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:

$A$ pendulum of length $l = 1\,m$ is released from $\theta_0 = 60^\circ$. The rate of change of speed of the bob at $\theta = 30^\circ$ is ........ $m/s^2$ $(g = 10\,m/s^2)$.

If a simple pendulum is released from the given position,find the velocity of the bob when it reaches the lowest position. (Take $g = 10 \ m/s^2$) (in $m/s$)

If the length of a simple pendulum is increased by $44\%$,then what is the percentage change in the time period of the pendulum (in $\%$)?

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:

If a body dropped freely from a height of $20 \,m$ reaches the surface of a planet with a velocity of $31.4 \,ms^{-1}$, then the length of a simple pendulum that ticks seconds on the planet is (in $\,m$)