ring about an axis passing through the center and perpendicular to the plane is-
(a) 1
(b) 2
(c) \(\frac { 7 }{ 5 }\)
(d) \(\frac { 3 }{ 2 }\)
Answer:
(a) 1

Question 80.
The ratio of K²/ R² of a thin uniform disc about an axis passing through the center and perpendicular to the plane is –
(a) 1
(b) 2
(c) \(\frac {1}{ 2 }\)
(d) \(\frac { 3 }{ 2 }\)
Answer:
(c) \(\frac {1}{ 2 }\)

Question 81.
When no external torque acts on the body, the net angular momentum of a rotating body.
(a) increases
(b) decreases
(c) increases or decreases
(d) remains constant
Answer:
(d) remains constant

Question 82.
Moment of inertia of a body is proportional to –
(a) ω
(b) \(\frac { 1 }{ ω }\)
(c) ω²
(d) \(\frac{1}{\omega^{2}}\)
Answer:
(b) \(\frac { 1 }{ ω }\)

Question 83.
When the hands are brought closer to the body, the angular velocity of the ice dancer –
(a) decreases
(b) increases
(c) constant
(d) may decrease or increase
Answer:
(b) increases

Question 84.
When the hands are stretched out from the body, the moment of inertia of the ice dancer –
(a) decreases
(b) increases
(c) constant
(d) may decrease or increase
Answer:
(b) increases

Question 85.
The work done by the torque is –
(a) F. ds
(b) F. dθ
(c) τ dθ
(d) r.dθ
Answer:
(c) τ dθ

Question 86.
Rotational Kinetic energy of a body is –
(a) \(\frac { 1 }{ 2 }\)mr
(b) \(\frac { 1 }{ 2 }\) Iω²
(c) \(\frac { 1 }{ 2 }\)Iv²
(d) \(\frac { 1 }{ 2 }\)mω²
Answer:
(b) \(\frac { 1 }{ 2 }\) Iω²

Question 87.
Rotational kinetic energy is given by –
(a) \(\frac { 1 }{ 2 }\)mr
(b) \(\frac { 1 }{ 2 }\)Iv²
(c) \(\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\)
(d) \(\frac{2 \mathrm{I}}{\mathrm{L}^{2}}\)
Answer:
(c) \(\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\)

Question 88.
If E is a rotational kinetic energy then angular momentum is-
(a) \(\sqrt{2 \mathrm{IE}}\)
(b) \(\frac{\mathrm{E}^{2}}{2 \mathrm{I}}\)
(c) \(\frac{2 \mathrm{I}}{\mathrm{E}^{2}}\)
(d) \(\frac{E}{I^{2} \omega^{2}}\)
Answer:
(a) \(\sqrt{2 \mathrm{IE}}\)

Question 89.
The product of torque acting on a body and angular velocity is –
(a) Energy
(b) power
(c) work done
(d) kinetic energy
Answer:
(b) power

Question 90.
The work done per unit time in rotational motion is given by –
(a) \(\vec{F}\) .v
(b) \(\frac {dθ}{dt}\)
(c) τ ω
(d) I ω
Answer:
(c) τ ω

Question 91.
While rolling, the path of center of mass of an object is –
(a) straight line
(b) parabola
(c) hyperbola
(d) circle
Answer:
(a) straight line

Question 92.
In pure rolling, the velocity of the point of the rolling object which comes in contact with the surface is –
(a) maximum
(b) minimum
(c) zero
(d) 2 V_CM
Answer:
(c) zero

Question 93.
In pure rolling velocity of center of mass is equal to –
(a) zero
(b) Rω
(c) \(\frac { ω }{ R }\)
(d) \(\frac { R }{ ω }\)
Answer:
(b) Rω

Question 94.
In pure rolling, rotational velocity of points at its edges is equal to-
(a) Rω
(b) velocity of center of mass
(c) transnational velocity
(d) all the above
Answer:
(a) Rω

Question 95.
Sliding of the object occurs when –
(a) V_trans < V_rot
(b) V_trans = V_rot
(c) V_trans > V_rot
(d) V_trans = 0
Answer:
(c) V_trans > V_rot

Question 96.
Sliding of the object occurs while –
(a) V_trans = V_rot
(b) V_CM = Rω
(c) V_CM < Rω
(d) V_CM > Rω
Answer:
(d) V_CM > Rω

Question 97.
Slipping of the object occurs when –
(a) V_trans < V_rot
(b) V_trans = V_rot
(c) V_trans > V_rot
(d) V_trans = 0
Answer:
(a) V_trans < V_rot

Question 98.
Slipping of the object occurs when –
(a) V_trans = V_rot
(b) V_CM = Rω
(c) V_CM < Rω
(d) V_CM > Rω
Answer:
(c) V_CM < Rω

Question 99.
In sliding, the resultant velocity of a point of contact acts along –
(a) forward direction
(b) backward direction
(c) either (a) or (b)
(d) tangential direction
Answer:
(a) forward direction

Question 100.
In slipping, the resultant velocity of a point of contact acts along –
(a) forward direction
(b) backward direction
(c) either (a) or (b)
(d) tangential direction
Answer:
(b) backward direction

Question 101.
When a solid sphere is undergoing pure rolling, the ratio of transnational kinetic energy to rotational kinetic – energy is –
(a) 2 : 5
(b) 5 : 2
(c) 1 : 5
(d) 5 : 1
Answer:
(b) 5 : 2

Question 102.
Time taken by the rolling object in inclined plane to reach its bottom is –

Answer:

Question 103.
The velocity of the rolling object on inclined plane at the bottom of inclined plane is –

Answer:

Question 104.
Moment of inertia of an annular disc about an axis passing through the centre and perpendicular to the plane of disc is –

Answer:

Question 105.
Moment of inertia of a cube about an axis passing through the center of mass and perpendicular to face is –
(a) \(\frac{\mathrm{Ma}^{2}}{6}\)
(b) \(\frac {1}{3}\) Ma²
(c) \(\frac {Ma}{6}\)
(d) \(\frac{\mathrm{Ma}^{2}}{12}\)
Answer:
(a) \(\frac{\mathrm{Ma}^{2}}{6}\)

Question 106.
Moment of inertia of a rectangular plane sheet about an axis passing through center of mass and perpendicular to side b in its plane is –
(a) \(\frac{\mathrm{Ml}^{2}}{12}\)
(b) \(\frac{\mathrm{Ma}^{2}}{12}\)
(c) \(\frac{\mathrm{Mb}^{2}}{12}\)
(d) \(\frac{\mathrm{Ml}^{2}}{6}\)
Answer:
(c) \(\frac{\mathrm{Mb}^{2}}{12}\)

Question 107.
Rotational kinetic energy can be calculated by using –
(a) \(\frac{1}{2}\) I ω²
(b) \(\frac{\mathrm{L}^{2}}{2I}\)
(c) \(\frac{1}{2}\) Lω
(d) all the above
Answer:
(b) \(\frac{\mathrm{L}^{2}}{2I}\) )

Question 108.
The radius of gyration of a solid sphere of radius r about a certain axis is r. The distance of that axis from the center of the sphere is –
(a) \(\frac{2}{5}\)r
(b) \(\sqrt{\frac{2}{5}}\)r
(c) \(\sqrt{0.6r}\)
(d) \(\sqrt{\frac{5}{3}}\)
Answer:
(c) \(\sqrt{0.6r}\)
From parallel axis theorem
I = I_G + Md²
mr² = \(\frac{2}{5}\) mr² + md²
d = \(\sqrt{\frac{3}{5}}\)r = \(\sqrt{0.6r}\)

Question 109.
A wheel is rotating with angular velocity 2 rad/s. It is subjected to a uniform angular acceleration 2 rad/s² then the angular velocity after 10 s is
(a) 12 rad/s
(b) 20 rad/s
(c) 22 rad/s
(d) 120 rad/s
Answer:
(c) 22 rad/s
ω = ω₀ + αt
Here ω₀ = 2 rad/s,
α = 2 rad/s2
ω = 10 s
ω = 2 + 2 x 10 = 22 rad/s

