Two simple pendulums have first $(A)$ bob of mass $M_1$ and length $L_1$,second $(B)$ of mass $M_2$ and length $L_2$. Given $M_1 = M_2$ and $L_1 = 2 L_2$. If their total energies are the same,then which of the following statements is correct?

In the case of a simple pendulum executing $SHM$,at $t=0$,the bob is not at the mean position. The graph drawn between the tension $(T)$ in the string and time $(t)$ is:

$A$ simple pendulum of time period $1\,s$ and length $l$ is hung from a fixed support at $O$. The bob is at a distance $H$ vertically above $A$ on the ground. The amplitude is $\theta_0$. The string snaps at $\theta = \frac{\theta_0}{2}$. Find the time taken by the bob to hit the ground and the distance from $A$ where the bob hits the ground. Assume $\theta_0$ to be small,so that $\sin \theta_0 \approx \theta_0$ and $\cos \theta_0 \approx 1$.

For a simple pendulum, the graph of time period $(T)$ versus length $(l)$ is depicted by:

$A$ simple pendulum of length $1\,m$ is allowed to oscillate with an amplitude of $2^o$. It collides elastically with a wall inclined at $1^o$ to the vertical. Its time period will be: (use $g = \pi^2$)

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 $\times$)