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 session**. **Visit 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.

**Answer :**

**(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/s^{2} 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 T_{1 }and T_{2} 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 T_{1 }and T_{2} be the time periods of the two pendulums of lengths l_{1} and l_{2}, 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

**Question 9. **How much time does the bob of a second’s pendulum take to move from one extreme to the other extreme of its oscillation to the other extreme ?

**Answer :**

Time period of a seconds’ pendulum = 2 s

Time taken to complete half oscillation, i.e. from one extreme to the other extreme = 1 s.

— : End of Measurements and Experimentation Exe-1C Measurement of Time and Simple Pendulum Numericals Questions :–

Return to **Concise Selina Physics ICSE Class-9 Solutions**

Thanks

Please share with your friends