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Law of Motion MCQ question and answer with solution | Physics Gk MCQs

Learn Law of Motion MCQ Questions and answers with easy and logical explanations in Physics Gk . Law of Motion MCQ question provides all type of technical mcq questions which is important for SSC-CGL, UPPSC, NDA, CDS and all govt examinations Page-3 section-2

Each section contains 100 MCQs on Law of Motion .
21
If the velocity of a car increases from 5 meters per second to 10 meters per second in 5 seconds, then what is its acceleration?
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Answer & Solution
Answer: Option D
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22
Two parallel train tracks run from North to South. On one track train A is moving with a speed 30 m/s from North to South. An the other track a train B is moving with speed 50 m/s from South to North. The relative velocity of train A with respect to train B is
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Answer & Solution
Answer: Option B
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23
A particle is moving in a uniform circular motion with constant speed v along a circle of radius r. The acceleration of the particle is
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Answer & Solution
Answer: Option D
Solution:
The question is asking about the acceleration of a particle moving in a circle at a constant speed.

Important Point: Even though the speed is constant, the velocity isn't! Velocity includes both speed AND direction. Since the particle is moving in a circle, its direction is constantly changing.

Because the direction is changing, the particle is accelerating. This acceleration is called centripetal acceleration, and it always points towards the center of the circle.

Now, the formula for centripetal acceleration is:

a = v2 / r

Where:
* a is the centripetal acceleration
* v is the speed of the particle
* r is the radius of the circle

So, let's look at the options:

Option A: zero - Incorrect because we know there *is* acceleration.
Option B: v/r - Incorrect. This is not the correct formula.
Option C: v/r2 - Incorrect. This is not the correct formula.
Option D: v2/r - Correct! This matches the formula for centripetal acceleration.

Therefore, the answer is D. The acceleration of the particle is v2/r.
24
A particle moves with uniform acceleration along a straight line from rest. The percentage increase in displacement during 6th compared to that in 5th is about
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Answer & Solution
Answer: Option B
Solution:
The question asks about the percentage increase in displacement of a particle moving with uniform acceleration.
It compares the distance covered in the 6th second to the distance covered in the 5th second.

Understanding the Concept:
Since the particle starts from rest and has uniform acceleration, it means its velocity increases at a constant rate.
The distance covered in each successive second will keep increasing.

Formula to Use:
The distance traveled in the *n*th second (Sn) can be calculated using the formula: Sn = u + (a/2)(2n - 1)
Where:
* u = initial velocity (0 in this case, since it starts from rest)
* a = acceleration
* n = the second you are interested in

Calculations:
Distance in 5th second (S5) = 0 + (a/2)(2*5 - 1) = (a/2) * 9 = 4.5a
Distance in 6th second (S6) = 0 + (a/2)(2*6 - 1) = (a/2) * 11 = 5.5a

Percentage Increase:
Percentage increase = [(S6 - S5) / S5] * 100
Percentage increase = [(5.5a - 4.5a) / 4.5a] * 100
Percentage increase = (a / 4.5a) * 100
Percentage increase = (1 / 4.5) * 100
Percentage increase ≈ 22.22%

Answer:
Therefore, the percentage increase is approximately 22%. So the correct answer is Option B: 22%
25
The slope of a distance-time graph shows . . . . . . . .
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Answer: Option D
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26
If the distance S covered by a moving car in rectilinear motion with a speed v in time t is given by s = vt, then the car undergoes
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Answer: Option C
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27
Which of the following statements is true for Newton's Third Law of Motion.
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Answer: Option C
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28
The rate of change of displacement is called-
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Answer: Option D
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29
The mechanical gain of lever is
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Answer: Option A
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30
Nuclear force is
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Answer: Option D
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