Motion Test

Result

Motion Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

Speed of a body depends on

A
Time

B
Mass of the body

C
Distance travelled by the body

D
Both A and C

Solution

Speed is defined as

$$Speed = \dfrac{\text{Distance travelled}}{\text{Time interval}}$$

Hence speed depends both on time and distance travelled by the body.
So both A and C are correct.
Question 2
1 / -0

If a body is thrown up with an initial velocity u and covers a maximum height of h, then h is equal to

A
$$\dfrac{u^2}{2g}$$

B
$$\dfrac{u}{2g}$$

C
$$2u^2g$$

D
None of these

Solution

At the maximum height , velocity of the ball is zero
$$v^2 = u^2 - 2gh $$
v $$ = 0$$
$$ u^2 = 2gh$$
$$h = \dfrac{u^2}{2g} $$
Question 3
1 / -0

Average speed of a body is

A
Total Distance $$\div $$ Total Time

B
Total Distance + Total Time

C
Both (a) and (b)

D
None of these

Solution

We know that,
The average speed of a body is the ratio of total distance and total time.
So, Average speed = $$\dfrac{\text{Total distance}}{\text{Total time}}$$
$$v_{av}=\dfrac{D}{T}$$
Here,
$$D$$ = Total distance
$$T$$ = Total time
Hence, Option A is the correct.
Question 4
1 / -0

A particle is projected up with a velocity of $$\sqrt{29}\ ms^{-1}$$ from a tower of height 10m. Its velocity on reaching the ground is $$ x $$ $$\ ms^{-1}$$. Find $$ x $$.

A
5

B
10

C
15

D
20

Solution
Question 5
1 / -0

$$N/kg$$ is the unit of :

A
Retardation

B
Acceleration

C
Rate of change of velocity

D
All of these

Solution

Acceleration or retardation $$a = \dfrac{Force}{Mass}$$
Thus $$N \ kg^{-1}$$ is the unit of retardation as well as acceleration.
Also, acceleration is defined as the rate of change of velocity.
Thus $$N \ kg^{-1}$$ is also the unit of rate of change velocity.
Hence Option D.
Question 6
1 / -0

In the graph provided, the velocity

A
increases between point 0 and A

B
increases between point A and B

C
decreases between points A and B

D
is zero throughout

Solution

Slope of velocity-time graph gives the acceleration of the body. Since the slope of the given v-t graph is negative, this means the body is decelerating. Hence, the velocity of the body decreases in going from A to B.
Question 7
1 / -0

A stone is projected up with a velocity of $$4.9\ m/s$$ from the top of a tower and it reaches the ground after 3 s. Then the height of that tower is

A
4.9 m

B
19.6 m

C
9.8 m

D
29.4 m

Solution

From second equation of motion, $$s=ut+at^2/2$$
Assume upward direction to be positive.
$$u=4.9\ m/s,\ t=3\ sec,\ a=-9.8\ m/s^2$$
$$s=4.9\times 3-9.8\times 3^2/2$$
$$s =-29.4\ m$$
Height = 29.4 m
Question 8
1 / -0

The relationship between average speed, time and distance is

A
Average speed = distance $$\times $$ time

B
Average speed $$= \cfrac{\text{total distance}}{\text{total time}}$$

C
$$\text{Time} = \cfrac{\text{average speed}}{\text{distance}}$$

D
$$\text{Distance} = \text{average speed } \times \text{time}$$

Solution

Average speed is the ratio of total distance to total time.
Question 9
1 / -0

Two balls are dropped from height $$h_1$$ and $$h_2$$ respectively. What is the ratio of their velocities on reaching the ground is?

A
$$\sqrt{\dfrac{h_2}{h_1}}$$

B
$$\sqrt{\dfrac{h_1}{h_2}}$$

C
$$\dfrac{h_2}{h_1}$$

D
$$\dfrac{h_1}{h_2}$$

Solution

From 3rd equation of motion,
$$v^2=u^2+2as$$
Given $$u=0$$ for both the balls since they are dropped.
Hence, $$v \propto \sqrt{s}$$
$$\dfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\dfrac { \sqrt { { h }_{ 1 } } }{ \sqrt { { h }_{ 2 } } } $$
Question 10
1 / -0

A freely falling body from rest acquires velocity V falling through a distance h. After the body falls through a further distance h velocity acquired by it is

A
2v

B
v

C
$$\sqrt{2}$$v

D
4v

Solution

Using the equations of motion
$$v^2 = u^2 + 2gh $$ (u is 0)
$$v^2 = 2gh $$
$$V = \sqrt{2gh} $$

$$v^2 = u^2 + 2gh_2 $$
$$V_2^2 = 4gh$$
$$V_2^2 = 4gh = 2V^2 $$
$$V_2 = \sqrt{2} V $$

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Re-Attempt Weekly Quiz Competition

Motion Test - 35

Result

Motion Test - 35

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Rank

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SHARING IS CARING

Question 1

Time

Mass of the body

Distance travelled by the body

Both A and C

Solution

Question 2

$$\dfrac{u^2}{2g}$$

$$\dfrac{u}{2g}$$

$$2u^2g$$

None of these

Solution

Question 3

Total Distance $$\div $$ Total Time

Total Distance + Total Time

Both (a) and (b)

None of these

Solution

Question 4

5

10

15

20

Solution

Question 5

Retardation

Acceleration

Rate of change of velocity

All of these

Solution

Question 6

increases between point 0 and A

increases between point A and B

decreases between points A and B

is zero throughout

Solution

Question 7

4.9 m

19.6 m

9.8 m

29.4 m

Solution

Question 8

Average speed = distance $$\times $$ time

Average speed $$= \cfrac{\text{total distance}}{\text{total time}}$$

$$\text{Time} = \cfrac{\text{average speed}}{\text{distance}}$$

$$\text{Distance} = \text{average speed } \times \text{time}$$

Solution

Question 9

$$\sqrt{\dfrac{h_2}{h_1}}$$

$$\sqrt{\dfrac{h_1}{h_2}}$$

$$\dfrac{h_2}{h_1}$$

$$\dfrac{h_1}{h_2}$$

Solution

Question 10

2v

v

$$\sqrt{2}$$v

4v

Solution

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