At a place,the length of the oscillating simple pendulum is made $\frac{1}{4}$ times keeping the amplitude the same. Then,the total energy will be:

The length of a seconds pendulum is $1 \,m$ on the earth. If the mass and diameter of a planet are double that of the earth, then the length of the seconds pendulum on the planet will be: (in $\,m$)

For a simple pendulum,the graph between $L$ and $T$ will be:

$A$ 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

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

The bob of a simple pendulum is a spherical hollow ball filled with water. $A$ plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation,till water is coming out,the time period of oscillation would