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.

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

The period of a simple pendulum is measured as $T$ in a stationary lift. If the lift moves upwards with an acceleration of $5g$,the new period will be:

$A$ small sphere oscillates simple harmonically in a watch glass whose radius of curvature is $1.6 \ m$. The period of oscillation of the sphere is (acceleration due to gravity $g = 10 \ m/s^2$) (in $\pi \ s$)

The ratio of the frequencies of two simple pendulums is $4: 3$ at the same place. The ratio of their respective lengths is

$A$ light balloon filled with helium of density $\rho_{He}$ is tied to a light string of length $L.$ The string is tied to the ground forming an "inverted" simple pendulum (figure). If the balloon is displaced slightly from equilibrium as in figure and released, the period of the motion is. Take the density of air to be $\rho_{air}$. Assume the air applies a buoyant force on the balloon but does not otherwise affect its motion.