Choose the correct length $(L)$ versus square of time period $(T^2)$ graph for a simple pendulum executing simple harmonic motion.

If the length of the oscillating simple pendulum is made $\frac{1}{3}$ times the original, keeping the amplitude the same, then the increase in its total energy at a place will be: (in $\times$)

$A$ simple pendulum is attached to the roof of a lift. If the time period of oscillation when the lift is stationary is $T$,then the frequency of oscillation when the lift falls freely will be

The period of a simple pendulum is doubled,when

Answer the following questions:
$(a)$ The time period of a particle in $SHM$ depends on the force constant $k$ and mass $m$ of the particle: $T=2 \pi \sqrt{\frac{m}{k}}$. $A$ simple pendulum executes $SHM$ approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
$(b)$ The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation,a more involved analysis shows that $T$ is greater than $2 \pi \sqrt{\frac{l}{g}}$. Think of a qualitative argument to appreciate this result.
$(c)$ $A$ man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
$(d)$ What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

$A$ second's pendulum is placed in a space laboratory orbiting around the earth at a height $3R$,where $R$ is the radius of the earth. The time period of the pendulum is