Question 110.
Two rotating bodies A and B of masses m and 2m with moments of inertia I_A and I_B (I_b > I_A) have equal kinetic energy of rotation. If L_A and L_B be their angular momenta respectively,
then,
(a) L_B > L_A
(b) L_A > L_B
(c) L_A = \(\frac{L_{B}}{2}\)
(d) L_A = 2L_B
Answer:
(a) L_B > L_A

Question 111.
Three identical particles lie in x, y plane. The (x, y) coordinates of their positions are (3, 2), (1, 1), (5, 3) respectively. The (x, y) coordinates of the center of mass are –
(a) (a, b)
(b) (1, 2)
(c) (3, 2)
(d) (2, 1)
Answer:
(c) The X and Y coordinates of the center of mass are

Question 112.
A solid cylinder of mass 3 kg and radius 10 cm is rotating about its axis with a frequency of 20/π. The rotational kinetic energy of the cylinder
(a) 10 π J
(b) 12 J
(c) \(\frac{6 \times 10^{2}}{\pi}\) J
(d) 3 J
Answer:
(b) 12 J
Given,
M = 3 kg
R = 0.1 m
v = 20 / π
Angular frequency ω = 2πv = \(\frac{2π x 20}{π}\) = 40 rad/s^-1
Moment of inertia of the cylinder about its axis = I = \(\frac{1}{2}\) mR² = \(\frac{1}{2}\) x 3 x (0.1)² = 0.015 kg m²
K.E. = \(\frac{1}{2}\) Iω² = \(\frac{1}{2}\) x 0.015 x (40)² = 12 J

Question 113.
A circular disc is rolling down in an inclined plane without slipping. The percentage of rotational energy in its total energy is
(a) 66.61%
(b) 33.33%
(c) 22.22%
(d) 50%
Answer:
(b) 33.33%
Rotational K.E. = \(\frac{1}{2}\)Iω² \(\frac{1}{2}\)(\(\frac{1}{2}\)MR²)ω² = \(\frac{1}{4}\) MR²ω²
Transnational K.E. = \(\frac{1}{2}\)MV² = \(\frac{1}{2}\)M(Rω)² = \(\frac{1}{2}\) MR²ω²
Total kinetic energy = E_rot + E_trans = \(\frac{1}{4}\) MR²ω²\(\frac{1}{2}\)M(Rω)² = \(\frac{3}{4}\) MR²ω²
% of E_rot = \(\frac{E_{\text {rot }}}{E_{\text {Tot }}}\) x 100% = 33.33%

Question 114.
A sphere rolls down in an inclined plane without slipping. The percentage of transnational energy in its total energy is
(a) 29.6%
(b) 33.4%
(c) 71.4%
(d) 50%
Answer:
(c) 71.4%
Rotational K.e. E_rot =

Question 115.
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the center of mass is –
(a) 30 m/s
(b) 20 m/s
(c) 10 m/s
(d) 5 m/s
Answer:
(c) According to law of conservation of linear momentum
MV = (M + M) V_CM
V_CM = \(\frac{MV}{M + M}\) = \(\frac{10 × 10}{10 + 4}\) = 10 ms^-1

Question 116.
A mass is whirled in a circular path with constant angular velocity and its angular momentum is L. If the string is now halved keeping the angular velocity the same, the angular momentum is –
(a) \(\frac{L}{4}\)
(b) \(\frac{L}{2}\)
(c) L
(d) 2L
Answer:
(a) \(\frac{L}{4}\)
We know that
angular momentum L = Mr²
Here, m and co are constants L α r²
If r becomes \(\frac{r}{2}\) angular momentum becomes \(\frac{1}{4}\) th of its initial value.

Question 117.
The moment of inertia of a thin uniform ring of mass 1 kg and radius 20 cm rotating about the axis passing through the center and perpendicular to the plane of the ring is –
(a) 4 x 10^-2 kg m²
(b) 1 x 10^-2 kg m²
(c) 20 x 10^-2 kg m²
(d) 10 x 10^-2 kg m²
Answer:
(b) Moment of inertia I = MR² = 1 x (10 x 10^-2)² = 1 x 10^-2 kg m².

Question 118.
A solid sphere is rolling down in the inclined plane, from rest without slipping. The angle of inclination with horizontal is 30°. The linear acceleration of the sphere is –
(a) 28 ms^-2
(b) 3.9 ms^-2
(c) \(\frac{25}{7}\)ms^-2
(d) \(\frac{1}{20}\)ms^-2
Answer:
(c) \(\frac{25}{7}\)ms^-2
We know that,a =

Question 119.
An electron is revolving in an orbit of radius 2 A with a speed of 4 x 10⁵ m /s. The angular momentum of the electron is [Me = 9 x 10^-31 kg]
(a) 2 x 10^-35 kg m² s^-1
(b) 72 x 10^-36 kg m² s^-1
(c) 7.2 x 10^-34 kg m² s^-1
(d) 0.72 x 10^-37 kg m² s^-1
Answer:
(b) Angular momentum L = mV x r = 9 x 10^-31 x 4 x 10⁵ x 2 x 10^-10 = 72 x 10^-36kg m² s^-1

Question 120.
A raw egg and hard boiled egg are made to spin on a table with the same angular speed about the same axis. The ratio of the time taken by the eggs to stop is –
(a) =1
(b) < 1
(c) > 1
(d) none of these
Answer:
(d) When a raw egg spins, the fluid inside comes towards its side.
∴ “1” will increase in – turn it decreases ω. Therefore it takes lesser time than boiled egg.
∴\(\frac {time fìr raw egg}{time for boiled egg}\) < 1

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions (1 Mark)

Question 1.
What is a rigid body?
Answer:
A rigid body is the one which maintains its definite and fixed shape even when an external force acts on it.

Question 2.
When an object will have procession? Give one example.
Answer:
The torque about the axis will rotate the object about it and the torque perpendicular to the axis will turn the axis of rotation when both exist simultaneously on a rigid body the body will have a procession.
Example:
The spinning top when it is about to come to rest.

Question 3.
Define angular momentum. Give an expression for it.
Answer:
The angular momentum of a point mass is defined as the moment of its linear momentum.
\(\vec{L}\) = \(\vec{r}\) x \(\vec{p}\) or L = rp sin θ

Question 4.
When an angular momentum of the object will be zero?
Answer:
If the straight path of the particle passes through the origin, then the angular momentum is zero, which is also a constant.

Question 5.
When an object be in-mechanical equilibrium?
Answer:
A rigid body is said to be in mechanical equilibrium when both its linear momentum and angular momentum remain constant.

Question 6.
Derive an expression for the power delivered by torque.
Answer:
Power delivered is the work done per unit time. IF we differentiate the expression for work done with respect to time, we get the instantaneous
power (P).
p = \(\frac{dw}{dt}\) = τ \(\frac{dθ}{dt}\)
p = τ dω

Question 7.
A boy sits near the edge of revolving circular disc

1. What will be the change in the motion of a disc?
2. If the boy starts moving from edge to the center of the disc, what will happen?

Answer:

1. As we know L = Iω = constant if the boy sits on the edge of revolving disc, its I will be increased in turn it reduces angular velocity.
2. If the boy starts moving towards the center of the disc, its I will decrease in turn that increases its angular velocity.

Question 8.
Are moment of inertia and radius of gyration of a body constant quantities?
Answer:
No, moment of inertia and radius of gyration depends on axis of rotation and also on the distribution of mass of the body about its axis.

.

Question 9.
A cat is able to land on its feet after a fall. Which principle of physics is being used? Explain.
Answer:
A cat is able to land on its feet after a fall. This is based on law of conservation of angular ~ momentum. When the cat is about to fall, it curls its body to decrease the moment of inertia and increase its angular velocity. When it lands it stretches out its limbs. By which it increases its moment of inertia and inturn it decreases its angular velocity. Hence, the cat lands safety.

Question 10.
About which axis a uniform cube will have minimum moment of inertia ?
Answer:
It will be about an axis passing through the center of the cube and connecting the opposite comers.

Question 11.
State the principle of moments of rotational equilibrium.
Answer:
∑ =\(\bar{\tau}\) = 0

Question 12.
Write down the moment of inertia of a disc of radius R and mass m about an axis in its plane at a distance R / 2 from its center.
Answer:
\(\frac { 1 }{ 2 }\) MR²

Question 13.
Can the couple acting on a rigid body produce translator motion ?
Answer:
No. It can produce only rotatory motion.

