# Measurements and Experimentation Exe-1C Numericals Class-9 ICSE Physics Selina Publishers

Measurements and Experimentation Exe-1C Measurement of Time and Simple Pendulum Numericals Questions Solutions for Class-9 ICSE Concise Physics. There is the solutions of Numericals Questions of simple pendulum of your latest textbook which is applicable in 2023-24 academic sessionVisit official Website CISCE for detail information about ICSE Board Class-9.

## Measurements and Experimentation Exe-1 C Measurement of Time and Simple Pendulum Numericals

(ICSE Class – 9 Physics Concise Selina Publishers)

 Board ICSE Class 9 Subject Physics Writer / Publication Concise selina Publishers Chapter-1 Measurements and Experimentation Exe – 1 C Measurement of Time and Simple Pendulum Topics Solution of Exe-1(C) Numericals Academic Session 2023-2024

### Numericals of Exe-1 C Measurement of Time and Simple Pendulum

Ch-1 Measurements and Experimentation Physics Class-9 ICSE Concise

Page 28

#### Question 1 . A simple pendulum completes 40 oscillations in a minute. Find its (a) Frequency and (b) Time period.

(a) Frequency = Oscillations per second

= (40/60) s-1

= 0.67 s-1

(b) Time period = 1/frequency

= (1/0.67) s

= 1.5 s

#### Question 2. The time period of a simple pendulum is 2 s. What is its frequency? What name is given to such a pendulum ?

Answer : Time period = 2 s

Frequency = 1/time period

= (½)s-1

= 0.5 s-1

Such a pendulum is called the seconds’ pendulum.

#### Question 3 . A seconds pendulum is taken to a place where acceleration due to the gravity falls to one-fourth. How is the time period of the pendulum affected, if at all? Give reasons. What will be its new time period?

Answer : Time period of ‘a’ is inversely proportional to the square root of acceleration due to gravity.

i.e. T∝ 1/√g

Now, if the acceleration due to gravity falls to one-fourth, the time period will be doubled.

Let the new time period be T’ and let g’ be the acceleration due to gravity.

Then,

#### Question 4. Find the length of a second’s pendulum at a place where g = 10 m s-2 (Take π = 3.14)

Answer : Given, g= 10 m/s2 and time period (T) = 2s

Let ‘l’ be the length of the seconds’ pendulum.

#### Question 5 . Compare the time periods of the two pendulums of lengths 1 m and 9 m, respectively.

Answer : Let Tand T2 be the time periods of the two pendulums of lengths 1m and 9m, respectively.

#### Question 6. A pendulum completes two oscillations in 5 s. What is its time period? If g = 9.8 m s-2, find its length.

Answer : Time period = Total time/total no. of oscillations

= (5/2) s

= 2.5 s

Let ‘l’ be the length. Then,

#### Question 7. The time periods of two simple pendulums at a place are in the ratio 2:1. What will be the ratio of their lengths?

Answer : Let Tand T2 be the time periods of the two pendulums of lengths l1 and l2, respectively.

Then, we know that the time period is directly proportional to the square root of the length of the pendulum.

#### Question 8. It takes 0.2 s for a pendulum bob to move from the mean position to one end. What is the time period of the pendulum?

Answer : Time period = Time taken to complete 1 oscillation

= (4 × 0.2) s

= 0.8 s