The matallic bob of a simple pendulum has the relative density `rho`. The time period of this pendulum is `T` it the metallic bob is immersed in water the new time period is given by

Options

A. (a)`2pisqrt((l)/(ng))`
B. (b)`2pisqrt((l)/((1-(1)/(n))g))`
C. (c)`2pisqrt((ln)/(g))`
D. (d)`2pisqrt((l)/((n-1)g))`

Explanation

To solve the problem of finding the new time period of a simple pendulum with a metallic bob immersed in water, 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 is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g'}} \]
where \( L \) is the length of the pendulum and \( g' \) is the effective acceleration due to gravity when the bob is immersed in a fluid.

2. **Determine the Effective Gravity**:
When the metallic bob is immersed in water, the effective acceleration due to gravity \( g' \) is modified by the buoyant force acting on the bob. The effective gravity can be expressed as:
\[ g' = g \left(1 - \frac{\rho}{\sigma}\right) \]
where:
- \( g \) is the acceleration due to gravity,
- \( \rho \) is the density of the liquid (water in this case),
- \( \sigma \) is the density of the bob material.

3. **Substituting \( g' \) into the Time Period Formula**:
Now, substituting \( g' \) into the time period formula, we get:
\[ T' = 2\pi \sqrt{\frac{L}{g \left(1 - \frac{\rho}{\sigma}\right)}} \]

4. **Simplifying the Expression**:
We can simplify the expression further:
\[ T' = 2\pi \sqrt{\frac{L}{g}} \cdot \frac{1}{\sqrt{1 - \frac{\rho}{\sigma}}} \]

5. **Relating to the Original Time Period**:
Since the original time period \( T \) is:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
We can express the new time period \( T' \) in terms of the original time period:
\[ T' = T \cdot \frac{1}{\sqrt{1 - \frac{\rho}{\sigma}}} \]

### Final Answer:
Thus, the new time period \( T' \) when the metallic bob is immersed in water is given by:
\[ T' = T \cdot \frac{1}{\sqrt{1 - \frac{\rho}{\sigma}}} \]