Question 14.
Which component of linear momentum does not contribute to angular momentum?
Answer:
Radial Component.

Question 15.
A system is in stable equilibrium. What can we say about its potential energy ?
Answer:
PE. is minimum.

Question 16.
Is radius of gyration a constant quantity ?
Answer:
No, it changes with the position of axis of rotation.

Question 17.
Two solid spheres of the same mass are made of metals of different densities. Which of them has a large moment of inertia about the diameter?
Answer:
Sphere of smaller density will have larger moment of inertia.

Question 18.
The moment of inertia of two rotating bodies A and B are I_A and I_B (I_A > I_B) and their angular momenta are equal. Which one has a greater kinetic energy ?
Answer:
K = \(\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\) ⇒ K_A > K_A

Question 19.
A particle moves on a circular path with decreasing speed. What happens to its angular momentum?
Answer:
As \(\vec{L}\) = \(\vec{r}\) x m\(\vec{v}\) i.e., \(\vec{L}\) magnitude decreases but direction remains constant.

Question 20.
What is the value of instantaneous speed of the point of contact during pure rolling ?
Answer:
Zero.

Question 21.
Which physical quantity is conserved when a planet revolves around the sun ?
Answer:
Angular momentum of planet.

Question 22.
What is the value of torque on the planet due to the gravitational force of sun ?
Answer:
Zero.

Question 23.
If no external torque acts on a body, will its angular velocity be constant ?
Answer:
No.

Question 24.
Why there are two propellers in a helicopter ?
Answer:
Due to conservation of angular momentum.

Question 25.
A child sits stationary at one end of a long trolley moving uniformly with speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, then what is the effect of the speed of the centre of mass of the (trolley + child) system ?
Answer:
No change in speed of system as no external force is working.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions (2 Marks)

Question 26.
State the factors on which the moment of inertia of a body depends.
Answer:

• Mass of body
• Size and shape of body
• Mass distribution w.r.t. axis of rotation
• Position and orientation of rotational axis

Question 27.
On what factors does radius of gyration of body depend?
Answer:
Mass distribution.

Question 28.
Why the speed of whirl wind in a Tornado is alarmingly high?
Answer:
In this, air from nearly regions get concentrated in a small space, so I decreases considerably. Since Iω = constant so ω increases so high.

Question 29.
Can a body be in equilibrium while in motion? If yes, give an example.
Answer:
Yes, if body has no linear and angular acceleration then a body in uniform straight line of motion will be in equilibrium.

Question 30.
There is a stick half of which is wooden and half is of steel, (i) it is pivoted at the wooden end and a force is applied at the steel end at right angle to its length (ii) it is pivoted at the steel end and the same force is applied at the wooden end. In which case is the angular acceleration more and why?
Answer:
I (first case) > 1 (Second case)
∴ τ r = l α
⇒ α (first case) < α (second case)

Question 31.
If earth contracts to half of its present radius what would be the length of the day at equator?
Answer:

Question 32.
An internal force cannot change the state of motion of center of mass of a body. Flow does the internal force of the brakes bring a vehicle to rest?
Answer:
In this case the force which bring the vehicle to rest is friction, and it is an external force.

Question 33.
When does a rigid body said to be in equilibrium? State the necessary condition for a body to be in equilibrium.
Answer:
For translation equilibrium
∑ F_ext  = 0
For rotational equilibrium
∑ \(\overline{\mathrm{τ}}\)_ext  = 0

Question 34.
How will you distinguish between a hard boiled egg and a raw egg by spinning it on a table top’
Answer:
For same external torque, angular acceleration of raw egg will be small than that of Hard boiled egg.

Question 35.
Equal torques are applied on a cylinder and a sphere. Both have same mass and radius. Cylinder rotates about its axis and sphere rotates about one of its diameter. Which will acquire greater speed and why?
Answer:
τ = I α α = \(\frac { τ }{ I }\)
α in cylinder, α_C = \(\frac{\tau}{I_{C}}\)
α in sphere, α_S = \(\frac{\tau}{I_{S}}\)

Question 36.
In which condition a body lying in gravitational field is in stable equilibrium?
Answer:
When vertical line through center of gravity passes through the base of the body.

Question 37.
Give the physical significance of moment of inertia. Explain the need of fly wheel in Engine.
Answer:
It plays the same role in rotatory motion as the mass does in translator y motion.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions (3 Marks)

Question 38.
Three mass point m₁, m₂, m₃ are located at the vertices of equilateral A of side ‘a’. What is the moment of inertia of system about an axis along the altitude of A passing through mi?
Answer:

Question 39.
A disc rotating about its axis with angular speed ω₀ is placed lightly (without any linear push) on a perfectly friction less table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in figure. Will the disc roll?
Answer:

For A V_A = R ω₀ in forward direction
For B = V_B = R ω₀ in backward direction R
For C, V_C = \(\frac {R}{2}\) ω₀ in forward direction disc will not roll.

Question 40.
Find the torque of a force 7\(\hat{i}\) – 3\(\hat{j}\) – 5\(\hat{k}\) about the origin which acts on a particle whose position vector is \(\hat{j}\) +\(\hat{j}\) – \(\hat{j}\)
Answer:

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Numericals

Question 41.
Three masses 3 kg, 4 kg and 5 kg are located at the comers of an equilateral triangle of side 1 m. Locate the center of mass of the system.
Answer:
(x,y) = (0.54 m, 0.36 m)

Question 42.
Two particles mass 100 g and 300 g at a given time have velocities 10\(\hat{j}\) – 7\(\hat{j}\) – 3\(\hat{j}\) and 7\(\hat{i}\) – 9 \(\hat{j}\) + 6\(\hat{k}\) ms^-1 respectively. Determine velocity of center of mass.
Answer:
Velocity of center of mass = \(\frac{31 \hat{i}-34 \hat{j}+15 \hat{k}}{2}\) ms^-1

Question 43.
From a uniform disc of radius R, a circular disc of radius R / 2 is cut out. The center of the hole is at R / 2 from the center of original disc. Locate the center of gravity of the resultant flat body.
Answer:
Center of mass of resulting portion lies at R/6 from the center of the original disc in a direction opposite to the center of the cut out portion.

Question 44.
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds,

1. What is its angular acceleration (assume the acceleration to be uniform)
2. How many revolutions does the wheel make during this time ?

Answer:
a = 4π rad s^-2
n = 576

Question 45.
A meter stick is balanced on a knife edge at its center. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm, what is the mass of the meter stick?
Answer:
m = 66.0 gm.

Question 46.
A solid sphere is rolling op a friction less plane surface about its axis of symmetry. Find ratio of its rotational energy to its total energy.
Answer:

Question 47.
Calculate the ratio of radii of gyration of a circular ring and a disc of the same radius with respect to the axis passing through their centers and perpendicular to their planes.
Answer:
2 : 1

Question 48.
Two discs of moments of inertia I₁ and I₂ about their respective axes (normal to the disc and passing through the center), and rotating with angular speed col and ω₂ are brought into contact face to face with their axes of rotation coincident,

1. What is the angular speed of the two – disc system ?
2. Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy ? Take ω₁ ≠ ω₂.

Answer:

1. Let co be the angular speed of the two-disc system. Then by conservation of angular momentum
   
2. Initial K.E. of the two discs.
   

Hence there is a loss of rotational K.E. which appears as heat.
When the two discs are brought together, work is done against friction between the two discs.

Question 49.
In the HCL molecule, the separating between the nuclei of the two atoms is about 1.27 Å (1Å = 10^-10m). Find the approximate location of the CM of the molecule, given that the chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in all its nucleus.
Answer:
As shown in Fig. suppose the H nucleus is located at the origin. Then,
x₁ = 0, x₂ = 1.27 Å, m₁ = 1, m₂ = 35.5
The position of the CM of HCl molecule is


Thus the CM of HCl is located on the line joining H and Cl nuclei at a distance of 1.235 Å from the H nucleus.

Question 50.
A child stands at the center of turn table with his two arms out stretched. The turn table is set rotating with an angular speed of 40 rpm. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/3 times the initial value?

