A simple pendulum with a metallic bob has a time preiod 10 s. The bob is now immersed in a non-viscous liquide of density 1/3 that of metal. the time period of the same pendulum becomes

Options

A. 10 s
B. `10/sqrt(2^s)`
C. `10sqrt(3/2)s`
D. `10 sqrt3 s`

Answer

To solve the problem, we need to find the new time period of a simple pendulum when its bob is immersed in a non-viscous liquid with a density that is one-third of the density of the metal bob.

### 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 acceleration due to gravity.

2. **Initial Conditions**:
The initial time period \( T_1 \) is given as 10 seconds. Therefore:
\[ T_1 = 10 \text{ s} \]

3. **Determine Effective Gravity When Immersed**:
When the bob is immersed in a liquid, the effective gravity \( g' \) acting on the bob changes due to the buoyant force. The buoyant force \( F_b \) is given by:
\[ F_b = \text{density of liquid} \times \text{volume of bob} \times g \]
Given that the density of the liquid is one-third that of the metal, if we denote the density of the metal as \( \rho \), the density of the liquid is:
\[ \text{density of liquid} = \frac{\rho}{3} \]

4. **Calculate the Buoyant Force**:
The weight of the bob is \( mg \) and the buoyant force is:
\[ F_b = \frac{\rho}{3} V g \]
where \( V \) is the volume of the bob. The apparent weight \( W' \) of the bob in the liquid is:
\[ W' = mg - F_b = mg - \frac{\rho}{3} V g \]
Since \( m = \rho V \), we can substitute:
\[ W' = \rho V g - \frac{\rho}{3} V g = \rho V g \left(1 - \frac{1}{3}\right) = \rho V g \cdot \frac{2}{3} \]

5. **Effective Gravity**:
The effective gravity \( g' \) in the liquid becomes:
\[ g' = \frac{2g}{3} \]

6. **Calculate the New Time Period**:
The new time period \( T_2 \) can now be calculated using the formula for the time period:
\[ T_2 = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' \):
\[ T_2 = 2\pi \sqrt{\frac{L}{\frac{2g}{3}}} = 2\pi \sqrt{\frac{3L}{2g}} \]

7. **Relate \( T_2 \) to \( T_1 \)**:
Since \( T_1 = 2\pi \sqrt{\frac{L}{g}} \), we can express \( T_2 \) in terms of \( T_1 \):
\[ T_2 = T_1 \sqrt{\frac{3}{2}} = 10 \sqrt{\frac{3}{2}} \]

8. **Final Calculation**:
Therefore, the new time period is:
\[ T_2 = 10 \cdot \sqrt{\frac{3}{2}} = 10 \cdot \frac{\sqrt{3}}{\sqrt{2}} = 10 \cdot \frac{\sqrt{6}}{2} = 5\sqrt{6} \text{ seconds} \]

### Conclusion:
The new time period of the pendulum when the bob is immersed in the liquid is \( 5\sqrt{6} \) seconds.