System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 35

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Question 1
1 / -0

An object will not undergo rotational motion when:

A
the forces are acting on it at different positions

B
every forces is creating different turning effects

C
every moment has the same amplitude

D
all the forces are acting at its centre of gravity

Solution

For an object to undergo rotational motion, a net torque must be exerted on the object. If all the forces are acting at its centre of gravity, net torque acting on the object is zero and so the object will not undergo any rotational motion.
Question 2
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What is the momentum of a $$6.0 kg$$ bowling ball with a velocity of $$2.2 {m}/{s}$$?

A
11.2

B
17.2

C
15.2

D
13.2

Solution

Given,

Mass, $$m=\,6\,kg$$

Velocity, $$v=2.2\,m{{s}^{-1}}$$

Momentum, $$P=mv=6\times 2.2\,=13.2\,Ns$$

Hence, momentum is $$13.2\,Ns$$
Question 3
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Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is:

A
4 : 1

B
2 : 1

C
1 : 2

D
1 : 16

Solution

Kinetic energy $$K = \dfrac{p^2}{2m}$$
Two masses are given as- $$m_1 = 1 \ gm$$ and $$m_2 = 4 \ gm$$
But $$K_1 = K_2$$
$$\therefore$$ $$ \dfrac{p^2_1}{2m_1} = \dfrac{P^2_2}{2m_2}$$
Or $$ \dfrac{P_1^2}{2(1)} = \dfrac{P^2_2}{2(4)}$$
$$\implies$$ $$P_1:P_2 = 1:2$$
Question 4
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If v, P and E denote the velocity, momentum and kinetic energy of the particle, then:

A
P = dE/dv

B
P = dE/dt

C
P = dv/dt

D
none of these.

Solution

Dimensions of momentum $$[P] = [MLT^{-1}]$$
Dimensions of velocity $$[v] = [LT^{-1}]$$
Dimensions of energy $$[E] = [ML^2T^{-2}]$$
Option A :
Dimensions of LHS $$ = [ML T^{-1}]$$
Dimensions of RHS $$ = \dfrac{ML^2 T^{-2}}{LT^{-1}} = MLT^{-1}$$
LHS = RHS
Thus option A is correct.
Question 5
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A man inside a freely falling box throws a heavy ball towards a side wall. The ball keeps on bouncing between the opposite walls of the box. We neglect air resistance and friction. Which of the following figures depicts the motion of the centre of mass of the entire system (man, the ball and the box) ?

A

B

C

D

Solution

Center of mass will go downward.The horizontal movement of ball will not create any change in movement of center of mass as there is no net force acting on body in horizontal direction.
Question 6
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Two bodies, $$A$$ and $$B$$ initially, at rest, move towards each other under mutual force of attraction. At the instant when the speed of $$A$$ is $$v$$ and that of $$B$$ is $$2v$$, the speed of the center of mass of the bodies is

A
$$3 v$$

B
$$2 v$$

C
$$1.5 v$$

D
Zero

Solution

Since no external forces act on the system, the center of mass keeps moving just as it did before the two objects started to exert forces on each other. And since the objects were at rest, their center of mass will remain at the same place. That is the speed of center of mass of the bodies is zero.
Question 7
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The radius of gyration of a disc of mass $$100 g$$ and radius $$5 cm$$ about an axis passing through its centre of gravity and perpendicular to the plane is (in cm)

A
$$0.5$$

B
$$2.5$$

C
$$3.54$$

D
$$6.54$$

Solution

Let radius of gyration be $$k$$
$$\therefore m{ k }^{ 2 }=m{ r }^{ 2 }\\ k=\sqrt { \dfrac { { R }^{ 2 } }{ 2 } } =\sqrt { \dfrac { 25 }{ 2 } } =\dfrac { 5 }{ \sqrt { 2 } } \\ =3.54cm$$
Question 8
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Let 'M' be the mass and 'L' be the length of a thin uniform rod. In first case, axis of rotation is passing through centre and perpendicular to the length of the rod. In second case axis of rotation is passing through one end and perpendicular to the length of the rod. The ratio of radius of gyration in first case to second case is

A
$$1$$

B
$$\dfrac{1}{2}$$

C
$$\dfrac{1}{4}$$

D
$$\dfrac{1}{8}$$

Solution

Moment of inertia of the rod about an axis passing through its centre and perpendicular to the length $$I_1 = \dfrac{1}{12}ML^2$$
Using $$Mk_1^2 = I_1 = \dfrac{1}{12}ML^2$$
We get radius of gyration $$k_1 = \dfrac{L}{\sqrt{12}}$$
Moment of inertia of the rod about an axis passing through its one end and perpendicular to the length $$I_2 = \dfrac{1}{3}ML^2$$
Using $$Mk_2^2 = I_2 = \dfrac{1}{3}ML^2$$
We get radius of gyration $$k_2 = \dfrac{L}{\sqrt{3}}$$
Thus ratio of radius of gyration $$\dfrac{k_1}{k_2} = \dfrac{L/\sqrt{12}}{L/\sqrt{3}}$$
$$\implies$$ $$\dfrac{k_1}{k_2} = \dfrac{1}{2}$$
Question 9
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A force-time graph for the motion of a body is shown in figure.
Change in linear momentum between 0 and 8 s is:

A
Zero

B
4 N-s

C
8 Ns

D
None

Solution

Area under the curve of Force-time graph gives the change in momentum of the body.
Area under the curve Area =$$\dfrac{1}{2}\times 1\times 8 = 4$$
Thus change in momentum of the body between $$0$$ to $$8$$ is $$4 \ Ns$$.
Question 10
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$$ML^2T^{-2}$$ is the dimensional formula for

A
moment of inertia

B
pressure

C
elasticity

D
couple acting on a body

Solution

$$\left[ MOI \right] =\left[ M{ R }^{ 2 } \right] =\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ 0 } \right] \\ \left[ Pressure \right] =\left[ N/{ M }^{ 2 } \right] =\left[ { M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 } \right] \\ \left[ Couple \right] =\left[ N.{ M } \right] =\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 } \right] $$