The time period of a sample pendulum is T. Now the bob is immersed in a liquid of density `sigma`. If density of material of bob is `rho`, what will be the time period of the pendulum.

Answer

To solve the problem of finding the new time period of a simple pendulum bob when it is immersed in a liquid, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Time Period of a Simple Pendulum**:
The time period \( T \) of a simple pendulum in air is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Identify the Changes When Immersed in Liquid**:
When the bob is immersed in a liquid, the effective gravitational acceleration acting on the bob changes due to the buoyant force exerted by the liquid. We need to find the new effective gravitational acceleration \( g' \).

3. **Calculate the Buoyant Force**:
The buoyant force \( F_b \) acting on the bob when submerged in the liquid is given by Archimedes' principle:
\[ F_b = \text{Volume of bob} \times \text{Density of liquid} \times g \]
If the density of the bob is \( \rho \) and the density of the liquid is \( \sigma \), then the weight of the bob is:
\[ F_g = \text{Volume of bob} \times \rho \times g \]

4. **Determine the Net Force**:
The net force \( F_{net} \) acting on the bob when submerged is:
\[ F_{net} = F_g - F_b = V \rho g - V \sigma g = V g (\rho - \sigma) \]
where \( V \) is the volume of the bob.

5. **Calculate the New Effective Gravitational Acceleration**:
The effective gravitational acceleration \( g' \) can be expressed as:
\[ g' = \frac{F_{net}}{m} = \frac{V g (\rho - \sigma)}{V \rho} = \frac{g (\rho - \sigma)}{\rho} \]

6. **Substitute \( g' \) into the Time Period Formula**:
The new time period \( T' \) of the pendulum when the bob is submerged in the liquid becomes:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{\frac{g (\rho - \sigma)}{\rho}}} \]

7. **Simplify the Expression**:
This can be simplified to:
\[ T' = 2\pi \sqrt{\frac{L \rho}{g (\rho - \sigma)}} \]
Recognizing that \( T = 2\pi \sqrt{\frac{L}{g}} \), we can express \( T' \) in terms of \( T \):
\[ T' = T \sqrt{\frac{\rho}{\rho - \sigma}} \]

### Final Result:
Thus, the new time period of the pendulum bob when immersed in the liquid is:
\[ T' = T \sqrt{\frac{\rho}{\rho - \sigma}} \]