1. Assume that the turn table rotates without friction
2. Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation.

How do you account for this increase in kinetic energy ?
Answer:
Here ω = 40 rpm, I₂ = \(\frac { 1 }{ 2 }\) I₁
By the principle of conservation of angular momentum,
I₁ω₁ = I_2ω2 or I₁ x 4o = \(\frac {2}{5}\) I₁ ω₁ or ω₂ = 100 rpm
(ii) Initial kinetic energy of rotation –

new kinetic energy of rotation –

Thus the child’s new kinetic energy of rotation is 2.5 times its initial kinetic energy of rotation. This increase in kinetic energy is due to the internal energy of the child which he uses in folding his hands back from the out stretched position.

Question 51.
To maintain a rotor at a uniform angular speed of 200 rad s^-1 an engine needs to transmit a torque of 180 N m. What is the power required by the engine? Assume that the engine is 100% efficient.
Here ω = 200 rad s^-1, τ = 180 N m
Power, P = τω = 180 x 200 = 36,000 W = 36 kW.

Question 52.
A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its center of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front and back wheel.
Answer:

For transnational equilibrium of car
N_F + N_B = W = 1800 x 9.8 = 17640 N
For rotational equilibrium of car
1.05 N_F = 0.75 N_B
1.05 N_F = 0.75(17640 – N_F )
1.8 N_F = 13230
N_F = 13230 / 1.8 = 7350 N
N_B = 17640 – 7350 = 10290 N
Force on each front wheel = \(\frac {7350}{ 2 }\) = 3675 N
Force on each back wheel = \(\frac {10290}{ 2 }\) = 5145 N

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Long Answer Questions (5 Marks)

Question 1.
Derive an expression for center of mass for distributed point masses.
Answer:
A point mass is a hypothetical point particle which has nonzero mass and no size or shape. To find the center of mass for a collection of n point masses, say,m₁ , m₂, m₃ ….. m_n  we have to first choose an origin and an appropriate coordinate system as shown in Figure. Let, x₁, x₂, x₃ …….. x_n be the X – coordinates of the positions of these point masses in the X direction from the origin.
The equation for the X coordinate of the center of mass is,

where, ∑ mi ¡s the total mass M of all the particles. ( ∑ mi = M).Hence,

Similarly, we can also find y and z coordinates of the center of mass for these distributed point masses as indicated in figure.

Hence, the position of center of mass of these point masses in a Cartesian coordinate system is (x_CM, y_CM z_CM). in general, the position of center of mass can be written in a vector form as,

where, is the position vector of the center of mass and \(\vec{r}_{i}\) = x_i \(\hat{j}\) + y_i\(\hat{j}\) + z_i\(\hat{k}\) is the position vector of the distributed point mass; where, \(\hat{i}\), \(\hat{j}\), and \(\hat{j}\) are the unit vectors along X, Y and Z-axis respectively.

Question 2.
Discuss the center of mass of two point masses with pictorial representation.
Answer:
With the equations for center of mass, let us find the center of mass of two point masses m₁ and m₂, which are at positions x₁ and x₂ respectively on the X – axis. For this case, we can express the position of center of mass in the following three ways based on the choice of the coordinate system.

(1) When the masses are on positive X-axis:
The origin is taken arbitrarily so that the masses m₁ and m₂ are at positions x₁ and x₂ on the positive X-axis as shown in figure (a). The center of mass will also be on the positive X- axis at x_CM as given by the equation,
\(x_{\mathrm{CM}}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}\)

(2) When the origin coincides with any one of the masses:
The calculation could be minimized if the origin of the coordinate system is made to coincide with any one of the masses as shown in figure (b). When the origin coincides with the point
mass m₁, its position x₁ is zero, (i.e. x₁ = 0). Then,
\(x_{\mathrm{CM}}=\frac{m_{1}(0)+m_{2} x_{2}}{m_{1}+m_{2}}\)
The equation further simplifies as,
x_CM = \(\frac{m_{2} x_{2}}{m_{1}+m_{2}}\)

(3) When the origin coincides with the center of mass itself:
If the origin of the coordinate system is made to coincide with the center of mass, then, x_CM = O and the mass rn1 is found to be on the negative X- axis as shown in figure (c). Hence, its position x1 is negative, (i.e. – x₁).

The equation given above is known as principle of moments.

Question 3.
Derive an expression for kinetic energy in rotation and establish the relation between rotational kinetic energy and angular momentum.
Answer:
Let us consider a rigid body rotating with angular velocity ω about an axis as shown in figure. Every particle of the body will have the same angular velocity ω and different tangential velocities v based on its positions from the axis of rotation. Let us choose a particle of mass m_i situated at distance r_i from the axis of rotation. It has a tangential velocity v_i given by the relation, v_i = r_i ω. The kinetic energy KE_i. of the particle is,
KE_i = \(\frac{1}{2} m_{i} v_{i}^{2}\)
Writing the expression with the angular velocity,
KE = \(\frac{1}{2}\) m_i(r_iω)² = \(\frac{1}{2} m_{i} r_{i}^{2}\)ω²

For the kinetic energy of the whole body, which is made up of large number of such particles, the equation is written with summation as,
KE = \(\frac{1}{2}\left(\sum m_{i} r_{i}^{2}\right)\)ω²
where, the term ∑ m_ir_i^r is the moment of inertia I of the whole body. ∑ m_ir_i^r
Hence, the expression for KE of the rigid body in rotational motion is –
KE = \(\frac{1}{2}\) Iω²
This is analogous to the expression for kinetic energy in transnational motion.
KE = \(\frac{1}{2}\) Mv²

Relation between rotational kinetic energy and angular momentum
Let a rigid body of moment of inertia I rotate with angular velocity ω.
The angular momentum of a rigid body is, L = Iω
The rotational kinetic energy of the rigid body is, KE = \(\frac{1}{2}\) Iω²
By multiplying the numerator and denominator of the above equation with I, we get a relation between L and KE as,

Question 4.
Discuss how the rolling is the combination of transnational and rotational and also be possibilities of velocity of different points in pure rolling.
Answer:
The rolling motion is the most commonly observed motion in daily life. The motion of wheel is an example of rolling motion. Round objects like ring, disc, sphere etc. are most suitable for rolling. Let us study the rolling of a disc on a horizontal surface. Consider a point P on the edge of the disc. While rolling, the point undergoes transnational motion along with its center of mass and rotational motion with respect to its center of mass.

Combination of Translation and Rotation:
We will now see how these transnational and rotational motions arc related in rolling. If the radius of the rolling object is R, in one full rotation, the center of mass is displaced by 2πR (its circumference). One would agree that not only the center of mass. but all the points Of l the disc are displaced by the same 2πR after one full rotation. The only difference is that the center of mass takes a straight path; but, all the other points undergo a path which has a combination of the transnational and rotational motion. Especially the point on the edge undergoes a path of a cyclonic as shown in the figure.

As the center of mass takes only a straight line path. its velocity v_CM is only transnational velocity v_TRANS (v_CM = v_TRANS). All the other points have two velocities. One is the transnational velocity v_TRANS (which is also the velocity of center of mass) and the other is the rotational velocity v_ROT (v_ROT = rω). Here, r ¡s the distance of the point from the center of mass and o is the angular velocity. The rotational velocity v_ROT is perpendicular to the instantaneous position vector from the center of mass as shown in figure (a). The resultant of these two velocities is v. This resultant velocity y is perpendicular to the position vector from the point of contact of the rolling object with the surface on which it is rolling as shown in figure (b).

We shall now give importance to the point of contact. In pure rolling, the point of the rolling object which comes in contact with the surface is at momentary rest. This is the case with every point that is on the edge of the rolling object. As the rolling proceeds, all’the points on the edge, one by one come in contact with the surface; remain at momentary rest at the time of contact and then take the path of the cycloid as already mentioned.
Hence, we can consider the pure rolling in two different ways.
(i) The combination of transnational motion and rotational motion about the center of mass.
(or)
(ii) The momentary rotational motion about the point of contact.
As the point of contact is at momentary rest in pure rolling, its resultant velocity v is zero (v = o). For example, in figure, at the point of contact, v_TRANS is forward (to right) and v_ROT is backwards (to the left).

