A simple pendulum with iron bob has a time period T. The bob is now immersed in a non-viscous liquid and oscillated, if the density of liquid is `(1)/(12)`th that of iron, then new time period will be

Options

A. `T sqrt((8)/(7))`
B. `T sqrt((12)/(13))`
C. `T sqrt((12)/(11))`
D. `T sqrt((6)/(5))`

Answer

To solve the problem, we need to determine the new time period of a simple pendulum bob when it is immersed in a non-viscous liquid with a density that is \( \frac{1}{12} \) of the density of iron.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the density of iron be \( \rho_i \).
- The density of the liquid is given as \( \rho_l = \frac{1}{12} \rho_i \).
- The time period of the pendulum in air is \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Calculate the Effective Gravity**:
- When the bob is immersed in the liquid, the effective gravity \( g_{\text{effective}} \) is modified due to the buoyant force acting on the bob.
- The buoyant force \( F_b \) can be calculated using Archimedes' principle: \( F_b = \rho_l \cdot V \cdot g \), where \( V \) is the volume of the bob.
- The weight of the bob \( W \) is given by \( W = \rho_i \cdot V \cdot g \).
- The effective weight acting on the bob when submerged is \( W - F_b = \rho_i \cdot V \cdot g - \rho_l \cdot V \cdot g \).

3. **Express the Effective Gravity**:
- The effective weight can be expressed as:
\[ \text{Effective Weight} = \left( \rho_i - \rho_l \right) V g \]
- Substituting \( \rho_l = \frac{1}{12} \rho_i \):
\[ \text{Effective Weight} = \left( \rho_i - \frac{1}{12} \rho_i \right) V g = \left( \frac{11}{12} \rho_i \right) V g \]
- The effective gravity \( g_{\text{effective}} \) is then:
\[ g_{\text{effective}} = \frac{11}{12} g \]

4. **Calculate the New Time Period**:
- The new time period \( T' \) of the pendulum in the liquid can be calculated using the formula:
\[ T' = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]
- Substituting \( g_{\text{effective}} = \frac{11}{12} g \):
\[ T' = 2\pi \sqrt{\frac{L}{\frac{11}{12} g}} = 2\pi \sqrt{\frac{12L}{11g}} = \sqrt{\frac{12}{11}} \cdot T \]

5. **Final Result**:
- Therefore, the new time period when the bob is immersed in the liquid is:
\[ T' = T \sqrt{\frac{12}{11}} \]