Laws of Motion Test

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Laws of Motion Test - 16

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Question 1
1 / -0

A uniform rod $$AB$$ is suspended from a point $$X$$, at a variable distance $$x$$ from $$A$$, as shown. To make the rod horizontal, a mass $$m$$ is suspended from its end $$A$$. A set of $$(m,x)$$ values is recorded. The appropriate variables that give a straight line, when plotted, are:

A
$$m,\cfrac { 1 }{ x } $$

B
$$m,\cfrac { 1 }{ { x }^{ 2 } } $$

C
$$m,x$$

D
$$m,{ x }^{ 2 }$$

Solution

Let, 'l' and 'M' be the length and mass of rod.
for equilibrium, taking moment about X, we get
$$m \times x = M \times (\dfrac{l}{2} - x)$$
this can be written as,
$$m = (\dfrac{Ml}{2}) \dfrac{1}{x} - M$$
comparing with straight line equation,
y = mx + c
we can say that,
$$m \propto \dfrac{1}{x}$$
Question 2
1 / -0

Two identical ladders are arranged as shown in the figure. Mass of each ladder is $$M$$ and length $$L$$. The system is in equilibrium. Find direction and magnitude of friction force acting at A or B.

A
$$ f=(\displaystyle \frac{M+m}{2})g\tan\theta$$ horizontally outwards

B
$$ f=(\displaystyle \frac{M+m}{2})g\cot\theta$$ horizontally inwards

C
$$ f=(\displaystyle \frac{M+m}{2})g\cos\theta$$ horizontally outwards

D
None of these.

Solution

Drawing the FBD of both rods
As the rods are in equilibrium

$$\Sigma F_{X}=0$$

$$\Sigma F_{y}=0$$

$$\tau_{\mathrm{n}\mathrm{e}\mathrm{t}}=0$$
For right rod

$$N_{2}+\displaystyle \frac{mg}{2}+Mg=N$$ (1)

$$N_{1}=f$$ (2)
For left rod

$$N_{2}+N=Mg+\displaystyle \frac{mg}{2}$$ (3)

From $$(1)\mathrm{a}\mathrm{n}\mathrm{d}(3)N_{1}=f$$

$$= N_{2}=0 = N=Mg+\displaystyle \frac{mg}{2}$$

Balancing torque about $$\mathrm{O}$$,

$$ Mg\displaystyle \frac{L}{2}\cos\theta +fL \sin\theta =NL\cos\theta$$

$$\displaystyle \frac{Mg\cos\theta}{2}+f\sin\theta =(Mg+\displaystyle \frac{mg}{2})\cos\theta$$

$$ f=(\displaystyle \frac{Mg}{2}+\frac{mg}{2})\cot\theta = f=(\displaystyle \frac{M+m}{2})g\cot\theta$$
Question 3
1 / -0

STATEMENT-1
It is easier to pull a heavy object than to push it on a level ground.
STATEMENT-2
The magnitude of frictional force depends on the nature of the two surfaces in contact.

A
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

B
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

C
STATEMENT -1 is True, STATEMENT-2 is False

D
STATEMENT -1 is False, STATEMENT-2 is True

Solution

Both statement are independently true but 2 is not the correct explanation of 1.
The reason of statement 1 is that while pulling, a component of force reduces the normal reaction and hence reduces the friction force.
Question 4
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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?

A
$$\sqrt{gR}$$

B
$$\sqrt{2gR}$$

C
$$\sqrt{3gR}$$

D
$$\sqrt{5gR}$$

Solution

To just complete the circle, at the highest point, tension is zero and the gravitational force provides the necessary centripetal force.
Hence, $$\dfrac{mv_{top}^2}{R}=mg.............(i)$$

Applying conservation of energy at the top and bottom points,
$$\dfrac{1}{2}mv_{bottom}^2=mg \times 2R + \dfrac{1}{2}mv_{top}^2.........(ii)$$

Substituting $$mv_{top}^2$$ from (i) in (ii) and solving,
$$v_{bottom}=\sqrt {5gR}$$
Question 5
1 / -0

A ball is suspended by a thread of length L at the point O on a wall which is inclined to the vertical by $$\alpha$$. The thread with the ball is displaced by a small angle $$\beta$$ away from the vertical and also away from the wall. If the ball is released, the period of oscillation of the pendulum when $$\beta > \alpha$$ will be

A
$$\displaystyle \sqrt{\frac{L}{g}}\left[\pi+2sin^{-1}\frac{\alpha}{\beta}\right]$$

B
$$\displaystyle \sqrt{\frac{L}{g}}\left[\pi-2sin^{-1}\frac{\alpha}{\beta}\right]$$

C
$$\displaystyle \sqrt{\frac{L}{g}}\left[2sin^{-1}\frac{\alpha}{\beta}-\pi\right]$$

D
$$\displaystyle \sqrt{\frac{L}{g}}\left[2sin^{-1}\frac{\alpha}{\beta}+\pi\right]$$

