Question

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

• A. $4\, s$
• B. $8\,s$
• C. $10\,s$
• D. $12\,s$

Answer 1

The Correct Option is C

Acceleration due to gravity on the surface of the moon is $\frac{1}{6}$ that of surface of the earth $\therefore g_{m}=\frac{1}{6}g_{e} \ldots\left(i\right)$ On earth, $T_{e}=2\pi\sqrt{\frac{L}{ge}}$ On moon, $T_{m}=2\pi\sqrt{\frac{L}{g_{m}}}$ $\therefore \frac{T_{m}}{T_{e}}=\sqrt{\frac{g_{e}}{g_{m}}}=\sqrt{6}$ (Using (i)) $T_{m}=\sqrt{6}T_{e}$ $=\sqrt{6}\times4\,s$ $=10\,s$