Questions :
1. In which of the following cases of motions, the distance moved and the magnitude of displacement are equal?
(a) If the car is moving on a straight road.
(b) If the car is moving in circular path.
(c) The pendulum is moving to and fro.
(d) The earth is revolving around the sun.
2. A particle is moving in a circular path of radius r, the displacement after half a circle would be:
(a) zero
(b) πr
(c) 2r
(d) 2πr
3. Slope of velocitytime graph gives
(a) the distance
(b) the displacement
(c) the acceleration
(d) the speed
4. A body is thrown vertically upward with velocity u, the greatest height ‘h’ to which it will rise is
(a) u/g
(b) u^{2}/2g
(c) u^{2}/g
(d) u/2g
5. If the displacement of an object is proportional to square of time, then the object moves with
(a) uniform velocity
(b) uniform acceleration
(c) increasing acceleration
(d) decreasing acceleration
6. From the vt graph, it can be inferred that the object is
(a) in uniform motion
(b) at rest
(c) in nonuniform motion
(d) moving with uniform acceleration
7. Suppose a boy is enjoying a ride on a merrygoround which is moving with a constant speed by 10 ms^{1}. It implies that the boy is
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity
8. A Circular motion is accelerated?
(a) Always
(b) When speed varies continue sly
(c) When circular motion stops
(d) When speed remains constant
9. 39. If the velocitytime graph of a particle is a curve (not straight line) then its acceleration is
(a) variable
(b) constant
(c) zero
(d) negative
10. Magnitude of displacement is equal to the distance covered by an object that means the particle is in
(a) circular motion
(b) zigzag motion
(c) straight line path
(d) parabolic path
11. 41. An object travels 16 m in 4s and then another 16 m in 2s. Its average speed is
(a) 6 ms^{1}
(b) 5 ms^{1}
(c) 8 ms^{1}
(d) 5.33 ms^{1}
12. The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of a trip. If the time taken is 10 h the average speed is
(a) 40 ms^{1}
(b) 40 kmh^{1}
(c) 50 kmh^{1}
(d) 13.9 ms^{1}
13. The distance covered by a Car when he starts running with acceleration of 0.5ms^{2}) till he achieve speed of 10m/s.
(a) 200m
(b) 120m
(c) 150m
(d) 100m
14. The distance covered by a Car when he starts running with acceleration of 0.5 ms^{2} till he achieve speed of 10m/s.
(a) 200m
(b) 120m
(c) 150m
(d) 100m
Chapter 8 Motion Quiz 3 Class 9th
Quizsummary
0 of 14 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
Information
Click on ‘Start Quiz’ to Take Test.
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 14 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
 Not categorized 0%
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 Answered
 Review

Question 1 of 14
1. Question
In which of the following cases of motions, the distance moved and the magnitude of displacement are equal?
Correct
Incorrect

Question 2 of 14
2. Question
A particle is moving in a circular path of radius r, the displacement after half a circle would be:
Correct
Displacement is the shortest distance between two points. So in circular motion after half a circle, the particle will reach diametrically opposite position
from where it started. Therefore, the displacement will equal to the length of the diameter i.e. 2r.
Incorrect
Displacement is the shortest distance between two points. So in circular motion after half a circle, the particle will reach diametrically opposite position
from where it started. Therefore, the displacement will equal to the length of the diameter i.e. 2r.

Question 3 of 14
3. Question
Slope of velocitytime graph gives
Correct
The slope of the graph is tan θ = perpendicular/base = velocity/time. So, it will give acceleration of the object.
Incorrect
The slope of the graph is tan θ = perpendicular/base = velocity/time. So, it will give acceleration of the object.

Question 4 of 14
4. Question
A body is thrown vertically upward with velocity u, the greatest height ‘h’ to which it will rise is
Correct
s = h, u = u, v = 0 and a = g
Here, the displacement will be maximum height. Acceleration will be downward direction = g and at maximum height, its final velocity will be zero.
s = (v^{2} – u^{2})/2a
so,h = (0 – u^{2}) / 2(g) ⇒ h = u^{2} / 2g
Incorrect
s = h, u = u, v = 0 and a = g
Here, the displacement will be maximum height. Acceleration will be downward direction = g and at maximum height, its final velocity will be zero.
s = (v^{2} – u^{2})/2a
so,h = (0 – u^{2}) / 2(g) ⇒ h = u^{2} / 2g

Question 5 of 14
5. Question
The numerical ratio of displacement to distance for a moving object is
Correct
Incorrect

Question 6 of 14
6. Question
If the displacement of an object is proportional to square of time, then the object moves with
Correct
Incorrect

Question 7 of 14
7. Question
From the vt graph, it can be inferred that the object is
Correct
Incorrect

Question 8 of 14
8. Question
Suppose a boy is enjoying a ride on a merrygoround which is moving with a constant speed by 10 ms^{1}. It implies that the boy is
Correct
Incorrect

Question 9 of 14
9. Question
A Circular motion is accelerated?
Correct
In a circular motion the direction of motion keep on changing,
Hence velocity keep on changing, therefor it is a accelerated motion always.
Incorrect
In a circular motion the direction of motion keep on changing,
Hence velocity keep on changing, therefor it is a accelerated motion always.

Question 10 of 14
10. Question
If the velocitytime graph of a particle is a curve (not straight line) then its acceleration is
Correct
Incorrect

Question 11 of 14
11. Question
Magnitude of displacement is equal to the distance covered by an object that means the particle is in
Correct
Incorrect

Question 12 of 14
12. Question
An object travels 16 m in 4s and then another 16 m in 2s. Its average speed is
Correct
Total distance covered = 16 + 16 = 32 m
Total time = 4 + 2 = 6 s
Average speed = 32/6 = 5.33 m/s
Incorrect
Total distance covered = 16 + 16 = 32 m
Total time = 4 + 2 = 6 s
Average speed = 32/6 = 5.33 m/s

Question 13 of 14
13. Question
The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of a trip. If the time taken is 10 h the average speed is
Correct
Total distance covered = 2400 – 2000 = 400 km
Time taken = 10 hrs
Average speed = 400/10 = 40 km/hr
Incorrect
Total distance covered = 2400 – 2000 = 400 km
Time taken = 10 hrs
Average speed = 400/10 = 40 km/hr

Question 14 of 14
14. Question
The distance covered by a Car when he starts running with acceleration of 0.5ms^{2} till he achieve speed of 10m/s.
Correct
Incorrect