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

The time period ($T$) of a simple pendulum is given by the formula:
$$T=2\pi \sqrt{\frac{l}{g}}$$

where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.

On the surface of the Moon, the acceleration due to gravity is approximately one-sixth of that on Earth:

$$g_{\text{moon}}=\frac{g_{\text{earth}}}{6}$$

Substituting this value into the expression for the time period on the Moon ($T_{\text{moon}}$):

$$T_{\text{moon}}=2\pi \sqrt{\frac{l}{g_{\text{earth}}/6}}=\sqrt{6}\cdot 2\pi \sqrt{\frac{l}{g_{\text{earth}}}}=\sqrt{6}{T}_{\text{earth}}$$

Since the time period increases by a factor of $\sqrt{6}$, the pendulum takes longer to complete each oscillation. Therefore, the clock ticks slower, meaning it runs $\sqrt{6}$ times slower.