Q1Multiple choice
A particle is moving in a circular path of radius r. The displacement after half a circle would be:
- (a)Zero
- (b)π r
- (c)2 r
- (d)2π r
Show answer
(c) 2 r
Chapter 8
Class 9 Science · 24 questions · 23 with answers
Questions
Q2Multiple choice
A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
- (a)u/g
- (b)u2/2g
- (c)u2/g
- (d)u/2g
Show answer
(b) u2/2g
Q3Multiple choice
The numerical ratio of displacement to distance for a moving object is
- (a)always less than 1
- (b)always equal to 1
- (c)always more than 1
- (d)equal or less than 1
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(d) equal or less than 1
Q4Multiple choice
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
Show answer
(b) uniform acceleration
Q5Multiple choice
From the given v – t graph (Fig. 8.1), it can be inferred that the object is
This question refers to a figure in the original PDF.
- (a)in uniform motion
- (b)at rest
- (c)in non-uniform motion
- (d)moving with uniform acceleration Fig. 8.1 16-04-2018
Show answer
(a) in uniform motion
Q6Multiple choice
Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 m s–1. It implies that the boy is
- (a)at rest
- (b)moving with no acceleration
- (c)in accelerated motion
- (d)moving with uniform velocity
Show answer
(c) in accelerated motion
Q7Multiple choice
Area under a v – t graph represents a physical quantity which has the unit
- (a)m2
- (b)m
- (c)m3
- (d)m s–1
Show answer
(b) m
Q8Multiple choice
Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in Fig. 8.2. Choose the correct statement
This question refers to a figure in the original PDF.
- (a)Car A is faster than car D.
- (b)Car B is the slowest.
- (c)Car D is faster than car C.
- (d)Car C is the slowest. Fig. 8.2
Show answer
(b) Car B is the slowest.
Q9Multiple choice
Which of the following figures (Fig. 8.3) represents uniform motion of a moving object correctly? Fig. 8.3
This question refers to a figure in the original PDF.
Show answer
Q10Multiple choice
Slope of a velocity – time graph gives
- (a)the distance
- (b)the displacement
- (c)the acceleration
- (d)the speed 16-04-2018
Show answer
(c) the acceleration
Q11Multiple choice
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 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
Show answer
(a) If the car is moving on straight road
Q12Short answer
The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Justify you answer.
Show answer
No. Though the moving object comes back to its initial position the distance travelled is not zero.
Q13Short answer
How will the equations of motion for an object moving with a uniform velocity change?
Show answer
Acceleration a = 0, v = u s = ut v2 – u2 = 0
Q14Short answer
A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement–time graph is shown in Fig.8.4. Plot a velocity–time graph for the same. Fig. 8.4
This question refers to a figure in the original PDF.
Q15Short answer
A car starts from rest and moves along the x-axis with constant acceleration 5 m s–2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?
Show answer
The distance travelled in first 8 s, x1= 0 + (5) (8)2 = 160 m. At this point the velocity v = u+ at = 0 + (5×8) =40 m s–1 Therefore, the distance covered in last four seconds, x 2 = (40 × 4) m =160 m Thus, the total distance x = x 1+x 2 = (160+ 160) m = 320 m x x
Q16Short answer
A motorcyclist drives from A to B with a uniform speed of 30 km h–1 and returns back with a speed of 20 km h–1. Find its average speed.
Show answer
Let AB = x, So t1=, and t 2 = 30 20 5x Total time = t 1 + t2 = h. Total distance 2 x Average speed for entire journey = = = 24 km h –1 Total Time 5 x
Q17Multiple choice
The velocity-time graph (Fig. 8.5) shows the motion of a cyclist. Find
This question refers to a figure in the original PDF.
- (i)its acceleration
- (ii)its velocity and
- (iii)the distance covered by the cyclist in 15 seconds. Fig. 8.5 MOTION 59 16-04-2018
Show answer
(i) its acceleration
(ii) its velocity and
(iii) the distance covered by the cyclist in 15 seconds. Fig. 8.5 MOTION 59 16-04-2018
Q18Short answer
Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.
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Long Answer Questions
Q19Long answer
An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time?
Show answer
Initial difference in height = (150 – 100) m = 50 m Distance travelled by first body in 2 s = h 1 = 0 + g (2)2 = 2 g Distance travelled by another body in 2 s = h2 = 0 + g (2)2 = 2 g After 2 s, height at which the first body will be = h 1′ = 150 – 2 g After 2 s, height at which the second body will be = h 2′ = 100 – 2 g Thus, after 2 s, difference in height = 150 – 2 g – (100 – 2 g) = 50 m = initial difference in height Thus, difference in height does not vary with time. ANSWERS 129 1 2 1
Q20Long answer
An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start.
Show answer
s1 = ut + at or 20=0+ a (2)2 or a= 10 m s–2, 2 2 v = u + at = 0 + (10 × 2) = 20 m s–1 1 1 s2 = 160 = vt′ + a′ (t′ )2 = (20 × 4) + ( a′ × 16) ⇒ a ′ =10 m s–2 2 2 Since acceleration is the same, we have v ′ =0+ (10 × 7) = 70 m s–1
Q21Long answer
Using following data, draw time - displacement graph for a moving object: Time (s) 0 2 4 6 8 10 12 14 16 Displacement (m) 0 2 4 4 4 6 4 2 0 Use this graph to find average velocity for first 4 s, for next 4 s and for last 6 s.
Show answer
Average velocity for first 4 s. Change in displa cement Average velocity = Total time taken 4−0 4 v= = = 1 m s–1 4−0 4 4−4 0 For next 4 s, v = = = 0 m s–1 8−4 4 (or as x remains the same from 4 to 8 seconds, velocity is zero) 0−6 For last 6 s, v = = –1 m s–1 16 − 10
Q22Multiple choice
An electron moving with a velocity of 5× 104 m s-1 enters into a uniform electric field and acquires a uniform acceleration of 104 m s–2 in the direction of its initial motion.
- (i)Calculate the time in which the electron would acquire a velocity double of its initial velocity.
- (ii)How much distance the electron would cover in this time?
Show answer
(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(ii) How much distance the electron would cover in this time?
Q23Long answer
Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.
Show answer
Using the equation of motion s = ut + at 1 2 Distance travelled in 5 s s = u ×5+ a × 5 or s =5u+ a ——(i) Similarly, distance travelled in 4 s, s′ = 4 u + a——(ii) Distance travelled in the interval between 4th and 5th second = (s – s′) = ( u + a) m u2 - v2
Q24Long answer
Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of u 12 : u 22 ( Assume upward acceleration is –g and downward acceleration to be +g ). 16-04-2018
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We know for upward motion, v 2 = u2 – 2 g h or h = 2g But at highest point v = 0 Therefore, h = u 2g u 12 For first ball, h1 = 2g u 22 and for second ball, h2= 2g u 12 2g 2 h1 u1 Thus h = u 2 = 2 or h 1 : h 2 = u 21 : u 22 2g u2 ANSWERS 131 C hapter 9 _lqucpq Multiple Choice Questions