Motion Test

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Motion Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground.
What is the position of the ball in T/3 seconds?

A
h/9 metre from the ground

B
7h/9 metre from the ground

C
8h/9 metre from the ground

D
17h/18 metre from the ground

Solution
Question 2
1 / -0

Unit of acceleration is

A
$$m/s$$

B
$$ms$$

C
$$m/s^2$$

D
none of these

Solution

acceleration $$ = \dfrac{dv}{dt}$$
Units of 'v' is m/s and 't' is seconds
Therefore units of acceleration is $$\dfrac{m}{s^2} $$
Question 3
1 / -0

A body moving along a circular path will have :

A
a constant speed

B
a constant velocity

C
no tangential velocity

D
a radial acceleration

Solution

Velocity is not constant since the direction of the ball changes every instance.
The ball has radial acceleration since there is a centripetal force acting on it towards the center.
Ball moving in a circular path has tangential velocity. Cutting the thread , the ball moves tangentially
Question 4
1 / -0

(1) In uniform circular motion, tangential acceleration is zero.
(2) In uniform circular motion, velocity is constant.

A
Both 1 and 2 are true and 2 is correct explanation of 1

B
Both 1 and 2 are true and 2 is not correct explanation of 1

C
1 is true and 2 is false

D
1 is false and 2 is true

Solution

In uniform circular motion tangential acceleration is zero because angular velocity of motion is constant.
And the speed is uniform. But velocity of the particle in uniform circular motion keeps on changing.
Question 5
1 / -0

Distance travelled by a freely falling body is proportional to

A
Mass of the body

B
Square of the acceleration due to gravity

C
Square of the time of fall

D
Time of fall

Solution

From the equations of motion
$$ s = ut +\dfrac{1}{2} at^2 $$
$$a=g$$
Assume $$u= 0$$
$$s = \dfrac{1}{2} gt^2$$
From the above equation it is clear that distance travelled is directly proportional to $$ t^2$$
Question 6
1 / -0

The rate of change of displacement with time is

A
speed

B
acceleration

C
retardation

D
velocity

Solution

Correct answer: Option D

Explanation:
Velocity is defined as the rate at which an object's displacement changes over time (displacement over time). Because it has both magnitude (called speed) and direction, velocity is a vector. It is written as,
$$velocity = \dfrac{{Change{\text{ }}in{\text{ }}displacement}}{{time}}$$
Hence, The rate of change of displacement with time is velocity
Question 7
1 / -0

A body dropped from a height reaches the ground in $$5\;s$$. The velocity with which it reaches the ground is ($$g=9.8\; ms^{-2}$$):

A
$$0 \;ms^{-1}$$

B
$$49 \;ms^{-1}$$

C
$$4.9 \;ms^{-1}$$

D
$$9.8 \;ms^{-1}$$

Solution

$$v = u + gt$$
$$t = 5\ sec$$
$$u = 0$$
$$|v| = 5g = 5 \times 9.8 = 49\ m/s$$
Question 8
1 / -0

A vehicle is moving on a road. Ink drops are falling, one at a time, on the road from the vehicle. After the vehicle has moved away, what one observes is shown (qualitatively) in the figure given below. From the figure we can conclude about the vehicle to be moving...

A
From left to right with increasing speed

B
From left to right with decreasing speed

C
From right to left with increasing speed

D
From right to left with decreasing speed

Solution

As we can see the curved surface is towards the left side so the vehicle is moving from left to right and the distance between drops decreases continuously so the speed of the vehicle decreasing.
Question 9
1 / -0

If the positive direction is downward, which situation involves negative velocity and zero acceleration?

A
A rocket slowing down as it moves upward

B
A rocket moving downward at a constant speed

C
A rocket moving upward at a constant speed

D
A rocket speeding up as it moves upward
Question 10
1 / -0

The time taken by a vertically projected body before reaching the ground is:

A
directly proportional to initial velocity

B
directly proportional to square of initial velocity

C
inversely proportional to square of initial velocity

D
inversely proportional to initial velocity

Solution

$$s = ut - \dfrac {1}{2} gt^2$$

For reaching the ground again, $$ s = 0$$
$$\Rightarrow ut - \dfrac {1}{2} gt^2 = 0$$

$$\Rightarrow t = \dfrac {2u}{g}$$

$$\Rightarrow t \propto u $$ since $$g$$ is constant.

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Re-Attempt Weekly Quiz Competition

Motion Test - 19

Result

Motion Test - 19

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SHARING IS CARING

Question 1

h/9 metre from the ground

7h/9 metre from the ground

8h/9 metre from the ground

17h/18 metre from the ground

Solution

Question 2

$$m/s$$

$$ms$$

$$m/s^2$$

none of these

Solution

Question 3

a constant speed

a constant velocity

no tangential velocity

a radial acceleration

Solution

Question 4

Both 1 and 2 are true and 2 is correct explanation of 1

Both 1 and 2 are true and 2 is not correct explanation of 1

1 is true and 2 is false

1 is false and 2 is true

Solution

Question 5

Mass of the body

Square of the acceleration due to gravity

Square of the time of fall

Time of fall

Solution

Question 6

speed

acceleration

retardation

velocity

Solution

Question 7

$$0 \;ms^{-1}$$

$$49 \;ms^{-1}$$

$$4.9 \;ms^{-1}$$

$$9.8 \;ms^{-1}$$

Solution

Question 8

From left to right with increasing speed

From left to right with decreasing speed

From right to left with increasing speed

From right to left with decreasing speed

Solution

Question 9

A rocket slowing down as it moves upward

A rocket moving downward at a constant speed

A rocket moving upward at a constant speed

A rocket speeding up as it moves upward

Question 10

directly proportional to initial velocity

directly proportional to square of initial velocity

inversely proportional to square of initial velocity

inversely proportional to initial velocity

Solution

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