If the length of a second's pendulum is decreased by $2\%$,how many seconds will it lose per day?

$A$ 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 ........ $m/s^2$ $(g = 10\,m/s^2)$.

The linear displacement $x$ of the bob of a simple pendulum from its mean position varies as $x = a \sin \left(\frac{\pi}{\sqrt{2}} t\right)$,where $a$ is its amplitude expressed in meters and $t$ is in seconds. The length of the simple pendulum is (Take $g = \pi^{2} \ m/s^{2}$): (in $m$)

$A$ pendulum suspended from the ceiling of a train has a period $T$,when the train is at rest. When the train is accelerating with a uniform acceleration $a$,the period of oscillation will

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:

$A$ simple pendulum with a metallic bob has a time period $T$. The bob is now immersed in a non-viscous liquid and oscillated. If the density of the liquid is $1/4$ that of the metal,the time period of the same pendulum will be ...........