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

$A$ simple pendulum with an iron 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 $\frac{1}{12}$th that of iron,then the new time period will be:

The length of a seconds pendulum is .... $cm$.

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 is oscillating without damping. When the displacement of the bob is less than maximum,its acceleration vector $\vec a$ is correctly shown in

Identify the correct statement.