$A$ simple pendulum is hanging from the ceiling of a lift. When the lift is at rest,the time period of the pendulum is $T$. If the resultant acceleration becomes $g/4$,then the new time period of the pendulum is: (in $, T$)

$A$ 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 $\cos \theta_0 \approx 1$.

$A$ simple pendulum of length $l_1$ has time period $T_1$. Another simple pendulum of length $l_2$ $(l_1 > l_2)$ has time period $T_2$. Then the time period of the pendulum of length $(l_1 - l_2)$ will be

At a given place,to increase the number of oscillations made by a simple pendulum in one minute from $72$ to $90$,the length of the pendulum is to be decreased by (in $\%$)

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 time period of a simple pendulum inside a stationary lift is $T$. When the lift starts accelerating upwards with an acceleration of $\frac{g}{3}$,the time period of the pendulum will be