The time period of oscillations of a simple pendulum is $1$ minute. If its length is increased by $44 \%$,then its new time period of oscillation will be ......... $s$.

$A$ pendulum bob has a speed of $3\, m/s$ at its lowest position. The pendulum is $0.5\, m$ long. The speed of the bob,when the length makes an angle of $60^o$ with the vertical,will be ..... $m/s$ (If $g = 10\, m/s^2$)

The time period of a simple pendulum will be doubled if we:

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

$A$ tunnel has been dug through the centre of the Earth and a ball is released in it. It will reach the other end of the tunnel after:

The time period of a simple pendulum of length $L$ is $T_1$. The time period of a uniform rod of the same length $L$ suspended from one end and oscillating in a vertical plane is $T_2$. The amplitude of oscillation is small in both cases. Then the ratio $\frac{T_1}{T_2}$ is: