Students can practice the best Class 9 Science MCQ and Class 9 Science Exploration Chapter 4 MCQ Online Test with Answers, Describing Motion Around Us for exam preparation.
Class 9 Science Chapter 4 Describing Motion Around Us MCQ
Describing Motion Around Us MCQ
Class 9 Science Chapter 4 MCQ Online Test
Question 1.
The sound of thunder is heard approximately 6 s after the flash of lightning is seen. If speed of sound in air is 346 m/s, then distance of point of lightning is
A. 1200 m
B. 2076 m
C. 3460 m
D. 965 m
Answer:
(B) Distance = Speed × Time
=346 × 6 = 2076m
Question 2.
In 10 min a car whose speed is 35 km/h travels a distance of
A. 5.8 km
B. 3.5 km
C. 14 km
D. 28 km
Answer:
(A) Speed =35 km/h
= 35 x \(\left(\frac{5}{18}\right)\) =9.72 m/s
Time= 10 min =(10 × 60)=600s
Distance = Speed × Time
=9.72 × 600 = 5832m
= 5.832 km
Question 3.
In which of the following cases of motions, the distance moved and the magnitude of displacement are equal? (NCERT Exemplar)
A. If the car 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
Answer:
(A) If the car is moving on a straight road, then the distance moved and the magnitude of displacement are both equal.
Question 4.
A car travels 5 km towards north, then turns right and travels 3 km further, the car again turns right and travel 1 km and comes to rest. What is the distance travelled and displacement of the car? (Competency-Based Question)
A. Distance: 5 km and Displacement: 9 kin
B. Distance: 9 km and Displacement: 5 km
C. Distance: 9 km and Displacement: 7 km
D. Distance: 7 km and Displacement: 9 km
Answer:
(B) Distance travelled = AB + BC + CD
AB = 5 km, BC = 3 km,CD = 1 km
Distance travelled 5 + 3 + 1 = 9 km
Displacement = AD
AD2 = AE2 + ED2 (By using Pythagoras’ theorem)
AE = AB-BE
AB = 5km
BE = CD = 1 km
AE = 5-1 = 4km
ED = BC =3 km
AD2 = 42+ 32
AD2 =16+9, AD2 =25
Taking the square root on both sides, AD = 5 km
Question 5.
The numerical ratio of displacement and distance for a moving object is (NCERT Exemplar)
A. always less than 1
B. always equal to 1
C. always more than 1
D. equal or less than 1
Answer:
(D) Displacement of an object can be less than or equal to the distance covered by the object, because the magnitude of displacement is not equal to distance. However, it can be so if the motion is along a straight line without any change in direction. So, the numerical ratio of displacement to distance is always equal to or less than 1.
Question 6.
A car travels 3 km of distance in 10 min to reach the destination. On the return journey, the car travels the same distance in 15 min. What is the average speed of car in entire journey?
A. 5 m/s
B. 4 m/s
C. 6 m/s
D. 3 m/s
Answer:
(B) Total distance travelled by the car =3 +3=6 km i.e., 6000 m
Total time taken for the whole joining = 10+15 = 25 min
i.e. = 1500 s [1 min = 60 s]
∴ Average speed =\(\frac{\text { Total distance }}{\text { Total time }}=\frac{6000}{1500} \) =4 m/s
Question 7.
A car moves towards South with a speed of 20 ms-1. It changes its direction to East and moves with speed of 20 ms-1 ‘.What is its velocity now? (Competency-Based Question)
A. 20 ms-1 South-East
B. 20\(\sqrt{2} \) ms South-East
C. 10ms-1 South-East
D. 40ms-1 South-East
Answer:
(B) A car moves towards South with speed 20 ms-1, then it changes its direction to East with speed 20 ms-1 as shown in figure below
Using Pythagoras’ theorem,
AE2 = AS2 + SE2
AE = \(\sqrt{A S^2+S E^2} \) =\(\sqrt{(20)^2+(20)^2} \) =20\(\sqrt{2} \)
AE = 20\(\sqrt{2} \) m/s, South-East
Thus, its velocity is 20\(\sqrt{2} \) m/s South-East.
Question 8.
A car travels on a straight road with a velocity of 30 km/h in first one hour and in the next one hour it changes its velocity to 40 km/h. What is the average velocity of the car?
