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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

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

Answer:

(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.

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

Answer:

(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.

Answer:

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.

Answer:

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.

Answer:

https://www.jagranjosh.com/imported/images/E/Articles/CBSE-9th-Science-Exam-2020-21-ch-8-answer-18.jpg

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

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

Answer:

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

Answer:

(a) Zero. 


NCERT Solutions for Class 9 Maths: Download chapter-wise solutions in PDF

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