The amplitude of a simple pendulum is $10 \ cm$. When the pendulum is at a displacement of $4 \ cm$ from the mean position,the ratio of kinetic and potential energies at that point is

$A$ cylindrical block of wood (density $= 650 \ kg \ m^{-3}$),of base area $30 \ cm^2$ and height $54 \ cm$,floats in a liquid of density $900 \ kg \ m^{-3}$. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length ..... $cm$ (nearly).

In the following table, the relation of the graph in column-$I$ and the shape of the graph in column-$II$ is shown. Match them appropriately.

Column-$I$	Column-$II$
$(a)$ $T^2 \to l$	$(i)$ Linear
$(b)$ $T^2 \to g$	$(ii)$ Parabolic
$(c)$ $T \to l$	$(iii)$ Hyperbolic

Let $l_1$ be the length of a simple pendulum. Its length changes to $l_2$ to increase the periodic time by $20 \%$. The ratio $\frac{l_2}{l_1}$ is:

$A$ simple pendulum having length $\ell$ has a speed of $\sqrt{2g\ell}$ at the bottom-most point of its trajectory. Its motion will be:

The ratio of the period of oscillation of a conical pendulum to that of a simple pendulum is: (Assume the strings are of the same length in both cases and $\theta$ is the angle made by the string with the vertical in the case of the conical pendulum.)