A. 5 km/h
B. 10 km/h
C. 35 km/h
D. 60 km/h
Answer:
(C) Given,v1 =30km/h
v2 = 40km/h
Average velocity = \(\frac{v_1+v_2}{2}=\frac{30+40}{2}=\frac{70}{2} \) =35 km/h
Question 9.
A person rides a motorbike at the speed of 30 m/s. The person applies the brake, and the velocity of motorbike comes down to 20 m/s in 3 s. What is the magnitude of acceleration of motorbike?
A. -3.3 m/s2
B. 6.6 m/s2
C. 10 m/s2
D. -6.6 m/s2
Answer:
(A) Acceleration = – \(\frac{\text { Final velocity }- \text { Initial velocity }}{\text { Time taken }}\)
= \(\frac{v-u}{t} \)
a = \(\frac{20-30}{3} \)
⇒ a = -3.3 m/s2
Question 10.
Four cars A, B, C, and D are moving on a levelled road. Their distance versus time graphs are shown in figure. Choose the correct statement. (Competency-Based Question)
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
Answer:
(B) The slope of distance-time graph represents the speed. From the given graph, it is clear that the slope of distance-time graph for car B is less than all other cars. So, the slope is minimum for car B. Hence, car B is the slowest.
Question 11.
Which of the following figures represents uniform motion of a moving object correctly? (Competency-Based Question)
Answer:
(A) For uniform motion, the distance-time graph is a straight line (because in uniform motion object covers equal distances in equal intervals of time).
Question 12.
Velocity-time graph for a moving object is found to be curved line, then its acceleration is
A. constant
B. variable
C. zero
D. None of these
Answer:
(B) Since the velocity-time graph for a moving object is curved line, this means that object is moving with non-uniformly accelerated motion, therefore body is moving with variable acceleration or non-uniform acceleration.
Question 13.
The slope of a velocity-time graph gives NCERT Exemplar
A . the distance
B. the displacement
C. The acceleration
D. the speed
Answer:
(C) Slope of velocity-time graph gives acceleration.
Because slope of the curve = \(\frac{\text { Velocity }}{\text { Time }}=\frac{v}{t}\)
Acceleration = \(\frac{v-u}{t}\).
Question 14.
In the following figure of velocity-time graph for the motion of the body, the total distance covered by the body from 3 s to 7 s is (Competency Based Question)
A. 28m
B. 56rn
C. 14m
D. 35m
Answer:
(A) Total distance moved by the body from 3s to 7s = Area of shape ABCD
= \(\frac{(A B+D C) \times A D}{2} \)
From the given graph, we have
AB = 4 m/s
DC = 10 m/s
AD = (7 -3) =4 s
= \(\frac{(4+10)(7-3)}{2}\)
= \( \frac{14 \times 4}{2}\) = 28 m
Question 15.
From the given v-t graph (see figure), it can be inferred that the object is (NCERT Exemplar)
A. in uniform motion
B. at rest
C. in non-uniform motion
D. Moving with uniform acceleration
Answer:
(A) In the given graph, we find that the velocity is constant throughout; hence, the object represented in v-t graph is said to be in uniform motion.
Question 16.
Area under a v-t graph represents a physical quantity which has the unit (NCERT Exemplar)
A. m2
B. m
C. m3
D. ms -1
Answer:
(B) Area under velocity-time graph represents Velocity × Time = Displacement, which has the unit as metre (m).
Question 17.
A body moving along a straight line at 10 m/s undergoes an acceleration of 2 m/s2. After 2 s, its speed will be
A. – 8 m/s
B. 12 m/s
C. 14 m/s
D. 16 m/s
Answer:
(C) u = 10 m/s
a = 2 m/s2
⇒ v = u + at = 10 + 2 × 2 = 10 + 4 = 14 m/s
Question 18.
A train starting from rest attains a velocity of 90 km/h in 2 mm, then the distance travelled by the train for attaining this velocity is
A. 1.5 km
B. 2 km
C. 2.5 km
D. 1.2 km
Answer:
(A) Given: initial velocity, u = 0
Final velocity, v = 90 km/h =90 × \(\frac{5}{18} \) = 25 m/s
Time t = 2 min = 2 × 60 = 120s
From the equation of motion,
v = u + at
⇒ a = \(\frac{v}{t} \) [∵ u = 0 ]
= \(\frac{25}{2 \times 60}=\frac{5}{24} \) m/s2
The distance covered is
s = ut +\(\frac{1}{2} \) at2 = 0 + \(\frac{1}{2} \) × \(\frac{5}{24} \) ×(120) 2
= \(\frac{1}{2} \) × \(\frac{5}{24} \) × 120 × 120 = 1500 m = 15 km
Question 19.
