$A$ rectangular block of mass $m$ and area of cross-section $A$ floats in a liquid of density $\rho$. If it is given a small vertical displacement from equilibrium,it undergoes simple harmonic motion with a time period $T$. Then:

$A$ uniform disc of mass $M$ and radius $R$ is suspended in a vertical plane from a point on its periphery. Its time period of oscillation is ........

$A$ small spherical ball of radius '$r$' is rolling on a curved surface which is frictionless and has a radius of curvature '$R$'. Its motion is simple harmonic. Then its time period of oscillation is proportional to ($g=$ acceleration due to gravity).

In both figures shown below, there is a hole along the diameter of the Earth. In the first figure, a particle is released from point $A$ and it oscillates with a time period $T_1$. In the second figure, the same particle is released from point $B$ and it oscillates with a time period $T_2$. Then, [$O$ is the centre of the Earth]

$A$ particle of mass $m$ undergoes oscillations about $x=0$ in a potential given by $V(x) = \frac{1}{2} k x^2 - V_0 \cos \left(\frac{x}{a}\right)$,where $V_0, k, a$ are constants. If the amplitude of oscillation is much smaller than $a$,the time period is given by

$9 \text{ kg}$ of solution is poured into a glass $U$-tube as shown in the figure. The tube's inner diameter is $2 \sqrt{\frac{\pi}{5}} \text{ m}$ and the solution oscillates freely up and down about its position of equilibrium $(x=0)$. The period of oscillation in seconds is (Given: $1 \text{ m}^3$ of solution has a mass $\rho=900 \text{ kg/m}^3$,$g=10 \text{ m/s}^2$,ignore frictional and surface tension effects).