Motion Test

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Motion Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

A wooden block of mass $$10$$ gm is dropped from the top of a cliff $$100$$ m high. Simultaneously a bullet of mass $$10$$ gm is fired from the foot of the cliff upward with a velocity $$100$$ m/s. The bullet and the wooden block will meet each other after a time of:

A
10 s

B
0.5 s

C
1 s

D
7 s
Question 2
1 / -0

A ball is thrown upwards. It takes 4 s to reach back to the ground. Find its initial velocity.

A
$$30 ms^{-1}$$

B
$$10 ms^{-1}$$

C
$$40 ms^{-1}$$

D
$$20 ms^{-1}$$

Solution

$$\text{Using}\ v=u+at$$
Time of Flight $$= \dfrac{2u}{g}$$
Where u is the initial velocity and g is acceleration due to gravity
$$4=\dfrac{2u}{10}$$
$$u = 20 \ ms^{-1}$$
Question 3
1 / -0

A boy standing at the top of a tower of 20 $$m$$ height drops a stone. Assuming $$g=10 ms^{-2}$$, the velocity with which it hits the ground is:

A
$$20 ms^{-1}$$

B
$$40 ms^{-1}$$

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

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

Solution

When there is a free fall, we can directly use the equation of motion :
$$v^{2} = u^{2} + 2gh$$
here $$u = 0$$
so, $$v = \sqrt{2gh}$$ = $$\sqrt{2 \times 10 \times 20} = 20\ m/s$$
Question 4
1 / -0

Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of $$10\ ms^{-1}$$. It implies that the boy is :

A
at rest

B
moving with no acceleration

C
in accelerated motion

D
moving with uniform velocity
Solution
As an object moves in a merry-go-round, the speed of the object may be constant, but the direction of the object changes with the motion of the object hence its velocity will also change.
As velocity is changing, therefore the boy will be in an accelerated motion.
Hence option C is correct
Question 5
1 / -0

A body is thrown vertically upwards with a velocity $$u$$, the greatest height $$h$$ to which it will rise is:

A
$$u/g$$

B
$$u^{2}/2g$$

C
$$u^{2}/g$$

D
$$u/2g$$

Solution

The body that is thrown vertically upward with velocity $$u$$ will have final velocity $$v=0$$ at the greatest height $$h$$.
Substituting the given values in the third equation of motion, $$v^2=u^2+2as$$
we have $$0=u^2-2gh$$. (taking $$g$$ in the upward direction)
or $$h=\dfrac{u^{2}}{2g}$$.
Question 6
1 / -0

The velocity of a bullet is reduced from $$100 \ m/s$$ to $$0 \ m/s$$ while travelling through a wooden block of thickness $$10\ cm$$. The retardation, assuming it to be uniform will be:

A
$$5 \times {10^4}\ m/{s^2}$$

B
$$12 \times {10^4}\ m/{s^2}$$

C
$$14 \times {10^4}\ m/{s^2}$$

D
$$1 \times {10^4}\ m/{s^2}$$

Solution

$$v^2 = u^2 +2as$$
$$\text{Substituting}\ v=0, u =100 m/s, s= 10\ cm = 0.1 \ m$$
$$a = - \dfrac {100^2}{0.1\times 2 }$$
$$a=- 5 \times 10^4 \ m/s^2 $$
Question 7
1 / -0

Two bodies, A(of mass 1kg) and B(of mass 3kg),are dropped from heights of 16 m and 25 m respectively. The ratio of the time taken by them to reach the ground is:-

A
$$\dfrac{5} {4}$$

B
$$\dfrac{12} {5}$$

C
$$\dfrac{5} {12}$$

D
$$\dfrac{4} {5}$$

Solution

from equation of motion we know that when a body is dropped
$$h = \dfrac{1}{2}gt^2$$

So $$\dfrac{t_1}{t_2} = \sqrt{\dfrac{h_1}{1_2}} = \sqrt{\dfrac{16}{25}} = \dfrac{4}{5}$$

Hence (D) is correct answer
Question 8
1 / -0

A particle is moving in a circular path of radius $$r$$. The magnitude of displacement after half a circle would be :

A
Zero

B
$$\pi r$$

C
$$2 r$$

D
$$2 \pi r$$

Solution

Displacement is the shortest distance traveled by an object. Here, the object goes around a circle and after half a circle, it would be 2r as it is the diameter of the circle. Thus, the displacement from A to B is 2r for half the circle.
Question 9
1 / -0

A ball is dropped from height $$h$$ and another from $$2h$$. The ratio of time taken by the two balls to reach ground is:

A
$$1:\sqrt{2}$$

B
$$\sqrt{2}:1$$

C
$$2:1$$

D
$$1:2$$

Solution

Using equation of motion
$$s = ut + \dfrac{1}{2}gt^{2}$$,
Given $$s = h, u = 0$$
$$\therefore t = \sqrt{\dfrac{2h}{g}}$$
So, ratio of time taken: $$\dfrac{t_1}{t_2}=\sqrt{\dfrac{h_1}{h_2}}=\sqrt{\dfrac{h}{2h}}$$$$ = $$$$\dfrac{1}{\sqrt 2}$$
Question 10
1 / -0

If a body is thrown up with the velocity of $$15 m/s$$, then maximum height attained by the body is $$(g=10 \ m/s^2)$$

A
$$11.25 m$$

B
$$16.2 m$$

C
$$24.5 m$$

D
$$7.62 m$$

Solution

$$v^2=u^2+2as$$
$$\text{Substituting}\ u =15\ m/s, v=0\ m/s, a=g=-10m/s^2,$$
$$15^2=2\times 10\times s$$
$$s=225/20=11.25m$$

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Re-Attempt Weekly Quiz Competition

Motion Test - 33

Result

Motion Test - 33

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SHARING IS CARING

Question 1

10 s

0.5 s

1 s

7 s

Question 2

$$30 ms^{-1}$$

$$10 ms^{-1}$$

$$40 ms^{-1}$$

$$20 ms^{-1}$$

Solution

Question 3

$$20 ms^{-1}$$

$$40 ms^{-1}$$

$$5 ms^{-1}$$

$$10 ms^{-1}$$

Solution

Question 4

at rest

moving with no acceleration

in accelerated motion

moving with uniform velocity

Solution

Question 5

$$u/g$$

$$u^{2}/2g$$

$$u^{2}/g$$

$$u/2g$$

Solution

Question 6

$$5 \times {10^4}\ m/{s^2}$$

$$12 \times {10^4}\ m/{s^2}$$

$$14 \times {10^4}\ m/{s^2}$$

$$1 \times {10^4}\ m/{s^2}$$

Solution

Question 7

$$\dfrac{5} {4}$$

$$\dfrac{12} {5}$$

$$\dfrac{5} {12}$$

$$\dfrac{4} {5}$$

Solution

Question 8

Zero

$$\pi r$$

$$2 r$$

$$2 \pi r$$

Solution

Question 9

$$1:\sqrt{2}$$

$$\sqrt{2}:1$$

$$2:1$$

$$1:2$$

Solution

Question 10

$$11.25 m$$

$$16.2 m$$

$$24.5 m$$

$$7.62 m$$

Solution

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