If the displacement of an object is proportional to square of time, then the object moves with (NCERT Exemplar)
A. uniform velocity
B. uniform acceleration
C. increasing acceleration
D. decreasing acceleration
Answer:
(B) Let object starts from rest, i.e., initial velocity
(u) = 0 and an acceleration (a) in time (t).
From second equation of motion,
s = ut+ \(\frac{1}{2} \) at2
Then, s = 0 × t + \(\frac{1}{2} \) at2 ⇒ s = \(\frac{1}{2} \) at2
s ∝ t2, if a = Constant
Thus, displacement (s) of an object is proportional to square of time t, then the object moves with constant or uniform acceleration.
Question 20.
A body is thrown vertically upward with velocity u; the greatest height h to which it will rise is (NCERT Exemplar)
A. \(\frac{u}{g} \)
B. \(\frac{u^2}{2 g}\)
C. \( \frac{u^2}{g}\)
D. \(\frac{u}{2 g} \)
Answer:
(B) Initial velocity = u
Final velocity at height h, v =0
From equation of motion v2 – u2 = 2gh
⇒ h = \( \frac{u^2}{2 g}\) (as a =-g)
Question 21.
A particle is moving in a circular path of radius r. The displacement after haifa circle would be (NCERT Exemplar)
A. zero
B. πr
C. 2r
D. 2πr
Answer:
(C) Given, after half the circle, the particle will reach the diametrically opposite point, i.e., from point A to point B, as shown in figure below. We know displacement is shortest path between initial and final point.
∴ Displacement after half circle = AB = OA + OB = r + r = 2r [∵ Given,OA and OB=r]
Hence, the displacement after half circle is 2r.
Question 22.
Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 ms-1. It implies that the boy is (NCEXT Exemplar)
A. at rest
B. moving with no acceleration
C. in accelerated motion
D. Moving with uniform velocity
Answer:
(C) In merry-go-round, the speed is constant, but velocity is not constant, because its direction goes on changing i.e. there is acceleration in the motion.
Question 23.
Which of the following graphs shows that the object is at rest/stationary? (NCEXT Exemplar)
A. (i) and (iii)
B. (ii) and (iv) both
C. Only (ii)
D. Only (iv)
Answer:
C. Only (ii)
(C) For an object at rest, displacement-time graph is a straight line parallel to time axis. Hence, only graph (ii) is correct.
Question 24.
From the given v-t graph (see Figure), it can be inferred that the object is (Competency Based Question)
A. in uniform motion
B. at rest
C. in non-uniform motion
D. Moving with uniform acceleration
Answer:
A. in uniform motion
(A) Since velocity of the object is constant with time, it is in uniform motion.
Question 25.
If a body is moving on a circular path of radius 21 m with velocity of 2 m/s, then time taken by the body to complete half a revolution is
A. 11 s
B. 22 s
C. 44 s
D. 33 s
Answer:
D. 33 s
(D) Given: radius of circular path, r = 21 m
Velocity of the body,v = 2 m/s
Distance travelled in half revolution,
s = πr = \(\frac{22}{7} \times 21\) = 66 m
Hence time taken t = \(\frac{s}{v}=\frac{66}{2} \) = 33 s
Question 26.
A bridge is 400 m long. A 150 m long train crosses the bridge at a speed of 50 m/s. Time taken by the train to cross it.
A. 5s
B.8s
C. 6s
D. 11 said
Answer:
D. 11 said
(D) Time taken by the train to cross will be the total time since the engine enters the bridge and the last coach leaves the bridge.
So, total distance travelled, d = length of bridge + length of train = 400 + 150 = 550 m
Hence, time taken t =\( \frac{d}{v}=\frac{550}{50}\) =11 s.
Question 27.
When two bodies moves uniformly towards each other, then they cross each other at the speed of 10 m/s. If both the bodies move in the same direction, then they cross each other at the speed of 6 m/s. The speed of both bodies are
A. 8 m/s, 2 m/s
B. 8 m/s, 4 m/s
C. 6m/s, 2m/s
D. 6 m/s,4 m/s
Answer:
A. 8 m/s, 2 m/s
(A) Let the speeds of both bodies are v1 and v2.
