The time period of a simple pendulum on a freely moving artificial satellite is

$A$ small sphere oscillates simple harmonically in a watch glass whose radius of curvature is $1.6 \ m$. The period of oscillation of the sphere is (acceleration due to gravity $g = 10 \ m/s^2$) (in $\pi \ s$)

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 period of a simple pendulum is measured as $T$ in a stationary lift. If the lift moves upwards with an acceleration of $5g$,the new period will be:

$A$ 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

$A$ simple pendulum keeps correct time at $0^{\circ}C$. At $25^{\circ}C$,it loses $12.5 \, s$ in a day. What is the coefficient of linear expansion of the metal of the pendulum?