That implies that, v_TRANS and v_ROT are equal in magnitude and opposite in direction (v = v_TRANS – v_ROT = 0). Hence, we conclude that in pure rolling, for all the points on the edge, the magnitudes of v_TRANS and v_ROT are equal (v_TRANS = v_ROT) As v_TRANS = v_CM and v_ROT = Rω, in pure rolling we have,
v_CM = Rω

We should remember the special feature of the above equation. In rotational motion, as per the relation v = rω, the center point will not have any velocity as r is zero. But in rolling motion, it suggests that the center point has a velocity v_CM given by above equation v_CM – Rω. For the topmost point, the two velocities v_TRANS and v_ROT are equal in magnitude and in the same direction (to the right). Thus, the resultant velocity v is the sum of these two velocities, v = v_TRANS + v_ROT In other form, v = 2 v_CM as shown in figure below.

Question 5.
Derive an expression for kinetic energy in pure rolling.
Answer:
As pure is the combination of transnational and rotational motion, we can write the total kinetic energy (KE) as the sum of kinetic energy due to transnational motion (KE_TRANS) and kinetic energy due to rotational motion (KE_ROT).
KE = KE_TRANS + KE_ROT ………(i)
If the mass of the rolling object is M, the velocity of center of mass is v_CM, its moment of inertia about center of mass is I_CM and angular velocity is ω, then

With center of mass as reference:
The moment of inertia (I_CM) of a rolling object about the center of mass is, I_CM = MK² and v_CM = Rω. Here, K is radius of gyration.

With point of contact as reference:
We can also arrive at the same expression by taking the momentary rotation happening with respect to the point of contact (another approach to rolling). If we take the point of contact as o, then,
KE = \(\frac {1}{2}\) I₀ω²

Here, I₀ is the moment of inertia of the object about the point of contact. By parallel axis
theorem, I₀ = I_CM + MK² Further we can write, I₀ MK² + MR². With v_CM = Rω or ω = \(\frac{v_{\mathrm{CM}}}{\mathrm{R}}\)

As the two equations (v) and (vi) are the same, it ¡s once again confirmed that the pure tolling problems could be solved by considering the motion as any one of the following two cases.
(i) The combination of transnational motion and rotational motion about the center of mass.
(or)
(ii) The momentary rotational motion about the point of contact.

Question 6.

1. Can a body in translator y motion have angular momentum? Explain.
2. Why is it more difficult to revolve a stone by tying it to a longer string than by tying it to a shorter string?

Answer:
(1) Yes, a body in translatory motion shall have angular momentum unless fixed point about which angular momentum is taken lies on the line of motion of body
\(|\overrightarrow{\mathrm{L}}|\) = rp sin θ
= 0 only when θ = O° or 180°

(2) MI of stone I = ml² (l – length of string) l is large, a is very small
τ = Iα
α = \(\frac {τ}{I}\) = \(\frac{\tau}{m l^{2}}\)
if l is large a is very small.
∴ more difficult to revolve.

Students can Download Bio Botany Chapter 7 Cell Cycle Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany 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 11th Bio Botany Solutions Chapter 7 Cell Cycle

Samacheer Kalvi 11th Bio Botany Cell Cycle Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
The correct sequence in cell cycle is …………… .
(a) S-M-G1-G2
(b) S-G1-G2-M
(c) G1-S-G2-M
(d) M-G-G2-S
Answer:
(c) G1-S-G2-M

Question 2.
If mitotic division is restricted in G1 phase of the cell cycle then the condition is known as …………… .
(a) S Phase
(b) G2 Phase
(c) M Phase
(d) G₀ Phase
Answer:
(d) G₀ Phase

Question 3.
Anaphase promoting complex APC is a protein degradation machinery necessary for proper mitosis of animal cells. If APC is defective in human cell, which of the following is expected to occur?
(a) Chromosomes will be fragmented
(b) Chromosomes will not condense
(c) Chromosomes will not segregate
(d) Recombination of chromosomes will occur
Answer:
(b) Chromosomes will not condense

Question 4.
In S phase of the cell cycle …………… .
(a) Amount of DNA doubles in each cell
(b) Amount of DNA remains same in each cell
(c) Chromosome number is increased
(d) Amount of DNA is reduced to half in each cell
Answer:
(a) Amount of DNA doubles in each cell

Question 5.
Centromere is required for …………… .
(a) Transcription
(b) Crossing over
(c) Cytoplasmic cleavage
(d) Movement of chromosome towards pole
Answer:
(d) Movement of chromosome towards pole

Question 6.
Synapsis occur between …………… .
(a) mRNA and ribosomes
(b) Spindle fibres and centromeres
(c) Two homologous chromosomes
(d) A male and a female gamete
Answer:
(c) Two homologous chromosomes

Question 7.
In meiosis crossing over is initiated at …………… .
(a) Diplotene
(b) Pachytene
(c) Leptotene
(d) Zygotene
Answer:
(b) Pachytene

Question 8.
Colchicine prevents the mitosis of the cells at which of the following stage …………… .
(a) Anaphase
(b) Metaphase
(c) Prophase
(d) Interphase
Answer:
(b) Metaphase

Question 9.
The paring of homologous chromosomes on meiosis is known as …………… .
(a) Bivalent
(b) Synapsis
(c) Disjunction
(d) Synergids
Answer:
(b) Synapsis

Question 10.
Anastral mitosis is the characteristic feature of …………… .
(a) Lower animals
(b) Higher animals
(c) Higher plants
(d) All living organisms
Answer:
(c) Higher plants

Question 11.
Write any three significance of mitosis.
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

1. Genetic stability – daughter cells are genetically identical to parent cells.
2. Repair of tissues – damaged cells must be replaced by identical new cells by mitosis.
3. Regeneration – Arms of star fish.

Question 12.
Differentiate between Mitosis and Meiosis.
Answer:
Difference Between Mitosis and Meiosis:

Difference Between Mitosis and Meiosis
Mitosis	Meiosis
1. One division	1. Two divisions
2. Number of chromosomes remains the same	2. Number of chromosomes is halved
3. Homologous chromosomes line up separately on the metaphase plate	3. Homologous chromosomes line up in pairs at the metaphase plate
4. Homologous chromosome do not pair up	4. Homologous chromosome pairup to form bivalent
5. Chiasmata do not form and crossing over never occurs	5. Chiasmata form and crossingover occurs
6. Daughter cells are genetically identical	6. Daughter cells are genetically different from the parent cells
7. Two daughter cells are formed	7. Four daughter cells are formed

Question 13.
Given an account of G₀ phase.
Answer:
Some cells exit G₁ and enters a quiescent stage called G₀, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G₀ phase. In G₀ cells cease growth with reduced rate of RNA and protein synthesis. The G₀ phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G₀ .Many cells in animals remains in G₀ unless called onto proliferate by appropriate growth factors or other extracellular signals. G₀ cells are not dormant.

Question 14.
Differentiate Cytokinesis in plant cells and animal cells.
Answer:
1. Cytokinesis in Plant Cells:
Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from centre towards lateral walls centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments and vesicles from golgi apparatus and ER. The golgi vesicles contains carbohydrates such as pectin, hemicellulose which move along the microtubule of the pharagmoplast to the equator fuse, forming a new plasma membrane and the materials which are placed their becomes new cell wall.

The first stage of cell wall construction is a line dividing the newly forming cells called a cell plate. The cell plate eventually stretches right across the cell forming the middle lamella. Cellulose builds up on each side of the middle lamella to form the cell walls of two new plant cells.

2. Cytokinesis in Animal Cells:
It is a contractile process. The contractile mechanism contained in contractile ring located inside the plasma membrane. The ring consists of a bundle of microfilaments assembled from actin and myosin. This fibril helps for the generation of a contractile force. This force draws the contractile ring inward forming a cleavage furrow in the cell surface dividing the cell into two.

Question 15.
Write about Pachytene and Diplotene of Prophase I.
Answer:
1. Pachytene: At this stage bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consists of 4 chromatids and 2 centromeres. Synapsis is completed and recombination nodules appear at a site where crossing over takes place between non – sister chromatids of homologous chromosome. Recombination of homologous chromosomes is completed by the end of the stage but the chromosomes are linked at the sites of crossing over. This is mediated by the enzyme recombinase.

2. Diplotene: Synaptonemal complex disassembled and dissolves. The homologous chromosomes remain attached at one or more points where crossing over has taken place. These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called Chiasmata. Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialised chromosomal structures that hold the homologous chromosomes together.

