Laws of Motion Test

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Laws of Motion Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

A stone, tied at the end of a string $$80$$ cm long, is whirled in a horizontal circle with a constant speed. If the stone makes $$14$$ revolutions in $$25$$ sec, what is the magnitude of acceleration of the stone.

A
$$680$$ cm$$/s^2$$

B
$$720$$ cm$$/s^2$$

C
$$860$$ cm$$/s^2$$

D
$$990$$ cm$$/s^2$$

Solution

Given length of string = 80 cm and angular velocity $$\omega $$ = 2$$\pi $$f=$$\dfrac { 88 }{ 25 } $$
And we know centripetal acceleration =$$r{ \omega }^{ 2 }$$
a=$$80\times { (\frac { 88 }{ 25 } ) }^{ 2 }$$
=991.232 cm/s
=990 cm/s
Question 2
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A bullet of mass $$50\ g$$ travelling at $$500\ m/s$$ penetrates $$100\ cm$$ into a wooden block. Find the average force exerted on the block.

A
$$0.625\times 10^{4}\ N$$

B
$$62.5\times 10^{4}\ N$$

C
$$6.25\times 10^{4}\ N$$

D
$$2.65\times 10^{4}\ N$$

Solution

Mass, $$m = 0.05\ kg$$
Initial velocity, $$u = 500\ m/s$$
Distance covered, $$S = 100 cm = 1\ m$$

Final velocity $$v=0$$

Now, from the equation of motion

$$ {{v}^{2}}-{{u}^{2}}=2as $$

$$ 0-{{\left( 500 \right)}^{2}}=2\times a\times 1 $$

$$ a=-125000\,m/{{s}^{2}} $$

Now, the average force exerted on the block

$$ F=ma $$

$$ F=0.05\times 125000 $$

$$ F=6.25\times {{10}^{3}} $$

$$ F=0.625\times {{10}^{4}}\,N $$

Hence, the average force is $$0.625\times {{10}^{4}}\,N$$
Question 3
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A field gun of mass 150 $$kg$$ fires a shell of mass 1 kg with velocity of 150 m/s. Calculate the velocity of the recoil of the gun.

A
1 m/sec

B
3 m/sec

C
1.5 m/sec

D
5 m/sec

Solution
Question 4
1 / -0

A ball of weight $$0.1$$ kg coming with a speed $$30$$ m/s strikes a bat and returns in opposite direction with speed of $$40$$ m/s, then impulse is

A
$$-0.1 \times 40 -0.1 \times 30 Ns$$

B
$$0.1 \times 40 -0.1 \times (-30) Ns$$

C
$$0.1 \times 40 +0.1 \times (-30) Ns$$

D
$$0.1 \times 40 -0.1 \times 20 Ns$$

Solution

Given,
$$m=0.1kg$$
$$u=30m/s$$
$$v=-40m/s$$
Impulse = change in momentum
$$ mv_{2}−mv_{1}= 0.1\times 40−0.1\times(−30)$$
The correct option is B
Question 5
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Velocity of a particle of mass $$2\ kg$$ varies with time $$t$$ according to the equation $$\vec {v} = (2t\hat {i} + 4\hat {j}) m/s$$. Here $$t$$ is in seconds. Find the impulse imparted to the particle in the time interval from $$t = 0$$ to $$t = 2s$$.

A
$$8\hat {i} N-s$$

B
$$10\hat {i} N-s$$

C
$$12\hat {i} N-s$$

D
$$16\hat {i} N-s$$

Solution

velocity vector $$=v=(2t\hat i+4\hat j)\ m/s$$

since acceleration is the rate of change in velocity,

$$\therefore a=\dfrac{dv}{dt}=\dfrac{d(2t\hat i+4\hat j)\ m/s}{dt}=2\hat i\ m/s^2$$

$$\therefore$$ particle has constant acceleration. This means particle is acts upon by constant force.

$$\therefore F=ma=2\ kg\times (2\hat i)\ m/s^2=4\hat i\ N$$

Since a constant force acts on the particle the impulse imparts to the particle is given by :-

Impulse $$=\hat i \Delta t=(4\hat i N)\times (2s-0s)=8\hat i\ N\ s$$

$$\therefore$$ option $$A$$ is correct.
Question 6
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Two balls of same mass are dropped from the same height h , on to the floor . the first ball bounces to a height h/4, after the collection & the second ball to a height h/16. the impulse applied by the first & second ball on the floor are $${I_1}$$ and $${I_2}$$ respectively . then

A
$$5{I_1} = 6{I_2}$$

B
$$6{I_1} = 5{I_2}$$

C
$$3{I_1} = 2{I_2}$$

D
$$2{I_1} = {I_2}$$

Solution

When the ball is dropped from height h its velocity $${v}_{1}=\sqrt{2gh}$$
Now velocity of the ball after the bounce, where height $$h=\dfrac{h}{4}$$,velocity$${v}_{1a}=\sqrt{2g\dfrac{h}{4}}$$
Now impulse $${I}_{1}={F}_{1}t=mat=m({v}_{1}-{v}_{1a})=m(\sqrt { 2gh } -\sqrt { 2g\dfrac { h }{ 4 } } )=m({ v }_{ 1 }-\dfrac { 1 }{ 2 } { v }_{ 1 })=\dfrac { 1 }{ 2 } m{ v }_{ 1 }$$
Similarly for the ball when reaches a height of $$\dfrac{h}{4}$$ after bounce,velocity$${v}_{2a}=\sqrt{2g\dfrac{h}{16}}$$ and velocity of the ball before
bounce$${v}_{2}=\sqrt{2gh}$$
Similarly the impulse of the second ball$${I}_{2}=\dfrac{3}{4}{v}_{2}m$$
Given mass of both the ball are same.
We can compare both the impulses as
$$\dfrac{4{I}_{2}}{3{v}_{2}}=\dfrac{2{I}_{2}}{{v}_{1}}$$
Now as both $${v}_{2}={v}_{1}=\sqrt{2gh}$$
$$3{I}_{1}=2{I}_{2}$$
Question 7
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Two particles A and B equal masses are respectively tied to the centre and one end of a string, whose other end 0 is fixed. Both the particles always revolve in concentric circles of centre O. The ratio of tensions in both the parts of the string will be

A
3:2

B
2:3

C
1:2

D
1:1

Solution
Question 8
1 / -0

A stone of mass $$1\ kg$$ is tied to a string $$4\ m$$ long and is rotated at constant speed of $$40\ ms^{-1}$$ in a vertical circle. The ration of the tension at the top and the bottom is

A
11:12

B
39:41

C
41:39

D
12:11

Solution
Question 9
1 / -0

Circular flexible current loop of radius R carrying current I is placed in an inward magnetic field B. If we spin the loop with angular speed $$/omega $$, then tension in string

A
Is zero

B
is more than iBR

C
is less than iBR

D
Does not depend on rotation

Solution
Question 10
1 / -0

A 40 N block is supported by two ropes. One rope is horizontal and the other makes an angle of $${30^ \circ }$$ with the ceiling. The tension in the rope attached to the ceiling is approximately:

A
80 N

B
40 N

C
34.6 N

D
46.2 N

Solution

In $$x-$$ direction:
$$T_{1}=T_{2}\cos 30^{o}\rightarrow (i)$$
In $$y-$$ direction
$$T_{2}\sin 30^{o}=40$$
$$T_{2}=80\ N$$
Option $$A$$