Laws of Motion Test

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Laws of Motion Test - 60

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Question 1
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A rubber ball of mass 50$$\mathrm { g }$$ falls from a height of $$\mathrm { m }$$ and rebounds to a height of .5$$\mathrm { m }$$ . The impulse between the ball and the ground if the time during which they are in contact was $$0,1$$ sec is

A
$$12.5 N - s$$

B
$$11.8 \mathrm { N } - \mathrm { s }$$

C
$$2.36 \mathrm { N } - \mathrm { s }$$

D
$$3.76 \mathrm { N } - s$$

Solution

Given,
mass of ball is $$50g$$
Height is $$1m$$
Rebound at height $$0.5m$$
When ball is falling from height $$1m$$ then initial velocity will be zero. when it strike on ground then Final velocity is ,
$$v^{ 2 }={ u }^{ 2 }+2gh\\ { v }^{ 2 }={ 0 }^{ 2 }+2\times 9.8\times 1\\ v=\sqrt { 19.6 } \\ v=4.4m/s$$
Now when ball rebound then final velocity will be zero therefore initial velocity is
$$v^{ 2 }={ u }^{ 2 }-2gh\\ { 0 }^{ 2 }={ u }^{ 2 }-2\times 9.8\times 0.5\\ u=\sqrt { 9.8 } \\ u=3.13m/s$$

Time of contact$$(t)$$ is $$0.1sec$$
Impulse is change in momentum
$$I=m(v-u)\\ I=\frac { 50 }{ 1000 } (4.4-(-3.13))\\ I=\frac { 50 }{ 1000 } (4.4+3.13)\\ I=0.376N-s$$
Force exerted is $$F=\frac { 0.376 }{ 0.1 } =3.76N$$
Question 2
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The minimum speed of a bucket full of water whirled in a vertical circle of radius $$10 \ m$$ at the highest point so that the water may not fall $$\left( g = 10 \mathrm { ms } ^ { - 2 } \right)$$

A
$$1 \ ms ^ { -1 }$$

B
$$2 \ ms ^ { -1 }$$

C
$$5 \ ms ^ { -1 }$$

D
$$10 \ ms ^ { -1 }$$

Solution

At the highest point the minimum speed should be equal to $$ \sqrt{rg}$$
here $$r= 10m , g= 10m/s^2$$
$$\implies v= \sqrt{10*10} = 10 m/s$$
Question 3
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A car running with a velocity 72 kmph on a level road, is stopped after travelling a distance of 30 m after disengaging its engine $$\left( g={ 10ms }^{ -2 } \right) $$. The coefficient of friction between the road and the tyres is.

A
0.33

B
4.5

C
0.67

D
0.8

Solution

Velocity of car=$$u= 72kmph= 20m/s$$
Stopping distance of car $$= s=30m$$
here accceleration $$(a)$$ is negative because frictional force will be the stopping force and equal to $$-\mu g$$
Final veocity$$v=0$$
Using $$v^2-u^2 =2as$$
$$\implies a=\dfrac{-u^2}{2s}$$
$$\implies -\mu g= \dfrac{-u^2}{2s}$$
$$\implies \mu =\dfrac{u^2}{ 2sg}=\dfrac{400}{600}=0.67$$
Question 4
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A pendulum bob is held stationary by a horizontal force H. The three forces acting on the bob are shown in the diagram.
The tension in the string of the pendulum is T. The weight of the pendulum bob is W. The string is held at an angle of $$30^o$$ to the vertical.
Which statement is correct?

A
$$H=T\cos 30$$

B
$$T=H\sin 30$$

C
$$W=T\sin 30$$

D
$$W=T\cos 30$$

Solution

Take the components of tension $$\vec T = T\cos30^\circ\hat j-T\sin30^\circ\hat i$$
Horizontal force is $$H\hat i$$ and Weight is $$-W\hat j$$
For equilibrium, $$\vec H+\vec W+\vec T=0$$
$$\Rightarrow(H-T\sin30^\circ)\hat i+(T\cos30^\circ-W)\hat j=0$$
$$\Rightarrow H=T\sin30^\circ, \ \ \ W=T\cos30^\circ$$
Question 5
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Impulse indicates.

