A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscous 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 oscillation in this liquid is :

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the 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,
- \( g \) is the acceleration due to gravity.

### Step 2: Identify Changes When Immersed in Liquid
When the pendulum bob is completely immersed in a non-viscous liquid, the effective acceleration due to gravity \( g' \) acting on the bob changes due to the buoyant force exerted by the liquid.

### Step 3: Calculate the Buoyant Force
The buoyant force \( F_b \) acting on the bob can be expressed as:
\[ F_b = V \cdot \rho_L \cdot g \]
where:
- \( V \) is the volume of the bob,
- \( \rho_L \) is the density of the liquid.

Given that the density of the liquid is \( \frac{1}{16} \) of the density of the bob (\( \rho \)):
\[ \rho_L = \frac{\rho}{16} \]

### Step 4: Determine the Effective Gravity
The effective weight of the bob when submerged is given by:
\[ F_{net} = mg - F_b = mg - V \cdot \rho_L \cdot g \]
Substituting \( \rho_L \):
\[ F_{net} = mg - V \cdot \left(\frac{\rho}{16}\right) g \]
The effective gravitational force can be expressed as:
\[ F_{net} = V \cdot \rho \cdot g' \quad \text{(where \( g' \) is the effective gravity)} \]
Equating the two expressions for force:
\[ V \cdot \rho \cdot g' = mg - V \cdot \left(\frac{\rho}{16}\right) g \]
Since \( m = V \cdot \rho \):
\[ V \cdot \rho \cdot g' = V \cdot \rho \cdot g - V \cdot \left(\frac{\rho}{16}\right) g \]
Dividing through by \( V \cdot \rho \):
\[ g' = g - \frac{g}{16} = g \left(1 - \frac{1}{16}\right) = g \left(\frac{15}{16}\right) \]

### Step 5: Substitute Effective Gravity into the Period Formula
Now we substitute \( g' \) back into the period formula:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' = \frac{15g}{16} \):
\[ T' = 2\pi \sqrt{\frac{L}{\frac{15g}{16}}} = 2\pi \sqrt{\frac{16L}{15g}} = 2\pi \sqrt{\frac{16}{15}} \sqrt{\frac{L}{g}} \]
Since \( T = 2\pi \sqrt{\frac{L}{g}} \):
\[ T' = \sqrt{\frac{16}{15}} T = \frac{4}{\sqrt{15}} T \]

### Final Answer
Thus, the period of oscillation of the pendulum bob when completely immersed in the liquid is:
\[ T' = \frac{4T}{\sqrt{15}} \]