A highly rigid cubical block A of small mass M and side L is fixed rigidly onto another cubical block B of the same dimensions and modulus of rigidity η such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force is applied perpendicular to one of the sides faces of A. After the force is withdrawn, block A execute small oscillations the time period of which is given by

The correct option is 2π √(M/ηL).

To find the time period of oscillations of block A, we need to determine the restoring force acting on it when it is displaced by a small horizontal distance.
Let us analyze the setup:
Block A has a mass $M$ and side length $L$. It is fixed rigidly on top of block B, which has the same dimensions (side length $L$) and a modulus of rigidity $\eta$.
Since block A is highly rigid, it does not deform. However, block B deforms when a force is applied to block A.

When block A is displaced horizontally by a small distance $x$, block B undergoes shear deformation. The upper face of block B (which is attached to block A) is displaced by $x$, while its lower face remains fixed to the horizontal surface.

The shear strain $\theta$ in block B is given by the ratio of displacement to the height of block B:
$\theta =\frac{x}{L}$

The cross-sectional area of the face of block B undergoing shear is:
$A=L^2$

By definition, the modulus of rigidity $\eta$ is the ratio of shear stress to shear strain:
$\eta =\frac{\text{Shear Stress}}{\text{Shear Strain}}=\frac{F/A}{\theta }$

Substituting the values of $A$ and $\theta$ into the expression:
$\eta =\frac{F/L^2}{x/L}=\frac{F}{Lx}$

From this, the restoring force $F$ developed due to the elasticity of block B is:
$F=-\eta Lx$

This restoring force is directly proportional to the displacement $x$ and acts in the opposite direction. This indicates that block A executes Simple Harmonic Motion (SHM). Comparing this with the standard SHM restoring force equation $F=-kx$, we find the effective force constant $k$ to be:
$k=\eta L$

The time period $T$ of a simple harmonic oscillator of mass $M$ is given by:
$T=2\pi \sqrt{\frac{M}{k}}$

Substituting the value of $k$ into the formula yields the final expression for the time period:
$T=2\pi \sqrt{\frac{M}{\eta L}}$