Sister chromatids remain closely associated whereas the homologous chromosomes tend to separate from each other but are held together by chiasmata. This substage may last for days or years depending on the sex and organism. The chromosomes are very actively transcribed in females as the egg stores up materials for use during embryonic development. In animals, the chromosomes have prominent loops called lampbrush chromosome.

Samacheer Kalvi 11th Bio Botany Cell Cycle Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Most of the neurons in the brain are in …………… stage.
(a) G₁
(b) S
(c) G₂
(d) G₀
Answer:
(d) G₀

Question 2.
Short, constricted region in the chromosome is …………… .
(a) Kinetochore
(b) Centromere
(c) Satellite
(d) Telomere
Answer:
(b) Centromere

Question 3.
Robert Brown discovered the nucleus in the cells of …………… roots.
(a) Mirabilas
(b) Orchid
(c) Moringa
(d) Oryza
Answer:
(b) Orchid

Question 4.
Scientist who described chromosomes for the first time is …………… .
(a) Robert Brown
(b) Anton van Leeuwenhoek
(c) Boveri
(d) Anton Schneider
Answer:
(d) Anton Schneider

Question 5.
Number of chromosomes in onion cell is …………… .
(a) 8
(b) 16
(c) 32
(d) 64
Answer:
(a) 16

Question 6.
Longest part of the cell cycle is …………… .
(a) Prophase
(b) G₁ Phase
(c) Interphase
(d) Sphase
Answer:
(c) Interphase

Question 7.
Eukaryotic cells divides every …………… .
(a) 12
(b) 24
(c) 1
(d) 6
Answer:
(b) 24

Question 8.
Cell cycle was discovered by …………… .
(a) Singer & Nicolson
(b) Prevost & Dumans
(c) Schleider & Schwann
(d) Boveri
Answer:
(b) Prevost & Dumans

Question 9.
G₀ stage is called as …………… stage.
(a) Quiescent
(b) Metabolically active
(c) Synthesis of DNA
(d) Replication
Answer:
(a) Quiescent

Question 10.
…………… protein acts as major check point in phase.
(a) Porins
(b) Kinases
(c) Cyclins
(d) Ligases
Answer:
(c) Cyclins

Question 11.
Replication of DNA occurs at …………… phase.
(a) G₀
(b) G₁
(c) S
(d) G₂
Answer:
(c) S

Question 12.
Condensation of interphase chromosomes into mitotic forms is done by …………… proteins.
(a) MPF
(b) APF
(c) AMF
(d) MAF
Answer:
(a) MPF

Question 13.
Which of the following is also called as direct division?
(a) Amitosis
(b) Meiosis
(c) Mitosis
(d) Reduction division
Answer:
(a) Amitosis

Question 14.
Cells of mammalian cartilage undergoes …………… .
(a) Amitosis
(b) Meiosis
(c) Mitosis
(d) Equational division
Answer:
(a) Amitosis

Question 15.
Yeast cells undergo …………… .
(a) Open mitosis
(b) Closed mitosis
(c) Amitosis
(d) Meiosis
Answer:
(b) Closed mitosis

Question 16.
…………… is the longest phase in mitosis.
(a) Anaphase
(b) Telophase
(c) Prophase
(d) Interphase
Answer:
(c) Prophase

Question 17.
The DNA protein complex present in the centromere is …………… .
(a) Cyclin
(b) Kinesis
(c) MPF
(d) Kinetochore
Answer:
(d) Kinetochore

Question 18.
…………… protein induces the break down of cohesion proteins leading to chromatid separation during mitosis.
(a) APC
(b) MPF
(c) Cyclin
(d) Kinetochore
Answer:
(a) APC

Question 19.
Regeneration of arms of star fish is due to …………… .
(a) Meiosis
(b) Amitosis
(c) Mitosis
(d) Budding
Answer:
(c) Mitosis

Question 20
…………… is called as reduction division.
(a) Meiosis
(b) Mitosis
(c) Amitosis
(d) Budding
Answer:
(a) Meiosis

Question 21.
Bivalents occur at …………… stage.
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene
Answer:
(a) Zygotene

Question 22.
Recombination of chromosomes occur at …………… .
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Question 23.
Terminalisation of chiasmata occurs at …………… .
(a) Zygotene
(b) Leptotene
(c) Diakinesis
(d) Pachytene
Answer:
(c) Diakinesis

Question 24.
Number of daughter cells formed at the end of Meiosis I is …………… .
(a) 2
(b) 4
(c) 1
(d) 0
Answer:
(a) 2

Question 25.
…………… division leads to genetic variability.
(a) Mitotic
(b) Amitotic
(c) Meiotic
(d) Equational
Answer:
(c) Meiotic

Question 26.
Crossing over occurs at …………… stage.
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Question 27.
Which of the following is not a mitogen?
(a) Giberellin
(b) Ethylene
(c) Kinetin
(d) Colchicine
Answer:
(d) Colchicine

Question 28.
In plants mitosis occurs at …………… cells.
(a) Sclerenchyma
(b) Meristem
(c) Xylem
(d) Parenchyma
Answer:
(b) Meristem

Question 29.
Which of the following alone is formed in the division of plant cells?
(a) Aster
(b) Centrioles
(c) Spindle
(d) Microtubules
Answer:
(c) Spindle

Question 30.
Amphiastral type cell division is seen in …………… cells.
(a) Fungal
(b) Algal
(c) Plant cells
(d) Animal
Answer:
(d) Animal

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Name the two types of nuclear division.
Answer:
The two types of nuclear division:

1. Mitosis and
2. Meiosis.

Question 2.
Define Cell Cycle.
Answer:
A series of events leading to the formation of new cell is known as cell cycle.

Question 3.
Who discovered the Cell Cycle?
Answer:
Prevost & Dumans in 1824.

Question 4.
Draw a tabular column showing the duration of various phase in the cell cycle of human cell.
Answer:
A tabular column showing the duration of various phase in the cell cycle of human cell:

Cell cycle of a proliferating human cell
Phase	Time Duration (in hrs)
1. G2	1. 11
2. S	2. 8
3. G2	3. 4
4. M	4. 1

Question 5.
Define C – Value.
Answer:
C – Value is the amount in picograms of DNA contained within a haploid nucleus.

Question 6.
Which is the longest phase of cell cycle? What happens during that phase?
Answer:
Interphase is the longest phase. Cells are metabolically active and involved in protein synthesis and growth.

Question 7.
Name the phases which comprises the Interphase.
Answer:
The phases which comprises the Interphase:

1. G₁ Phase
2. S Phase and
3. G₂ Phase.

Question 8.
Name the proteins involved in the activation of genes & their proteins to perform cell division.
Answer:
Kinases & Cyclins.

Question 9.
What do you mean by G₀ stage?
Answer:
G₀ stage is called as quiescent stage, where the cells remain metabolically active without proliferation.

Question 10.
What is the role of MPF in Cell cycle?
Answer:
Maturation Promoting Factor (MPF) brings about condensation of interphase chromosomes into the mitotic form.

Question 11.
Distinguish between Karyokinesis & Cytokinesis.
Answer:
Between Karyokinesis & Cytokinesis:

• Karyokinesis: Karyokinesis refers to the nuclear division.
• Cytokinesis: Cytokinesis refers to the cytoplasmic division.

Question 12.
Point out any two cell – types which remain G₀ phase.
Answer:
Mature neurons and Skeletal muscle cells.

Question 13.
Why amitosis is called as incipient cell division?
Answer:
Amitosis is also called incipient cell division. Since there is no spindle formation and chromatin material does not condense.

Question 14.
List out the disadvantages of Amitosis.
Answer:
The disadvantages of Amitosis:

• Causes unequal distribution of chromosomes.
• Can lead to abnormalities in metabolism and reproduction.

Question 15.
Mitosis also called as equational division – Justify.
Answer:
At the end of mitosis the number of chromosomes in the parent and the daughter (Progeny) cells remain the same so it is also called as equational division.

Question 16.
Enumerate the stages of mitosis.
Answer:
Mitosis is divided into four stages prophase, metaphase, anaphase and telophase.

