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

Answer: (A)

1 / 6 m

Answer 2

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

$\Rightarrow l \propto g;$ ⇒ $\frac{l_{m}}{1} = \frac{1}{6}\frac{g}{g} \Rightarrow l_{m} = \frac{1}{6}m$