A
The momentum generated in the direction of force

B
The combined effect of mass and velocity

C
The main characteristics of particle nature

D
Both (b) and (c) are correct

Solution

Impulse is a term that quantifies the overall effect of a force acting over time. It is conventionally expressed in Newton-seconds.
Momentum is a measure of strength and a measure of how difficult it is to stop an object. An object that is not moving has zero momentum. A slow-moving, large object has a large momentum. A fast-moving, small object also has a large momentum. For example, if a ping-pong ball and a bowling ball have the same velocity, then the bowling ball has a greater momentum because it is more massive than the ping-pong ball. Following equation shows the momentum;
p = mv
In this formula, Momentum (p) equals Mass (m) times velocity (v). Momentum is a vector that is equal to the product of mass and velocity (which is also a vector).
from this, it is clear that impulse indicates the momentum generated in the direction of the force.
Thus Option A is the right answer.
Question 6
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The circumference of a circular track is $$400\pi/m$$. If the optimum speed with which a vehicle can be driven along it is to be $$20\ m/s$$, the angle of banking of the road must be about

A
$$\tan^{-1} (0.02)$$

B
$$\tan^{-1} (0.4)$$

C
$$\tan^{-1} (0.8)$$

D
$$\tan^{-1} (0.2)$$

Solution

Given that circumference of circle, $$=400\pi $$ m

So the radius of circle , $$r=\dfrac{400\pi}{2\pi} = 200$$ m

Given that Optimum speed , $$v= 20 m/s$$
Let $$\theta $$ be the angle of banking

$$tan \theta = \dfrac{v^2}{rg}= \dfrac{400}{2000}= \dfrac15 = 0.02$$

$$\theta = tan^{-1} 0.02$$
Question 7
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As shown in figure, ball of of mass $$m \,kg$$ is suspended by a string $$\ell \,cm$$ long. Keeping the string always taut, the ball describes a horizontal circle of radius $$\dfrac{\ell}{2}$$. Calculate the angular speed.

A
$$\sqrt{\dfrac{2g}{3\ell}}$$

B
$$\sqrt{\dfrac{2g}{\sqrt{3}\ell}}$$

C
$$\sqrt{\dfrac{3g}{2\ell}}$$

D
None of these

Solution

Let $$\theta$$ be the angle made by the string and with the vertical
$$\implies tan \theta= \dfrac1 {\sqrt{3}}$$
Let $$T$$ be the tension in the string and $$\omega $$ is the angular speed.

From Free Body Diagram
$$T cos \theta =mg$$
$$T sin \theta =m\omega^2 \dfrac l2 $$

Form above two equations we get

$$\omega =\sqrt{\dfrac{2tan \theta g}{l}}$$
using the value of $$ tan \theta $$

Angular speed , $$\omega= \sqrt{\dfrac{2 g}{\sqrt{3}l}}$$
Question 8
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If $$F = F_{0} (1 - e^{-t/\lambda})$$, the $$F-t$$ graph is

A

B

C

D

Solution
Question 9
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A road of width $$20$$m forms an arc of radius $$15$$ m, its outer edge is $$2$$m higher than its inner edge. For what velocity the road is banked?

A
$$\sqrt{10}$$ m/s

B
$$\sqrt{14.7}$$ m/s

C
$$\sqrt{9.8}$$ m/s

D
None of these

Solution
Question 10
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A cyclist goes round a circular path of length $$400$$m in $$20$$ second. The angle through which he bends from vertical in order to maintain the balance is?

A
$$\sin^{-1}(0.64)$$

B
$$\tan^{-1}(0.64)$$

C
$$\cos^{-1}(0.64)$$

D
None of these

Solution

Option B is correct