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

9th Grade > Physics

MOTION MCQs

Total Questions : 58 | Page 6 of 6 pages

Question 51.

What is the acceleration of a van that starts from rest and attains a speed of $20 m s^{- 1}$ in $25 s$ ?

$2 m s^{- 2}$
$0.8 m s^{- 2}$
$7.5 m s^{- 2}$
$4 m s^{- 2}$

Discuss Question

Answer: Option B. ->

0.8 m s^{- 2}

:
B
Given:
Initial velocity,

u = 0 m s^{- 1}

Final velocity,

v = 20 m s^{- 1}

Time,

t = 25 s

From the first equation of motion,

v = u + a t

20 = 0 + a \times 25

\Rightarrow a = \frac{20}{25} = 0.8 m s^{- 2}

Question 52.

A runner completes one round of a circular track of diameter 400 m in 40 s. What will be the distance covered and the displacement respectively at the end of 2 minutes 20 s?

4400 m, 200 m
4400 m, 400 m
2200 m, 400 m
2200 m, 800 m

Discuss Question

Answer: Option B. -> 4400 m, 400 m
:
B

A Runner Completes One Round Of A Circular Track Of Diameter...

Given:
Diameter of the circle,

d = 400 m

Time taken to complete one circle,

T = 40 s

Total time of motion,

T_{2} = 2 min 20 s

Circumference of circle,

C = π d

Speed is defined as the ratio of distance covered to time taken.
Hence, speed of the runner

v = \frac{C}{T} = \frac{π d}{T}

Total time of motion,

T_{2} = 2 min 20 s = 140 s

Total distance covered is given by:

D = v T_{2}

D = π d \frac{T_{2}}{T}

D = π \times 400 \times \frac{140}{40}

D \approx 4400 m

No. of rounds completed:

n = \frac{Total distance covered}{Circumference of the circle}

n = \frac{D}{π d}

n = \frac{T_{2}}{T}

n = \frac{1400}{40} = 3.5

This means that he will be at the diametrically opposite end, i.e., he starts at A and ends at B.
Hence displacement,

s = 2 r = 400 m

Question 53.

A block is sliding down on an inclined plane. The velocity changes at a constant rate from $10 c m s^{- 1}$ to $15 c m s^{- 1}$ in $2$ seconds. What is its acceleration?

$2.5 c m s^{- 2}$
$1.5 c m s^{- 2}$
$4.5 c m s^{- 2}$
$1.25 c m s^{- 2}$

Discuss Question

Answer: Option A. ->

2.5 c m s^{- 2}

:
A

The motion is depicted below

Let $u$ be the initial velocity, $v$ be the final velocity and $t$ be the time taken.
$v = 15 c m s^{- 1}, u = 10 c m s^{- 1}, t = 2 s$
We know that acceleration, $a$ is given by:
$a = \frac{v - u}{t}$
$a = \frac{15 - 10}{2}$ $c m s^{- 2}$
$a = 2.5 c m s^{- 2}$

Question 54.

What is the distance travelled by a body and its displacement when it completes one revolution around a circular track with radius r?

Distance and displacement are both same and equal to 0
Distance and displacement are both same and equal to 2πr
Distance = 0 and displacement = 2πr
Distance = 2πr and displacement = 0

Discuss Question

Answer: Option D. -> Distance = 2πr and displacement = 0
:
D

Distance only depends on the magnitude and is equal to the total path covered.
Here, distance = perimeter of the circular track = 2πr

Displacement depends on both magnitude and direction.
It depends only on the initial and final positions of the body and when you join initial and final points with a straight line, it gives the displacement of the body.
Here initial and final positions are the same, hence displacement is zero.

Question 55.

A bus decreases its speed from $80 k m p h$ to $60 k m p h$ in 5 seconds. Find the acceleration of the bus.

$2.22 m s^{- 2}$
$2.11 m s^{- 2}$
$1.11 m s^{- 2}$
0

Discuss Question

Answer: Option C. ->

1.11 m s^{- 2}

:
C
Initial speed,

u = 80 k m p h

80 \times \frac{1000}{60 \times 60} = \frac{800}{36} m s^{- 1}

Final speed,

v = 60 k m p h

60 \times \frac{1000}{60 \times 60} = \frac{600}{36} m s^{- 1}

Time taken

t = 5 s

From the first equation of motion,

a = \frac{v - u}{t}

= \frac{600 - 800}{36 \times 5}

= - 1.11 m s^{- 2}

.
Here sign of acceleration is negative which shows that the speed is decreasing and acceleration is in the opposite direction of the motion.

Question 56.

Which of the following statement is correct for a round trip along the same route?

The average speed is zero.
The average velocity is zero.
Both the average velocity and average speed are non-zero.
The average velocity is half of the average speed.

Discuss Question

Answer: Option B. -> The average velocity is zero.
:
B
The average speed is given by the ratio of the total distance travelled to the total time taken.
During a round trip,
The total distance travelled = length of the path

\neq 0

∴

the average speed

\neq 0

Also, the object comes back to the same point from where it has started the journey. So, the total displacement of the object is zero. The average velocity is given by the ratio of the total displacement to the total time taken.

∴

the average velocity

= 0

Hence, the average velocity is zero for a round trip.

Question 57.

A boy covers 6 metres in the first 10 seconds and another 6 metres in the next 10 seconds at same speed in the same direction. What can you surely tell about the motion of the boy?

Curvilinear motion
Uniform motion
Non-uniform motion
The boy is at rest

Discuss Question

Answer: Option B. -> Uniform motion
:
B
Uniform motion is when object covers equal distances in equal intervals of time in the same direction.
In this case, the boy covers equal distances in equal intervals of time because his speed does not change.
Also, his direction does not change.
Therefore, the motion of the boy is an example for uniform motion.

Question 58.

In a uniform circular motion, the direction of velocity at any point is _____.

towards the centre
along the circumference
random
along the tangent

Discuss Question

Answer: Option D. -> along the tangent
:
D
Uniform circular motion is described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. At all instances, the object is moving tangent to the circle. Since the direction of the velocity is the same as the direction of the object's motion, the velocity is directed tangent to the circle as well.