Question

The length of the second pendulum on the surface of earth is 1 m. The length of seconds pendulum on the surface of moon, where g is 1/6th value of g on the surface of earth, is

• A. 1/6 m
• B. 6 m
• C. 1/36 m
• D. 36 m

Answer 1

The correct option is A. 1/6 m.

We know that the time period of a pendulum clock is given by,

T = $2\pi \sqrt{\frac{l}{g}}$ = constant

where $l$ is the length of the pendulum clock,

$g$ is the acceleration due to gravity at that place.

$\Rightarrow \sqrt{\frac{l}{g}}=\text{constant}$ for a second pendulum.

$\frac{l_m}{g_m}=\frac{l_e}{g_e}$

$⟹l_m=\frac{l_e}{g_e}\times g_m=\frac{1}{g}\times \frac{1}{6}\times g=\frac{1}{6}m$.