Physics-Time Period & Frequency
NEET Physics Time Period & Frequency MCQ Question
Type: MCQ-numerical-Hard-Class 11
A simple pendulum has a time period of 2 s on the surface of Earth. What would be its time period on the surface of the Moon, where the acceleration due to gravity is 1.7 m/s²?
A
4.8 s
B
5.1 s
C
3.5 s
D
2.5 s
Correct Answer
Option A
Detailed Explanation
The time period of a pendulum is given by T = 2π√(L/g). On the Moon, g is smaller, so the time period increases. Using the given values and the formula, T_moon = 2π√(L/1.7), which results in approximately 4.8 s.
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