The time period of a simple pendulum in air is `T`. Now the pendulum is submerged in a liquid of density `(rho)/(16)` where `rho` is density of the bob of the pendulum. The new time period of oscillation is.

Options

A. `(4)/(sqrt(15))T`
B. `sqrt((4)/(15))T`
C. `sqrt((15)/(4))T`
D. `(sqrt(15))/(3)T`

Answer

To solve the problem, we need to determine the new time period of a simple pendulum when it is submerged in a liquid. The steps are as follows:

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

### Step 2: Identify Changes When Submerged in Liquid
When the pendulum is submerged in a liquid, the length \( L \) remains the same, but the effective gravitational acceleration \( g' \) changes due to the buoyant force acting on the pendulum bob.

### Step 3: Calculate the Effective Gravity \( g' \)
The effective gravitational force \( g' \) can be calculated by considering the forces acting on the bob:
- The weight of the bob acting downward: \( mg \)
- The buoyant force acting upward: \( F_b = \text{Volume} \times \text{Density of liquid} \times g \)

Given that the density of the bob is \( \rho \) and the density of the liquid is \( \frac{\rho}{16} \):
- The buoyant force becomes:
\[ F_b = V \left(\frac{\rho}{16}\right) g \]

The net downward force \( F_{\text{net}} \) when submerged is:
\[ F_{\text{net}} = mg - F_b = mg - V \left(\frac{\rho}{16}\right) g \]

### Step 4: Express Mass in Terms of Density
Using the relationship \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \), we can express the mass \( m \) as:
\[ m = V \rho \]

### Step 5: Substitute and Simplify
Substituting \( m \) into the net force equation:
\[ F_{\text{net}} = V \rho g - V \left(\frac{\rho}{16}\right) g \]
Factoring out \( Vg \):
\[ F_{\text{net}} = Vg \left(\rho - \frac{\rho}{16}\right) = Vg \left(\frac{15\rho}{16}\right) \]

### Step 6: Calculate Effective Gravity
Now, the effective gravitational acceleration \( g' \) is given by:
\[ g' = \frac{F_{\text{net}}}{m} = \frac{Vg \left(\frac{15\rho}{16}\right)}{V\rho} = g \left(\frac{15}{16}\right) \]

### Step 7: Substitute \( g' \) into the Time Period Formula
Now we can find the new time period \( T' \):
\[ T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{g \left(\frac{15}{16}\right)}} \]
This simplifies to:
\[ T' = 2\pi \sqrt{\frac{16L}{15g}} = \frac{4}{\sqrt{15}} \cdot 2\pi \sqrt{\frac{L}{g}} = \frac{4}{\sqrt{15}} T \]

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