Question 17.
Define an aster.
Answer:
In animal cell the centrioles extend a radial array of microtubules towards the plasma membrane when they reach the poles of the cell. This arrangement of microtubules is called an aster. Plant cells do not form asters.

Question 18.
What is metaphase plate?
Answer:
The alignment of chromosome into compact group at the equator of the cell is known as metaphase plate.

Question 19.
What is Kinetochore?
Answer:
Kinetochore is a DNA – Protein complex present in the centromere DNA, where the microtubules are attached. It is a trilaminar disc like plate.

Question 20.
How will you calculate the length of the S period.
Answer:
Length of the S period = Fraction of cells in DNA replication × generation time.

Question 21.
Which type of cell division occurs in reproductive cells? What will be the result?
Answer:
Meiosis takes place in the reproductive organs. It results in the formation of gametes with half the normal chromosome number.

Question 22.
Define Synapsis.
Answer:
In Zygotene, pairing of homologous chromosomes takes place and it is known as synapsis.

Question 23.
What do you understand by independent assortment?
Answer:
The random distribution of homologous chromosomes in a cell in Metaphase I is called independent assortment.

Question 24.
Define Mitogen. Give an example.
Answer:
The factors which promote cell cycle proliferation is called mitogen.
Example: gibberellin. These increase mitotic rate.

Question 25.
What are mitotic poisons.
Answer:
Certain chemical components act as inhibitors of the mitotic cell division and they are called mitotic poisons.

Question 26.
Distinguish between Anastral & Amphiastral.
Answer:
Between Anastral & Amphiastral:
Anastral:

1. This is present only in plant cells.
2. No asters or centrioles are formed only spindle fibres are formed during cell division.

Amphiastral:

1. This is found in animal cells.
2. Aster and centrioles are formed at each pole of the spindle during cell division.

Question 27.
Draw a simple diagram to show the Amitosis.
Answer:
The Amitosis:

III. Short Answer Type Questions (3 Marks)

Question 1.
What is the role of nucleus in the cell?
Answer:
The role of nucleus in the cell:

• Control activities of the cell.
• Genetic information copied from cell to cell while the cell divides.
• Hereditary characters are passed onto new individuals when gametic cells fuse together in sexual reproduction.

Question 2.
What are restriction points? Mention its role in Cell cycle.
Answer:
The checkpoint called the restriction point at the end of G₁ it determines a cells fate whether it will continue in the cell cycle and divide or enter a stage called G₀ as a quiescent stage and probably as specified cell or die.

Question 3.
Point out the reasons responsible for the arresting of the cell in G₁ phase?
Answer:
Cells are arrested in G₁ due to:

• Nutrient deprivation
• Lack of growth factors or density dependant inhibition
• Undergo metabolic changes and enter into G₀ state.

Question 4.
Write a note on G₀ phase.
Answer:
Some cells exit G₁ and enters a quiescent stage called G₀, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G₀ phase. In G₀ cells cease growth with reduced rate of RN A and protein synthesis. The G₀ phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G₀. Many cells in animals remains in G₀ unless called onto proliferate by appropriate growth factors or other extracellular signals. G₀ cells are not dormant.

Question 5.
List out the events taking place in S – Phase.
Answer:
S Phase – Synthesis phase – cells with intermediate amounts of DNA Growth of the cell continues as replication of DNA occur, protein molecules called histones are synthesised and attach to the DNA. The centrioles duplicate in the cytoplasm. DNA content increases from 2C to 4C.

Question 6.
Distinguish between Karyokinesis & Cytokinesis.
Answer:
Karyokinesis:

1. Involves division of nucleus.
2. Nucleus develops a constriction at the center and becomes dumbellshaped.
3. Constriction deepens and divides the nucleus into two.

Cytokinesis:

1. Involves division of cytoplasm.
2. Plasma membrane develops a constriction along nuclear constriction.
3. It deepens centripetally and finally divides the cell into two cells.

Question 7.
Explain the differences between closed and open mitosis.
Answer:
Between closed and open mitosis:

1. In closed mitosis, the nuclear envelope remains intact and chromosomes migrate to opposite poles of a spindle within the nucleus. Example: Yeast and slime molds.
2. In open mitosis, the nuclear envelope breaks down and then reforms around the 2 sets of separated chromosome. Example: Most plants and animals cells.

Question 8.
What happens to plant cells at the end of Telophase in Mitosis?
Answer:
In plants, phragmoplast are formed between the daughter cells. Cell plate is formed between the two daughter cells, reconstruction of cell wall takes place. Finally the cells are separated by the distribution of organelles, macromolecules into two newly formed daughter cells.

Question 9.
Bring out the significance of Meiosis.
Answer:
The significance of Meiosis:

• Meiosis maintains a definite constant number of chromosomes in organisms.
• Crossing over takes place and exchange of genetic material leads to variations among species. These variations are the raw materials to evolution. Meiosis leads to genetic variability by partitioning different combinations of genes into gametes through independent assortment.
• Adaptation of organisms to various environmental stress.

Question 10.
Differentiate between the mitosis of Plant Cell & Animal Cell.
Answer:
Plants:

1. Centrioles are absent
2. Asters are not formed
3. Cell division involves formation of a cell plate
4. Occurs mainly at meristem.

Animals:

1. Centrioles are present
2. Asters are formed
3. Cell division involves furrowing and cleavage of cytoplasm
4. Occurs in tissues throughout the body.

Question 11.
Explain briefly about Endomitosis.
Answer:
The replication of chromosomes in the absence of nuclear division and cytoplasmic division resulting in numerous copies within each cell is called endomitosis. Chromonema do not separate to form chromosomes, but remain closely associated with each other. Nuclear membrane does not rupture. So no spindle formation. It occurs notably in the salivary glands of Drosophila and other flies. Cells in these tissues contain giant chromosomes (polyteny), each consisting of over thousands of intimately associated, or synapsed, chromatids. Example: Polytene chromosome.

Question 12.
How G₀ cells help in Closing Technology?
Answer:
Since the DNA of cells in G₀, do not replicate. The researcher are able to fuse the donor cells from a sheep’s mammary glands into G₀, state by culturing in the nutrient free state. The G₀, donor nucleus synchronised with cytoplasm of the recipient egg, which developed into the clone Dolly.

IV. Long Answer Type Questions (5 Marks)

Question 1.
Draw and label the various stages of Prophase I.
Answer:
Label the various stages of Prophase I:

Question 2.
Explain in detail about the various stages of Prophase I.
Answer:
The various stages of Prophase I:
1. Prophase I – Prophase I is of longer duration and it is divided into 5 substages – Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

2. Leptotene – Chromosomes are visible under light microscope. Condensation of chromosomes takes place. Paired sister chromatids begin to condense.

3. Zygotene – Pairing of homologous chromosomes takes place and it is known as synapsis. Chromosome synapsis is made by the formation of synaptonemal complex. The complex formed by the homologous chromosomes are called as bivalent (tetrads).

4. Pachytene – At this stage bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consists of 4 chromatids and 2 centromeres. Synapsis is completed and recombination nodules appear at a site where crossing over takes place between non – sister chromatids of homologous chromosome. Recombination of homologous chromosomes is completed by the end of the stage but the chromosomes are linked at the sites of crossing over. This is mediated by the enzyme recombinase.

5. Diplotene – Synaptonemal complex disassembled and dissolves. The homologous chromosomes remain attached at one or more points where crossing over has taken place. These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called.

6. Chiasmata: Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialised chromosomal structures that hold the homologous chromosomes together. Sister chromatids remain closely associated whereas the homologous chromosomes tend to separate from each other but are held together by chiasmata. This substage may last for days or years depending on the sex and organism. The chromosomes are very actively transcribed in females as the egg stores up materials for use during embryonic development. In animals, the chromosomes have prominent loops called lampbrush chromosome.

7. Diakinesis – Terminalisation of chiasmata. Spindle fibres assemble. Nuclear envelope breaks down. Homologous chromosomes become short and condensed. Nucleolus disappears.

Question 3.
Describe the process of Cytokinesis in Plant cell & Animal Cell.
Answer:
1. Cytokinesis in Plant Cell: Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from centre towards lateral walls – centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments and vesicles from golgi apparatus and ER. The golgi vesicles contains carbohydrates such as pectin, hemicellulose which move along the microtubule of the pharagmoplast to the equator fuse, forming a new plasma membrane and the materials which are placed there becomes new cell wall.

