A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscours liquid. The density of the liquid is `(1)/(16)th` of the material of the bob. If the bob is inside liquid all the time, its period of scillation in this liquid is :

Options

A. `2Tsqrt((1)/(10))`
B. `2Tsqrt((1)/(14))`
C. `4Tsqrt((1)/(15))`
D. `4Tsqrt(1)/(14)`

Answer

To solve the problem, we need to find the new period of oscillation \( T' \) of a simple pendulum bob that is completely immersed in a non-viscous liquid, given that the density of the liquid is \( \frac{1}{16} \) of the density of the bob.

### Step-by-Step Solution:

1. **Understanding the Period of a Simple Pendulum:**
The 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. **Forces Acting on the Bob in Liquid:**
When the bob is immersed in the liquid, the forces acting on it include:
- The weight of the bob \( mg \) acting downwards.
- The buoyant force \( F_b \) acting upwards, which can be calculated using Archimedes' principle.

3. **Calculating the Buoyant Force:**
The buoyant force \( F_b \) is given by:
\[ F_b = \text{Volume of bob} \times \text{Density of liquid} \times g \]
Since the volume of the bob \( V \) is the same when it is in air and in the liquid, we can express the buoyant force as:
\[ F_b = V \cdot \rho_{liquid} \cdot g \]
Given that the density of the liquid \( \rho_{liquid} = \frac{1}{16} \rho_{bob} \), we can substitute this into the equation.

4. **Expressing the Mass of the Bob and Liquid:**
The mass of the bob is:
\[ m = V \cdot \rho_{bob} \]
The mass of the liquid displaced by the bob is:
\[ m_{liquid} = V \cdot \rho_{liquid} = V \cdot \left(\frac{1}{16} \rho_{bob}\right) \]

5. **Net Force on the Bob:**
The net force acting on the bob when it is submerged is:
\[ F_{net} = mg - F_b = mg - m_{liquid}g = mg - \left(\frac{1}{16} mg\right) = mg \left(1 - \frac{1}{16}\right) = mg \left(\frac{15}{16}\right) \]

6. **Effective Gravity in Liquid:**
The effective gravitational force acting on the bob in the liquid can be defined as:
\[ g' = g \left(\frac{15}{16}\right) \]

7. **Finding the New Period of Oscillation:**
The new period \( T' \) of the pendulum in the liquid can be expressed as:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' \):
\[ T' = 2\pi \sqrt{\frac{L}{g \left(\frac{15}{16}\right)}} = 2\pi \sqrt{\frac{16L}{15g}} = \frac{2\pi}{\sqrt{15}} \sqrt{16L} \]
Since \( T = 2\pi \sqrt{\frac{L}{g}} \), we can express \( T' \) in terms of \( T \):
\[ T' = T \cdot \frac{4}{\sqrt{15}} \]

8. **Final Result:**
Thus, the period of oscillation of the pendulum bob in the liquid is:
\[ T' = 4T \cdot \frac{1}{\sqrt{15}} \]

### Final Answer:
\[ T' = 4T \cdot \frac{1}{\sqrt{15}} \]