Solveeit Logo

Question

Physics Question on Gravitation

The time period of a simple pendulum on the surface of the earth is 4s4 \,s. Its time period on the surface of the moon is

A

4s4\, s

B

8s8\,s

C

10s10\,s

D

12s12\,s

Answer

10s10\,s

Explanation

Solution

Acceleration due to gravity on the surface of the moon is 16\frac{1}{6} that of surface of the earth gm=16ge(i)\therefore g_{m}=\frac{1}{6}g_{e} \ldots\left(i\right) On earth, Te=2πLgeT_{e}=2\pi\sqrt{\frac{L}{ge}} On moon, Tm=2πLgmT_{m}=2\pi\sqrt{\frac{L}{g_{m}}} TmTe=gegm=6\therefore \frac{T_{m}}{T_{e}}=\sqrt{\frac{g_{e}}{g_{m}}}=\sqrt{6} (Using (i)) Tm=6TeT_{m}=\sqrt{6}T_{e} =6×4s=\sqrt{6}\times4\,s =10s=10\,s