Question

A pendulum clock that keeps the correct time on the earth is taken to the moon. It will run

• A. at correct rate
• B. 6 times faster
• C. $\surd 6$ times faster
• D. $\surd 6$ times slower

Answer 1

The correct option is D

$\surd 6$ times slower

Step (1) Expression for the time period of the pendulum on the earth:

On the earth, time period, $T=2\pi \sqrt{\frac{l}{g}}$

Step (2) Expression for the time period of the pendulum on the moon:

As we know that the value of $g$ on the moon is $\frac{1}{6}$th of that of the value on the earth.

On the moon, the time period of the pendulum,

$$\begin{gathered} T’=2\pi \sqrt{\frac{l}{{\displaystyle \frac{g}{6}}}} \\ =2\pi \sqrt{\frac{6l}{g}} \end{gathered}$$

Step (3) Calculating the ratio of T' to T:

$$\begin{gathered} \left(\frac{T’}{T}\right)=\frac{\left(2\pi \sqrt{{\displaystyle \frac{6l}{g}}}\right)}{\left(2\pi \sqrt{{\displaystyle \frac{l}{g}}}\right)} \\ T\text{'}=\sqrt{6}T \end{gathered}$$

Hence, the answer is (D) $\surd 6$ times slower