Let $l_1$ be the length of a simple pendulum. Its length changes to $l_2$ to increase the periodic time by $20 \%$. The ratio $\frac{l_2}{l_1}$ is:

$A$ 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 angular velocity and the amplitude of a simple pendulum are $\omega$ and $a$ respectively. At a displacement $X$ from the mean position,if its kinetic energy is $T$ and potential energy is $V$,then the ratio of $T$ to $V$ is:

Two pendulums have time periods $T$ and $\frac{5T}{4}$. They start $S.H.M.$ at the same time from the mean position. What will be the phase difference between them after the bigger pendulum has completed one oscillation (in $^o$)?

If a simple pendulum is taken to a place where $g$ decreases by $4\%$,then the time period

$A$ small sphere of radius $r$ is placed on a concave surface of radius of curvature $R$ a little away from the center. When the sphere is released,it oscillates. Assuming the oscillation to be simple harmonic motion and $r << R$,then the time period is