Motion(9th Grade > Physics ) Questions and answers for exam Preparation

9th Grade > Physics

MOTION MCQs

Total Questions : 58 | Page 3 of 6 pages

Question 21. A ship was moving at a speed of

500 m s^{- 1}

and it attained speed of 600

m s^{- 1}

in 5 seconds. What is the acceleration of the ship?

20 ms−2
30 ms−2
10 ms−2
50 ms−2

Discuss Question

Answer: Option A. -> 20 ms−2
:
A
Given, initial velocity,

u = 500 m s^{- 1}

, final velocity,

v = 600 m s^{- 1}

and time taken, t = 5 s
Let acceleration be 'a'.
From first equation of motion,
v = u + at

600 = 500 + a \times 5

\Rightarrow a = \frac{600 - 500}{5}

20 m s^{- 2}

Question 22. Average velocity of a particle moving in a straight line, with constant acceleration

a

and initial velocity

u

in first

t

seconds is:

u
u+at
u+at2
u+at2

Discuss Question

Answer: Option C. -> u+at2
:
C
Given, initial velocity is

u

, acceleration is

a

and total time taken is

t

.
Let

s

be the displacement of the particle.
From second equation of motion,

s = u t + \frac{1}{2} a t^{2}

A v e r a g e V e l o c i t y = \frac{T o t a l d i s p l a c e m e n t}{T o t a l t i m e}

∴

A v e r a g e v e l o c i t y = \frac{u t + \frac{{a t}^{2}}{2}}{t}

= u + \frac{a t}{2}

Question 23. If a body is thrown up with an initial velocity

u

and covers a maximum height of

h

, then

h

is equal to

\frac{u^{2}}{2 g}

True
False
It is a scalar quantity.
It is always less than or equal to the distance travelled by the object.

Discuss Question

Answer: Option A. -> True
:
A
Let theinitial velocity be

u

, final velocity be

v

, acceleration be

a

and distance covered be

s

.
Using the equation ,

v^{2}

u^{2} + 2 a s

,
Here

a = - g

[

∵

acceleration due to gravity is acting opposite to the direction of motion]

s = h

At the highest point,

v = 0

∴

0 = u^{2} - 2 g h

\Rightarrow

h = \frac{u^{2}}{2 g}

Question 24. A ball thrown vertically upwards with a speed of

20 m s^{- 1}

from the top of a tower reaches the earth in

8 s

. Find the height of the tower. Take

g = 10 m s^{- 2}

120 m
100 m
60 m
160 m

Discuss Question

Answer: Option D. -> 160 m
:
D
Given:
Initial velocity,

u = 20 m s^{- 1}

Time taken,

t = 8 s

Let the height of the tower be

h

.
Assume the upwards direction to be positive.
The magnitude of the displacement is the height of the tower. Final position (ground)is in downwards (or negative) direction with respect to the initial position (top of tower).

∴ s = - h

Acceleration isin downwards direction.

∴ a = - g = - 10 m s^{- 2}

From the second equation of motion,

s = u t + \frac{1}{2} a t^{2}

\Rightarrow - h = u t - \frac{1}{2} {g t}^{2}

\Rightarrow - h = 20 \times 8 - (\frac{1}{2} \times 10 \times 8^{2})

\Rightarrow h = 160 m .

therefore, height of tower =

160 m

Question 25. Motion under gravity is an example of:

motion with uniform velocity
non-uniform motion
uniform motion
motion with non-uniform acceleration

Discuss Question

Answer: Option B. -> non-uniform motion
:
B
In motion under gravity, there is a constant acceleration. Hence, the velocity changes with time. Thus, the body covers unequal distances in equal intervals of time. This makes it a non-uniform motion.

Question 26. An object starts from rest and moves with uniform acceleration. The displacement of the object is proportional to ___ of time.

thrice
quad
square
twice

Discuss Question

Answer: Option C. -> square
:
C
Let 'u' be the initial velocity, 't' be the time taken, 's' be the displacement and 'a' be the acceleration.
From second equation of motion, we know,

s = u t + \frac{1}{2} a t^{2}

Since the object starts from rest, u = 0
Therefore,

s = \frac{1}{2} a t^{2}

The motion is uniform accelerated motion. Therefore 'a' is constant.

\Rightarrow s \propto t^{2}

Question 27. Choose the correct statement.

The average velocity of a moving body can be zero.
The average velocity of a moving body cannot be zero.
The average speed of a moving body can be zero.
The average speed of a moving body is always less than the average velocity.

Discuss Question

Answer: Option A. -> The average velocity of a moving body can be zero.
:
A
The average velocity of a moving body can be zero when the displacement is zero.

Average velocity = \frac{Total displacement}{Total time}

Average speed can never be zero because the body will cover some distance while moving.

Average speed = \frac{Total distance}{Total time}

Since displacement is the shortest path between the initial and final point, it can never be greater than the distance.
So, the average speed is always greater than or equal to theaverage velocity.

Question 28. A body goes from C to D with a velocity of

20 m s^{- 1}

and comes back from D to C with a velocity of

30 m s^{- 1}

. The average velocity of the body during the whole journey is:

25ms−1
0
20ms−1
26ms−1

Discuss Question

Answer: Option B. -> 0
:
B
Given:
Initial velocity,

u = 20 m s^{- 1}

Final velocity,

v = - 30 m s^{- 1}

Final point of motion is the same as the initial point of motion.
As the body comes back to the initial position, the totaldisplacement is

0

.
Velocity is defined as the rate of change of displacement.
Therefore, average velocity is

0

Question 29.

Which of the following statement(s) is/are true about the displacement of an object?

It can be zero.
It can be greater than the distance travelled by the object.
It is a scalar quantity.
It is always less than or equal to the distance travelled by the object.

Discuss Question

Answer: Option A. -> It can be zero.
:
A and D
The displacement of an object can be zero, for example, when a person moves in a circular path, he essentially comes back to the starting point after completing one full trip, in which case, his displacement is zero.

The magnitude of displacement is always less than or equal to the distance travelled, and would never be greater. It specifies a direction of motion, and so it is a vector quantity, not a scalar.

Question 30.

If a body is thrown up with an initial velocity $u$ and covers a maximum height of $h$ , then $h$ is equal to $\frac{u^{2}}{2 g}$ .

True
False
It is a scalar quantity.
It is always less than or equal to the distance travelled by the object.

Discuss Question

Answer: Option A. -> True
:
A

Let the initial velocity be $u$ , final velocity be $v$ , acceleration be $a$ and distance covered be $s$ .
Using the equation , $v^{2}$ = $u^{2} + 2 a s$ ,
Here $a = - g$
[ $∵$ acceleration due to gravity is acting opposite to the direction of motion]
$s = h$