If the time period of a two meter long simple pendulum is $2\, s$, the acceleration due to gravity at the place where pendulum is executing $S.H.M.$ is

The acceleration due to gravity on the surface of moon is $1.7 \;ms ^{-2}$. What is the time period of a simple pendulum on the surface of moon if its time period (in $sec$) on the surface of earth is $3.5\; s ?( g$ on the surface of earth is $9.8\; ms ^{-2} )$

On a planet a freely falling body takes $2 \,sec$ when it is dropped from a height of $8 \,m$, the time period of simple pendulum of length $1\, m$ on that planet is ….. $\sec$

The bob of a pendulum was released from a horizontal position. The length of the pendulum is $10 \mathrm{~m}$. If it dissipates $10 \%$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is : [Use, $\mathrm{g}: 10 \mathrm{~ms}^{-2}$ ]

A pendulum has time period $T$. If it is taken on to another planet having acceleration due to gravity half and mass $ 9 $ times that of the earth then its time period on the other planet will be