Simple Pendulum

The simple pendulum period from NCERT Class 11, Chapter 14 (Oscillations) is $T=2\pi \sqrt{\frac{L}{g}}$, showing that the period depends on the length $L$ and the local gravitational acceleration $g$, but not on the mass of the bob. Taking the pendulum to the Moon leaves $L$ unchanged while $g$ falls to about $g/6$. The ratio of periods is $\frac{T_{moon}}{T_{earth}}=\sqrt{\frac{g}{g/6}}=\sqrt{6}\approx 2.45$. So the period increases by roughly two and a half times, meaning the clock runs slow. The option of decreasing to one-sixth wrongly assumes period scales directly with $g$ rather than with $\sqrt{1/g}$. The option that it is unchanged ignores the $g$ dependence entirely. The option of a factor of six confuses the ratio of $g$ with the ratio of periods, forgetting the square root. Because the pendulum bob's mass cancels out of the equation of motion, only the geometry (length) and the gravitational field strength govern the timing, which is why the same clock behaves differently on the Moon without any change to the pendulum itself. In practical terms, a clock that gained its timekeeping from this pendulum would run about $2.45$ times too slow on the Moon, losing more than half its expected count of swings in a given interval. A magnitude check: weaker gravity gives a gentler restoring pull, so a slower swing and longer period make physical sense, and $\sqrt{6}\approx 2.45$ is the correct scaling with the ratio being dimensionless as required.