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`1/16th` 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. `4Tsqrt(1/15`
C. `4Tsqrt(1/14`
D. `4Tsqrt(1/14`

Explanation

To find the period of oscillation of a simple pendulum bob completely immersed in a non-viscous liquid, we can follow these steps:

### Step 1: Understand the Forces Acting on the Bob
When the pendulum bob is in air, it experiences its weight acting downwards, which is given by:
\[ W = mg \]
where \( m \) is the mass of the bob and \( g \) is the acceleration due to gravity.

When the bob is submerged in a liquid, it experiences two forces:
1. The weight of the bob acting downwards: \( W = mg \)
2. The buoyant force acting upwards, which is given by Archimedes' principle:
\[ F_b = V \cdot \rho_{liquid} \cdot g \]
where \( V \) is the volume of the bob and \( \rho_{liquid} \) is the density of the liquid.

### Step 2: Calculate the Buoyant Force
Given that the density of the liquid is \( \frac{1}{16} \) of the density of the bob material, we can express the buoyant force as:
\[ \rho_{liquid} = \frac{\rho_{bob}}{16} \]
Thus, the buoyant force becomes:
\[ F_b = V \cdot \left(\frac{\rho_{bob}}{16}\right) \cdot g \]

### Step 3: Determine the Effective Weight of the Bob
The effective weight of the bob when submerged in the liquid is the actual weight minus the buoyant force:
\[ W_{effective} = mg - F_b \]
Substituting the expression for the buoyant force:
\[ W_{effective} = mg - V \cdot \left(\frac{\rho_{bob}}{16}\right) \cdot g \]

### Step 4: Relate Volume and Mass
Since the mass of the bob can be expressed in terms of its volume and density:
\[ m = V \cdot \rho_{bob} \]
We can substitute this into the effective weight equation:
\[ W_{effective} = V \cdot \rho_{bob} \cdot g - V \cdot \left(\frac{\rho_{bob}}{16}\right) \cdot g \]
Factoring out \( Vg \):
\[ W_{effective} = Vg \left(\rho_{bob} - \frac{\rho_{bob}}{16}\right) \]
\[ W_{effective} = Vg \cdot \left(\frac{16\rho_{bob} - \rho_{bob}}{16}\right) \]
\[ W_{effective} = Vg \cdot \left(\frac{15\rho_{bob}}{16}\right) \]

### Step 5: Calculate the Effective Acceleration Due to Gravity
The effective gravitational acceleration \( g_{effective} \) can be defined as:
\[ g_{effective} = \frac{W_{effective}}{m} = \frac{Vg \cdot \left(\frac{15\rho_{bob}}{16}\right)}{V \cdot \rho_{bob}} \]
This simplifies to:
\[ g_{effective} = g \cdot \frac{15}{16} \]

### Step 6: Calculate the Period of Oscillation
The period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
When the pendulum is submerged in the liquid, the period \( T_{liquid} \) becomes:
\[ T_{liquid} = 2\pi \sqrt{\frac{L}{g_{effective}}} \]
Substituting \( g_{effective} \):
\[ T_{liquid} = 2\pi \sqrt{\frac{L}{\frac{15g}{16}}} \]
\[ T_{liquid} = 2\pi \sqrt{\frac{16L}{15g}} \]
Since the period in air is \( T = 2\pi \sqrt{\frac{L}{g}} \), we can express the new period in terms of \( T \):
\[ T_{liquid} = T \cdot \sqrt{\frac{16}{15}} \]

### Final Result
Thus, the period of oscillation of the pendulum bob in the liquid is:
\[ T_{liquid} = T \cdot \sqrt{\frac{16}{15}} \]

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