A highly rigid cubical block A of mass M and side L is fixed rigidly on the another cubical block of same dimensions and of modulus of rigidity `rho` 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 A small force F is applied perpendicular to one of the side faces of A. After the force is withdrawn, block A executes small oscillations, the time period of which is given by

Options

A. `2 pi sqrt(M rho L )`
B. `2pi sqrt((M rho //L)`
C. `2pi sqrt((M L //rho)) `
D. `2 pi sqrt((M//rhoL))`

Explanation

To solve the problem, we need to determine the time period of the oscillations of block A after the force F is withdrawn. The key parameters involved are the mass of the block (M), the side length of the block (L), and the modulus of rigidity (ρ).

### Step-by-Step Solution:

1. **Identify the System**:
- We have two cubical blocks, A and B, both with mass M and side length L. Block A is fixed on top of block B, which is held rigidly on a horizontal surface.

2. **Force Application**:
- A small force F is applied perpendicular to one of the side faces of block A. This force causes a displacement in block A.

3. **Restoring Force**:
- Once the force F is removed, block A will experience a restoring force due to the deformation of block B. The restoring force is proportional to the displacement and is governed by the modulus of rigidity (ρ).

4. **Determine the Effective Spring Constant (k)**:
- The effective spring constant k can be expressed in terms of the modulus of rigidity (ρ) and the dimensions of the blocks. The relationship can be derived from the definition of shear modulus:
\[ k = \frac{\rho A}{L} \]
- Here, A is the area of the face of block A that is in contact with block B, which is \(L^2\). Thus, we can write:
\[ k = \frac{\rho L^2}{L} = \rho L \]

5. **Using the Formula for Time Period (T)**:
- The time period T of oscillation for a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
- Substituting our expression for k:
\[ T = 2\pi \sqrt{\frac{M}{\rho L}} \]

6. **Final Expression for Time Period**:
- Therefore, the time period of oscillation of block A is:
\[ T = 2\pi \sqrt{\frac{M}{\rho L}} \]

### Conclusion:
The time period of the oscillations of block A after the force is withdrawn is given by:
\[ T = 2\pi \sqrt{\frac{M}{\rho L}} \]