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 amplitude of an oscillating simple pendulum is $10 \ cm$ and its period is $4 \ s$. Its speed $1 \ s$ after it passes its equilibrium position is ... $m/s$.

$A$ simple pendulum of time period $1\,s$ and length $l$ is hung from a fixed support at $O$. The bob is at a distance $H$ vertically above $A$ on the ground. The amplitude is $\theta_0$. The string snaps at $\theta = \frac{\theta_0}{2}$. Find the time taken by the bob to hit the ground and the distance from $A$ where the bob hits the ground. Assume $\theta_0$ to be small,so that $\sin \theta_0 \approx \theta_0$ and $\cos \theta_0 \approx 1$.

If the time period of a $2 \ m$ long simple pendulum is $2 \ s$,the acceleration due to gravity at the place where the pendulum is executing $S.H.M.$ is

$A$ hollow sphere is filled with water through a small hole in it. It is then hung by a long thread and made to oscillate. As the water slowly flows out of the hole at the bottom, the period of oscillation will

$A$ simple pendulum with a bob of mass $m$ oscillates from $A$ to $C$ and back to $A$ such that $PB$ is $H$. If the acceleration due to gravity is $g$,then the velocity of the bob as it passes through $B$ is