The first stage of cell wall construction is a line dividing the newly forming cells called a cell plate. The cell plate eventually stretches right across the cell forming the middle lamella. Cellulose builds up on each side of the middle lamella to form the cell walls of two new plant cells.

2. Cytokinesis in Animal Cells:
It is a contractile process. The contractile mechanism contained in contractile ring located inside the plasma membrane. The ring consists of a bundle of microfilaments assembled from actin and myosin. This fibril helps for the generation of a contractile force. This force draws the contractile ring inward forming a cleavage furrow in the cell surface dividing the cell into two.

Question 4.
What are the significances of Mitosis.
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

1. Genetic stability – Daughter cells are genetically identical to parent cells.
2. Growth – As multicellular organisms grow, the number of cells making up their tissue increases. The new cells must be identical to the existing ones.
3. Repair of tissues – Damaged cells must be replaced by identical new cells by mitosis.
4. Asexual reproduction – Asexual reproduction results in offspring that are identical to the parent. Example Yeast and Amoeba.
5. In flowering plants, structure such as bulbs, corms, tubers, rhizomes and runners are produced by mitotic division. When they separate from the parent, they form a new individual. The production of large numbers of offsprings in a short period of time, is possible only by mitosis. In genetic engineering and biotechnology, tissues are grown by mitosis (i.e. in tissue culture).
6. Regeneration – Arms of star fish

Question 5.
Explain the various phases in Cell Cycle.
Answer:
The different phases of cell cycle are as follows:
1. Interphase: Longest part of the cell cycle, but it is of extremely variable length. At first glance the nucleus appears to be resting but this is not the case at all. The chromosomes previously visible as thread like structure, have dispersed. Now they are actively involved in protein synthesis, at least for most of the interphase. C – Value is the amount in picograms of DNA contained within a haploid nucleus.

2. G₁ Phase: The first gap phase – 2C amount of DNA in cells of G₁ The cells become metabolically active and grows by producing proteins, lipids, carbohydrates and cell organelles including mitochondria and endoplasmic reticulum. Many checkpoints control the cell cycle. The checkpoint called the restriction point at the end of G₁ it determines a cells fate whether it will continue in the cell cycle and divide or enter a stage called G₀ as a quiescent stage and probably as specified cell or die. Cells are arrested in G₁ due to:

3. Nutrient deprivation: Lack of growth factors or density dependant inhibition. Undergo metabolic changes and enter into G₀ state. Biochemicals inside cells activates the cell division. The proteins called kinases and cyclins activate genes and their proteins to perform cell division. Cyclins act as major checkpoint which operates in G₁ to determine whether or not a cell divides.

4. G₀ Phase: Some cells exit G₁ and enters a quiescent stage called G₀, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G₀ phase. In G₀ cells cease growth with reduced rate of RNA and protein synthesis. The G₀ phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G₀. Many cells in animals remains in G₀ unless called on to proliferate by appropriate growth factors or other extracellular signals. G₀ cells are not dormant.

5. S phase – Synthesis phase – cells with intermediate amounts of DNA. Growth of the cell continues as replication of DNA occur, protein molecules called histones are synthesised and attached to the DNA. The centrioles duplicate in the cytoplasm. DNA content increases from 2C to 4C.

6. G₂ – The second Gap phase – 4C amount of DNA in cells of G₂ and mitosis. Cell growth continues by protein and cell organelle synthesis, mitochondria and chloroplasts divide. DNA content remains as 4C. Tubulin is synthesised and microtubules are formed. Microtubles organise to form spindle fibre. The spindle begins to form and nuclear division follows.

One of the proteins synthesized only in the G₂ period is known as Maturation Promoting Factor (MPF). It brings about condensation of interphase chromosomes into the mitotic form. DNA damage checkpoints operates in G₁ S and G₂ phases of the cell cycle.

Question 6.
List out the important features of Chromosomes.
Answer:
The four important features of the chromosome are:
1. The shape of the chromosome is specific: The long, thin, lengthy structured chromosome contains a short, constricted region called centromere. A centromere may occur any where along the chromosome, but it is always in the same position on any given chromosome. The number of chromosomes per species is fixed: For example the mouse has 40 chromosomes, the onion has 16 and humans have 46.

2. Chromosomes occur in pairs: The chromosomes of a cell occur in pairs, called homologous pairs. One of each pair come originally from each parent. Example, human has 46 chromosomes, 23 coming originally from each parent in the process of sexual reproduction. Chromosomes are copied: Between nuclear divisions, whilst the chromosomes are uncoiled and cannot be seen, each chromosome is copied. The two identical structures formed are called chromatids.

V. Higher Order Thinking Skills (HOTs)

Question 1.
Given that the average duplication time of E.coli is 20 minutes. How much time will two E.coli cells takes to become 32 cells?
Answer:

One cells takes 80 minutes to form 16 cells. If 2 cells undergoes division simultaneously, it take 160 minutes (2 hours 40 minutes) to form 32 cells.

Question 2.
Complete the cell cycle by filling the gaps with respective phases.
Answer:
X= S phase or Synthesis phase
Y= M phase or Mitosis phase
Z= G₀ phase

Question 3.
Telophase is reverse of prophase – Comment
Answer:
Events in Prophase:

1. Nuclear membrane disappears
2. Nucleolus disappear
3. Spindle fibre begins to form
4. Chromosomes threads condeme to form chromosomes

Events in Telophase:

1. Nuclear membrane reappears
2. Nucleolus reappears
3. Spindle fibre disappears
4. Chromosomes decondeme to form chromosomes

Question 4.
Name the pathological condition when uncontrolled cell division occurs.
Answer:
Uncontrolled cell division & abnormal growth of cells leads to the pathological condition called tumor or cancer.

Question 5.
Microspores are produced in the multiples of four, why?
Answer:
Microspores are haploid spores produced from diploid microspores mother cells. Each microspores mother cell (2n) undergoes meiosis producing four Microspores (n). Because a complete meiotic division yields 4 cells. Thus microspores are produced in multiples of four.

Question 6.
Between Prokaryotes & Eukaryotes, which cell has a shorter cell division time.
Answer:
Prokaryotes like bacteria undergo simple form of cell division called binary fission which will get completed with in a hour, whereas Eukaryotic cell division (mitosis) takes nearly 24 hours to get completed. Hence Prokaryotes have shorter cell division time.

Question 7.
Though Prokaryotic cell division differs from Eukaryotic cell division, both show certain common aspects during cell division. Explain.
Answer:
Whether a cell is Prokaryote or Eukaryote, while undergoing division, the following events must occur in common.

1. Replication of DNA.
2. Cytokinesis at the end of cell division.

Question 8.
An anther has 1204 pollen grains. How many Pollen mother cells must have been there to produce them?Explain.
Answer:
301 – Pollen mother cells: 301 Pollen mother cells undergo meiosis producing 1204 pollen grains. Because at the end of meiosis, each pollen mother cells produces 4 pollen grains.

Question 9.
A cell has 32 chromosomes. It undergoes mitosis. What will be the chromosome number during metaphase?
Answer:
During S phase of interphase, the genetic material of the cell is duplicated. So during metaphase, the chromosome number(chromatid number) will be doubled thus 64 chromosomes (chromatids) will be present.

Question 10.
Why sibilings show disimilarities?
Answer:
Though born to same parents, siblings show dissimilarities and variation due to the crossing over and recombination of chromosomes during meiosis.

Question 11.
Ramu’s met with an accident while riding cycle and got wounded in his leg. After few days, the wound was healed and the skin becomes normal. How?
Answer:
Ramu’s wound was healed because of the mitotic division. As a result of mitosis, new cells are produced and damaged tissues were repaired resulting the damaged skin to become normal.

Question 12.
A flower of tomato plant following the process of sexual reproduction produces 240 viable seeds. What is the minimum number of microspore mother cells involved in this process?
Answer:
60 microspore mother cells are involved in providing 240 pollen grains. Because each microspore mother cell undergoes meiosis producing four pollen grains (i.e. 60 × 4 = 240). Each pollen grain produces two male gametes of which one undergoes true fertilization of ovule producing seeds. Other male gamete participate in double fertilization.