According to question, when they cross each other, their relative speed is 10 m/s
i.e. v1+ v2 = 10 ……………………….. (i)
When they are moving in same direction, their relative speed is 6 m/s, i.e.
v1 – v2= 6 …………………………….. (ii)
Adding Eqs. (i) and (ii), we get
2v1 =16
⇒ v1 =8 m/s
So weget, v2 = 2 m/s
Describing Motion Around Us Class 9 Assertion and Reason Questions
Directions (Q.Nos. 1- 10) In each of the following questions, a statement of Assertion is given by the corresponding statement of Reason. Of the given statements, mark the correct answer as
A. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C. If Assertion is true but Reason is false.
D. If Assertion is false but Reason is true.
Question 1.
Assertion (A) Displacement of a moving body may be zero when distance travelled by it is not zero.
Reason (R): The displacement is the longer distance between initial and final positions.
Answer:
(C) Displacement may be positive, negative or zero while distance is always positive. The shortest distance between two points is called displacement. Hence Assertion is true, but Reason is false.
Question 2.
Assertion (A): The displacement of an object can be either positive, negative, or zero.
Reason (R): Displacement has both magnitude and direction.
Answer:
(A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion. Displacement may be positive, negative or zero as displacement is a vector quantity.
Question 3.
Assertion (A): Motion with uniform velocity is always along a straight-line path.
Reason (R): In such a motion, speed is the magnitude of the velocity and is equal to the instantaneous velocity.
Answer:
(A) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion. Uniform velocity means that speed and direction remains unchanged.
Question 4.
Assertion (A): Acceleration of a moving body is always positive.
Reason (R): Acceleration of a moving body is the rate of change in velocity with respect to time.
Answer:
(D) Assertion is false because acceleration may be positive, negative, or zero. Hence, Assertion is false, but Reason is true.
Question 5.
Assertion (A): A body can have acceleration even its speed is constant.
Reason (R): In uniform circular motion, speed of body is constant, but its velocity continuously changes.
Answer:
(A) In uniform circular motion, speed of body remains same while its velocity continuously changes due to change in its direction at each point. Hence, body moves with acceleration. Therefore, Assertion and Reason both are true, and Reason is the correct explanation of Assertion.
Question 6.
Assertion (A): Acceleration and displacement are in the opposite direction during retardation.
Reason (R): Acceleration is given as the change in velocity per unit time.
Answer:
(B) Acceleration is detìncd as the rate of change of velocity with respect to time, and negative acceleration is called retardation. Displacement is length of the shortest path measured in direction from initial position to final position of the object. Both Assertion and Reason are true, and Reason is the not correct explanation of Assertion.
Question 7.
Assertion (A): The graph between two physical quantities P and Q is straight line, when P/Q is constant.
Reason (R): The straight line graph means that P is proportional to Q or P is equal to constant multiplied by Q.
Answer:
(A) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Equation of straight line, y = mx,
P = mQ
\(\frac{P}{Q}\) = constant
Therefore, P ∝ Q or P = mQ So, option (A) is correct.
Question 8.
Assertion (A) Distance travelled by a body may be positive, negative or zero.
Reason (R) Shortest distance travelled by the body between two points is called displacement.
Answer:
(D) Distance travelled by a body is always positive. It can never be zero (if body is in motion) or negative. Displacement is defined as the shortest distance between two points. Hence, Assertion is wrong and Reason is true.
Question 9.
Assertion (A) Acceleration of a body can be calculated from velocity-time graph.
Reason (A) Area of velocity-time graph gives displacement of an body.
Answer:
(B) Acceleration is equal to the slope of v-t graph, and the area under v – t graph gives us the displacement. Hence, Assertion and Reason both are true, but Reason is not a correct explanation of Assertion.
Question 10.
Assertion (A): A body performing uniform circular motion with constant speed may have acceleration.
Reason (R): When speed of a body remains constant, then its acceleration is always zero
Answer:
(C) A body in uniform circular motion has constant speed, but it has variable velocity since its direction always keeps changing. So, it has acceleration associated with it. Constant speed doesn’t guarantee a zero acceleration in every case, just like the case above. Hence, Assertion is true, but Reason is wrong.