Solution

The motion of the string can be represented as an angular displacement from the mean position,given as: $$\theta=\beta sin(\omega t)$$
Time taken from displacement from mean position to $$\beta$$ and back to mean position=$$t_1=\dfrac{T}{2}$$
where $$T$$ is the time period of the oscillation without the wall barrier.
Thus $$t_1=\dfrac{1}{2}\times 2\pi\sqrt{\dfrac{L}{g}}=\pi\sqrt{\dfrac{L}{g}}$$
Time taken to displace from mean position to $$\alpha$$ can be found by:
$$\alpha=\beta sin(\omega t)$$
$$\implies t=\dfrac{1}{\omega}sin^{-1}\dfrac{\alpha}{\beta}$$
$$=\sqrt{\dfrac{L}{g}}sin^{-1}\dfrac{\alpha}{\beta}$$
Thus time taken to displace from mean position to $$\alpha$$ and back to mean position = $$t_2=2\sqrt{\dfrac{L}{g}}sin^{-1}\dfrac{\alpha}{\beta}$$
Thus the time period of the oscillation = $$t_1+t_2=\sqrt{\dfrac{L}{g}}[\pi+2sin^{-1}\dfrac{\alpha}{\beta}]$$
Question 6
1 / -0

An object placed on a ground is in stable equilibrium. If the objects is given a slight push, then initially the position of centre of gravity:

A
moves nearer to ground

B
rises higher above the ground

C
remains as such

D
may remain at same level

Solution

A stable equilibrium is a state in which a body tends to return to its original position after being disturbed.
At the instant of slight pushing the body rises higher above the ground due to increase in its potential energy , but it again tends to lose the potential energy to attain the equilibrium position.
Question 7
1 / -0

The combined effect of mass and velocity is taken into account by a physical quantity called:

A
Moment of momentum

B
Moment of force

C
Momentum

D
None of these

Solution

The force required to stop a moving body depends on the velocity as well as its mass. This combined quantity of mass and velocity is defined as momentum.
$$Momentum = mass \times velocity$$
$$\overrightarrow{P} = m \times \overrightarrow{v}$$
Question 8
1 / -0

When an athlete takes part in a long jump he runs for a while and then jumps. This is because:

A
inertia of motion helps him to take a longer jump.

B
running for a time makes him alert.

C
it helps him to focus on the point of jump.

D
he can apply more force for the jump after running.

Solution

The inertia helps him to retain his horizontal speed and hence the athlete can jump with a greater horizontal velocity to cover a longer distance.
Question 9
1 / -0

Statement A: If the force varies with time then the net force is measured by the total change in momentum of the body.
Statement B: Change in momentum and impulsive force are numerically equal.

A
A & B are correct

B
A & B are wrong

C
A is correct and B is wrong

D
A is wrong and B is correct

Solution

Force is a rate of change of momentum.
$$Force$$ is measured as $$rate\ of\ change\ of\ momentum.$$ It is not equal to change of momentum.
$$F=\dfrac{dM}{dt}$$ or, $$F\cdot dt=dM$$
Impuse of force, $$J=F\cdot dt=dM$$
Hence, Change in momentum and impulsive force are numerically equal.
A is wrong and B is correct.
Question 10
1 / -0

Which of following Newton's laws reveals the underlying symmetry in the forces that occur in nature?

A
First law

B
Second law

C
Third law

D
Law of gravitation

Solution

Newton's third law of motion states that all the forces in nature have an equal and opposite reaction force present. This explains the underlying symmetry in the forces that occur in nature.

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Laws of Motion Test - 16

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Laws of Motion Test - 16

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Question 1

$$m,\cfrac { 1 }{ x } $$

$$m,\cfrac { 1 }{ { x }^{ 2 } } $$

$$m,x$$

$$m,{ x }^{ 2 }$$

Solution

Question 2

$$ f=(\displaystyle \frac{M+m}{2})g\tan\theta$$ horizontally outwards

$$ f=(\displaystyle \frac{M+m}{2})g\cot\theta$$ horizontally inwards

$$ f=(\displaystyle \frac{M+m}{2})g\cos\theta$$ horizontally outwards

None of these.

Solution

Question 3

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

STATEMENT -1 is True, STATEMENT-2 is False

STATEMENT -1 is False, STATEMENT-2 is True

Solution

Question 4

$$\sqrt{gR}$$

$$\sqrt{2gR}$$

$$\sqrt{3gR}$$

$$\sqrt{5gR}$$

Solution

Question 5

$$\displaystyle \sqrt{\frac{L}{g}}\left[\pi+2sin^{-1}\frac{\alpha}{\beta}\right]$$

$$\displaystyle \sqrt{\frac{L}{g}}\left[\pi-2sin^{-1}\frac{\alpha}{\beta}\right]$$

$$\displaystyle \sqrt{\frac{L}{g}}\left[2sin^{-1}\frac{\alpha}{\beta}-\pi\right]$$

$$\displaystyle \sqrt{\frac{L}{g}}\left[2sin^{-1}\frac{\alpha}{\beta}+\pi\right]$$

Solution

Question 6

moves nearer to ground

rises higher above the ground

remains as such

may remain at same level

Solution

Question 7

Moment of momentum

Moment of force

Momentum

None of these

Solution

Question 8

inertia of motion helps him to take a longer jump.

running for a time makes him alert.

it helps him to focus on the point of jump.

he can apply more force for the jump after running.

Solution

Question 9

A & B are correct

A & B are wrong

A is correct and B is wrong

A is wrong and B is correct

Solution

Question 10

First law

Second law

Third law

Law of gravitation

Solution

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