NEET Physics Time Period & Frequency MCQ Question with Solution

A simple pendulum has a time period of 3.5 s on Earth. What is its time period on the Moon, where the acceleration due to gravity is 1.7 m/s²?

A

8.5 s

B

4.9 s

C

1.2 s

D

6.0 s

Correct Answer

Option B

Detailed Explanation

The time period of a simple pendulum is given by T = 2π√(L/g). On the moon, with g = 1.7 m/s², T_moon = T_earth * √(g_earth/g_moon) = 3.5 * √(9.8/1.7) ≈ 4.9 s.

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