What happens to the time period of a simple pendulum when it is taken to the moon's surface from the earth's surface?

The graph between the time period $(T)$ and the length $(l)$ of a simple pendulum is

If the length of the oscillating simple pendulum is made $\frac{1}{3}$ times the original, keeping the amplitude the same, then the increase in its total energy at a place will be: (in $\times$)

$A$ simple pendulum of length $l_1$ has time period $T_1$. Another simple pendulum of length $l_2$ $(l_1 > l_2)$ has time period $T_2$. Then the time period of the pendulum of length $(l_1 - l_2)$ will be

The time period of a simple pendulum is $2 \, sec$. If its length is increased $4$ times,then its period becomes ..... $\sec$.

$A$ simple pendulum of period $T$ has a metal bob which is negatively charged. If it is allowed to oscillate above a positively charged metal plate,its period will