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

9th Grade > Physics

MOTION MCQs

Total Questions : 58 | Page 4 of 6 pages

Question 31.

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 32.

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 is in 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 33.

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 the average velocity.

Question 34.

In a uniform circular motion, which of the following is true?

Velocity is constant and speed is changing.
Velocity and speed both remain constant.
Speed remains constant and velocity changes continuously.
Neither velocity nor speed remains constant.

Discuss Question

Answer: Option C. -> Speed remains constant and velocity changes continuously.
:
C

In a uniform circular motion, the speed remains constant. However, the direction of the body keeps changing. This is due to an acceleration that is perpendicular to the direction of velocity at every point. Hence, the velocity does not remain constant.

Question 35.

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:

$25 m s^{- 1}$
0
$20 m s^{- 1}$
$26 m s^{- 1}$

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 total displacement is

0

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

0

Question 36.

Average velocity of a particle moving in a straight line, with constant acceleration $a$ and initial velocity $u$ in first $t$ seconds is:

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

Discuss Question

Answer: Option C. ->

u + \frac{a t}{2}

:
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 37.

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 38.

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?

$20 m s^{- 2}$
$30 m s^{- 2}$
$10 m s^{- 2}$
$50 m s^{- 2}$

Discuss Question

Answer: Option A. ->

20 m s^{- 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 39.

A coin is thrown in a vertically upward direction with a velocity of $5 m s^{- 1}$ . If the acceleration of the coin during its motion is $10 m s^{- 2}$ in the downward direction, what will be the height attained by the coin and how much time will it take to reach there ?

$s = 1.5 m, t = 0.25 s$
$s = 0.25 m, t = 1.5 s$
$s = 1.25 m, t = 0.75 s$
$s = 1.25 m, t = 0.5 s$

Discuss Question

Answer: Option D. ->

s = 1.25 m, t = 0.5 s

:
D

Given, initial velocity, $u = 5 m s^{- 1}$ , acceleration, $a = - 10 m s^{- 2}$ .
Let ' $v$ ' be the final velocity and ' $s$ ' be the height attained.
At the highest point of motion, $v = 0$
Using third equation of motion, $v^{2} = u^{2} + 2 a s,$
$0 = 5^{2} + [2 \times (- 10) \times s]$
$\Rightarrow s = 1.25 m$
Hence the height attained by the coin is $1.25 m$ .
Let the time taken be $t$ .
Using the first equation of motion,
$v = u + a t$
$0 = 5 - 10 t$
$\Rightarrow t = 0.5 s$
$∴$ Time taken by the coin is $0.5 s$ .

Question 40.

A particle moves 3 m north, then 4 m east and finally 6 m south. What is the distance travelled and the displacement of the particle?

The distance travelled is 13 m.
The distance travelled is 12 m.
The displacement is 13 m.
The displacement is 5 m.

Discuss Question

Answer: Option A. -> The distance travelled is 13 m.
:
A and D

Let the particle start from A and finally reach B as shown. Distance is the length of the actual path taken by it.
In this case,
distance = 3 m + 4 m + 6 m = 13 m
Displacement is the shortest distance between the initial and final position.
Here, displacement = AB
Using Pythagoras theorem,
AB $= \sqrt{(B C^{2} + A C^{2})}$
$= \sqrt{(3^{2} + 4^{2})} m = 5 m$ .