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 $m/s$)

$A$ cylindrical block of wood (density $= 650 \ kg \ m^{-3}$),of base area $30 \ cm^2$ and height $54 \ cm$,floats in a liquid of density $900 \ kg \ m^{-3}$. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length ..... $cm$ (nearly).

$A$ simple pendulum has a time period $T_1$. The point of suspension is now moved upward according to the equation $y = kt^2$,where $k = 1\,m/s^2$. If the new time period is $T_2$,then the ratio $\frac{T_1^2}{T_2^2}$ will be:

$A$ simple pendulum consisting of a ball of mass $m$ tied to a thread of length $l$ is made to swing on a circular arc of angle $\theta$ in a vertical plane. At the end of this arc,another ball of mass $m$ is placed at rest. The momentum transferred to this ball at rest by the swinging ball 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

The time period of a simple pendulum of length $L$ as measured in an elevator descending with acceleration $\frac{g}{3}$ is