# CBSE Class 9 Science Exam 2020-21: NCERT Based Important Extra Questions (With Answers) Chapter 8 - Motion

Check NCERT based important extra questions & answers of CBSE Class 9 Science (Chapter 8 Motion). These questions and answers are important for the preparation of upcoming CBSE Class 9 Science Exam 2020-21

Created On: Sep 25, 2020 17:54 IST CBSE Class 9 Science Exam 2020-21: NCERT Based Important Extra Questions (With Answers) Chapter 8 - Motion

Check NCERT based important extra questions & answers of CBSE Class 9 Science (Chapter 8 Motion). These questions and answers are important for the preparation of the upcoming CBSE Class 9 Science Exam 2020-21. These questions can be asked in the exam with slight modifications.

CBSE Class 9 Science Exam 2020-21: NCERT Based Important Extra Questions (With Answers) Chapter 8 - Motion

Question: In which of the following cases of motions, the distance moved and the magnitude of displacement are equal?

(a) A boy is moving on a straight road

(b) A bicycle is moving in circular path

(c) The earth is revolving around the Sun

(d) The pendulum is moving to and fro

(a) A boy is moving on a straight road.

CBSE Class 9 Revised Syllabus for 2020-2021

Question: What is the slope of a velocity – time graph gives?

Answer: The slope of a velocity – time graph gives acceleration.

Question: The displacement of a moving ball in a given interval of time is zero.

Would the distance travelled by the object also be zero? Justify your answer.

No because it might be possible that the moving object returns to its initial position. So the distance travelled is not zero.

Question: The numerical ratio of displacement to distance for a moving object is

(a) always less than one

(b) always equal to one

(c) always more than one

(d) equal or less than one

(d) equal or less than one.

As, Distance ≥ Displacement.

Question: A ball starting from rest travels 20 m in the first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start.

s1 = ut + (½) at2 or 20 = 0 + (½) a (2)2 or a = 10 m s–2

v = u + at = 0 + (10 × 2) = 20 m s–1

s2 = 160 = vt′ + (½) a’ (t’)2 = (20 × 4) + (½) a’ x 16 ⇒ a’ = 10 m s–2

As acceleration is the same, we have v′ = 0 + (10 × 7) = 70 m s–1

Question: What is uniform circular motion?

Answer: If an object moves in a circular path with uniform speed, its

motion is called uniform circular motion.

Question: Fill in the blanks

The __________ of an object is the distance covered per unit time, and ______ is the displacement per unit time.

The speed of an object is the distance covered per unit time, and velocity is the displacement per unit time.

Question: Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.

Question: Obtain a relation for the distance travelled by an object moving with a

uniform acceleration in the interval between 4th and 5th seconds.

Using the equation of motion, s = ut + (½) at2

Distance travelled in 5 s, s = u × 5 + (½)  a x 52 or s = 5 u + (25/2)a ....(i)

Similarly, distance travelled in 4 s, s′ = 4 u + (16/2) a …..(ii)

Distance travelled in the interval between 4th and 5th second = (s – s′) = [u + (9/2) a] unit.

Question: A particle is moving in a circular path of radius r. The displacement after traversing a complete circle would be:

(a) Zero

(b) π r

(c) 2 r

(d) 2π r