A pendulum clock that keeps correct time on the earth is taken to the moon. It will run
[Take $ {g_{moon}} = \dfrac{1}{6}{g_{earth}} $ ]
(A) At correct rate
(B) $ 6 $ times faster
(C) $ \sqrt 6 $ times faster
(D) $ \sqrt 6 $ times slower

Hint: This question can be solved by application of Newton’s law of gravity. The square of the time period depends inversely on the acceleration due to gravity and depends on the separation from the centre.

Formula used: The formulae used in the solution are given here.
The time period of a pendulum is given by, $ T = 2\pi \sqrt {\dfrac{L}{g}} $ where $ L $ is the length of the pendulum from the point of suspension to the centre of the bob and $ g $ is the acceleration due to gravity.

Complete step by step solution:
It has been given that, $ {g_{moon}} = \dfrac{1}{6}{g_{earth}} $ . The significance of this statement is that the value of acceleration due to gravity on the moon is one-sixth the value on earth.
The time period of a simple pendulum is defined as the time taken by the pendulum to finish one full oscillation and is denoted by “ $ T $ ”.
For a pendulum whose length, from the point of suspension to the centre of the bob is $ L $ , and acceleration due to gravity is given by $ g $ ,
Time period is given by the formula, $ T = 2\pi \sqrt {\dfrac{L}{g}} $ .
Since it has been given that, $ {g_{moon}} = \dfrac{1}{6}{g_{earth}} $ ,
The time of the pendulum on the Earth surface = $ {T_{Earth}} = 2\pi \sqrt {\dfrac{L}{{{g_{earth}}}}} $ .
The time of the pendulum on the surface of the moon = $ {T_{Moon}} = 2\pi \sqrt {\dfrac{L}{{{g_{moon}}}}} $ where $ {g_{moon}} $ is the acceleration due to gravity on the moon.
 $ {T_{Moon}} = 2\pi \sqrt {\dfrac{L}{{{g_{moon}}}}} = 2\pi \sqrt {\dfrac{L}{{\dfrac{1}{6}{g_{earth}}}}} $
Simplifying the equation,
 $ {T_{Moon}} = 2\pi \sqrt {\dfrac{{6L}}{{{g_{earth}}}}} $ .
Since, the time period of a pendulum is inversely proportional to the square root of the acceleration due to gravity, thus on decrease of the value of acceleration due to gravity, the time period increases.
This implies that the pendulum becomes slower.
Thus, $ {T_{Moon}} = 2\pi \sqrt {\dfrac{{6L}}{{{g_{earth}}}}} = \sqrt 6 {T_{Earth}} $ .
Thus, the correct answer is Option D.

Note:
The Moon's surface gravity is weaker because it is far less massive than Earth. A body's surface gravity is proportional to its mass, but inversely proportional to the square of its radius.