The metallic bob of a simple pendulum has a relative density of $5$. The time period of this pendulum is $10 \, s$. If the metallic bob is immersed in water,then the new time period becomes $5 \sqrt{x} \, s$. The value of $x$ will be.

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

$A$ simple pendulum has a time period $T$ in air. Its time period when it is completely immersed in a liquid of density one-eighth the density of the material of the bob is:

The time period of a simple pendulum is $T$. When the length is increased by $10 \ cm$, its period is $T_1$. When the length is decreased by $10 \ cm$, its period is $T_2$. Then, the relation between $T, T_1$, and $T_2$ is

$A$ simple pendulum of length $1\, m$ is oscillating with an angular frequency $10\, rad/s$. The support of the pendulum starts oscillating up and down with a small angular frequency of $1\, rad/s$ and an amplitude of $10^{-2}\, m$. The relative change in the angular frequency of the pendulum is best given by

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 $\times$)