Question

If the ratio of the acceleration due to gravity on the surface of earth to that on the surface of the moon is 6:1, then the ratio of the periods of a simple pendulum on their surfaces is

• A. 1:1
• B. 1:6
• C. 1:3
• D. 1:√6
• E. 1:√3

Answer 1

Answer: (E)

The period \( T \) of a simple pendulum is given by the formula: 

\[ T = 2\pi \sqrt{\frac{L}{g}} \] Where: - \( T \) is the period, - \( L \) is the length of the pendulum, - \( g \) is the acceleration due to gravity. We are given that the ratio of the acceleration due to gravity on the surface of Earth to that on the surface of the Moon is \( 6:1 \). This means: \[ \frac{g_{\text{Earth}}}{g_{\text{Moon}}} = 6 \] The ratio of the periods of the pendulum on the Earth and the Moon is given by: \[ \frac{T_{\text{Earth}}}{T_{\text{Moon}}} = \frac{2\pi \sqrt{\frac{L}{g_{\text{Earth}}}}}{2\pi \sqrt{\frac{L}{g_{\text{Moon}}}}} \] Simplifying: \[ \frac{T_{\text{Earth}}}{T_{\text{Moon}}} = \sqrt{\frac{g_{\text{Moon}}}{g_{\text{Earth}}}} = \sqrt{\frac{1}{6}} = \frac{1}{\sqrt{6}} \] Therefore, the ratio of the periods is \( 1 : \sqrt{6} \).

Correct Answer:

Correct Answer: (E) \( 1 : \sqrt{3} \)