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 of low 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 A small force F is applied perpendicular to one of the side faces of A. After the force is withdrawn block A executes small oscillations. The time period of which is given by

The correct option is 2π √(M/Lη).

Step-by-step Explanation:

Let us analyze the system consisting of the two cubical blocks, A and B.
Block A is highly rigid and has mass ^M and side length ^L.
Block B has the same dimensions (side length ^L) but has a low modulus of rigidity $\eta$.
The lower face of B is rigidly fixed to a horizontal surface, and the lower face of A is rigidly fixed to the upper face of B.

When a small force ^F is applied horizontally to block A, block B undergoes shear strain. Let ^x be the horizontal displacement of block A (and the upper face of B) relative to the fixed lower face of B.
Since the height of block B is ^L, the shear strain $\theta$ in block B is given by:
$\theta =\frac{x}{L}$

The modulus of rigidity $\eta$ is defined as the ratio of shear stress to shear strain:
$\eta =\frac{\text{Shear Stress}}{\text{Shear Strain}}=\frac{F_{\text{restoring}}/A}{\theta }$
where ^A is the area of the face parallel to the shearing force. For a cube of side length ^L, the surface area of the face is:
$A=L^2$

Substituting the area ^A and shear strain $\theta$ into the formula for $\eta$:
$\eta =\frac{F_{\text{restoring}}}{L^2}·\frac{L}{x}=\frac{F_{\text{restoring}}}{Lx}$

From this, the magnitude of the restoring force $F_{\text{restoring}}$ acting on block A is:
$F_{\text{restoring}}=\eta Lx$
Since this is a restoring force, it acts in the direction opposite to the displacement ^x:
$F=-\left(\eta L\right)x$

Comparing this with the equation for simple harmonic motion ($F=-kx$), the effective force constant (spring constant) ^k of the system is:
$k=\eta L$

The time period ^T of the small oscillations of block A of mass ^M is given by:
$T=2\pi \sqrt{\frac{M}{k}}$

Substituting the value of ^k into the time period formula:
$T=2\pi \sqrt{\frac{M}{L\eta }}$
Hence, the time period of the oscillations is 2π √(M/Lη).