A highly rigid cubical block A of small mass `M` and slide `L` is fixed rigidly on to 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 A small force `F` is applied perpendicular to one of the sides faces of A. After the force is withdrawn, block A executes small oscillations the time period of which is given by.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 eta L)`
B. `(2 pi) sqrt (M eta)/(L)`
C. `(2 pi) sqrt (M L)/(eta)`
D. `(2 pi)sqrt (M)/(eta L)`

Explanation

To solve the problem, we need to determine the time period of the small oscillations of block A after a force F is applied and then withdrawn. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Setup
We have two cubical blocks, A and B. Block A is rigid and has a small mass M, while block B has a low modulus of rigidity η. Block A completely covers block B, which is fixed to a horizontal surface.

### Step 2: Apply the Force
A small force F is applied perpendicular to one of the side faces of block A. This force causes block B to deform slightly due to its low modulus of rigidity. The deformation leads to a shear strain in block B.

### Step 3: Relate Force to Deformation
The shear stress (τ) in block B due to the applied force can be expressed as:
\[ \tau = \frac{F}{A} \]
where A is the area of the face of block A in contact with block B, which is \(L^2\) (since the side length of the cube is L).

The shear strain (γ) is given by:
\[ \gamma = \frac{\Delta L}{L} \]
where \(\Delta L\) is the amount of deformation in block B.

### Step 4: Use Modulus of Rigidity
The modulus of rigidity (η) relates shear stress and shear strain:
\[ \eta = \frac{\tau}{\gamma} \]
Substituting the expressions for shear stress and shear strain:
\[ \eta = \frac{F / (L^2)}{\Delta L / L} \]
Rearranging gives:
\[ F = \eta \frac{L^2}{\Delta L} \cdot \Delta L = \eta L \Delta L \]

### Step 5: Relate to Simple Harmonic Motion
When the force F is removed, block A will oscillate about its equilibrium position. The restoring force can be modeled as:
\[ F = -K \Delta L \]
where K is the effective spring constant. From our previous equation, we can identify:
\[ K = \frac{\eta L}{\Delta L} \]

### Step 6: Find Angular Frequency
In simple harmonic motion, the force can also be expressed as:
\[ F = M \omega^2 x \]
where \(x\) is the displacement from the equilibrium position. Setting the two expressions for force equal gives:
\[ \eta L \Delta L = M \omega^2 \Delta L \]
Cancelling \(\Delta L\) (assuming it is not zero):
\[ \eta L = M \omega^2 \]
Thus, we can solve for \(\omega\):
\[ \omega = \sqrt{\frac{\eta L}{M}} \]

### Step 7: Calculate Time Period
The time period \(T\) of the oscillation is related to the angular frequency by:
\[ T = \frac{2\pi}{\omega} \]
Substituting for \(\omega\):
\[ T = 2\pi \sqrt{\frac{M}{\eta L}} \]

### Final Result
Thus, the time period of the oscillations of block A is given by:
\[ T = 2\pi \sqrt{\frac{M}{\eta L}} \]