An object of density `sigma(ltrho)` is dropped from rest at a height h into a liquid of density `rho` kept in a tall vertical cylindrical container. Neglect all dissipative effects and assume that there is no spilling of the liquid.
The time taken by the object to reach the starting point (from which it is dropped) is

Options

A. `2sqrt((2h)/g)`
B. `4sqrt((2h)/g)`
C. `3sqrt((2h)/g)`
D. it will never reach starting point

Explanation

To solve the problem of determining the time taken by an object of density \(\sigma\) (which is less than the density \(\rho\) of the liquid) to reach the starting point from which it is dropped, we can break it down into several steps.

### Step-by-Step Solution:

1. **Understanding the Initial Drop**:
The object is dropped from a height \(h\) into a liquid. Since the object is less dense than the liquid, it will initially fall under the influence of gravity until it reaches the surface of the liquid.

2. **Calculating the Time to Fall to the Liquid Surface**:
The time \(t_1\) taken to fall a distance \(h\) under gravity can be calculated using the equation of motion:
\[ h = \frac{1}{2} g t_1^2 \]
Rearranging this gives:
\[ t_1 = \sqrt{\frac{2h}{g}} \]

3. **Behavior of the Object in the Liquid**:
Once the object enters the liquid, it experiences a buoyant force. The net force acting on the object can be expressed as:
\[ F = V \sigma g - V \rho g = V g (\sigma - \rho) \]
where \(V\) is the volume of the object. The acceleration \(a\) of the object in the liquid can be derived from Newton's second law:
\[ ma = V g (\sigma - \rho) \implies a = g \left(1 - \frac{\rho}{\sigma}\right) \]

4. **Calculating the Time to Reach the Bottom of the Liquid**:
The object will continue to move downwards until it reaches the bottom of the liquid. The time \(t_2\) taken to reach the bottom can be calculated using the same equation of motion, but now with the effective acceleration:
\[ h = \frac{1}{2} a t_2^2 \]
Substituting \(a\) gives:
\[ h = \frac{1}{2} g \left(1 - \frac{\rho}{\sigma}\right) t_2^2 \]
Rearranging this gives:
\[ t_2 = \sqrt{\frac{2h}{g \left(1 - \frac{\rho}{\sigma}\right)}} \]

5. **Calculating the Time to Return to the Starting Point**:
After reaching the bottom, the object will rise back to the surface of the liquid. The time taken to rise back to the surface will be the same as the time taken to fall to the bottom, which is \(t_2\).

6. **Total Time Calculation**:
The total time \(T\) taken for the object to return to the starting point is the sum of the time taken to fall to the liquid surface, the time taken to reach the bottom, and the time taken to rise back to the surface:
\[ T = t_1 + t_2 + t_2 = t_1 + 2t_2 \]
Substituting the values of \(t_1\) and \(t_2\):
\[ T = \sqrt{\frac{2h}{g}} + 2\sqrt{\frac{2h}{g \left(1 - \frac{\rho}{\sigma}\right)}} \]

### Final Expression:
The total time taken by the object to reach the starting point from which it is dropped is:
\[ T = \sqrt{\frac{2h}{g}} + 2\sqrt{\frac{2h}{g \left(1 - \frac{\rho}{\sigma}\right)}} \]