A simple pendulum oscillating in air has a time period of `sqrt(3)`s. Now the bob of the pendulum is completely immersed in a non-viscous liquid whose density is equal to `(1)/(4)`th that of the material of the bob. The new time period of simple pendulum will be

Options

A. `sqrt(2) s`
B. `sqrt(3) s`
C. `2 s`
D. `sqrt((7)/(2)) s`

Explanation

To solve the problem, we need to find the new time period of a simple pendulum when its bob is immersed in a non-viscous liquid. Here are the steps to find the solution:

### Step 1: Understand the initial conditions
The time period of the pendulum in air is given as \( T = \sqrt{3} \) seconds. The formula for the time period of a simple pendulum is:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

### Step 2: Determine the effective gravitational acceleration in the liquid
When the bob is immersed in a liquid, it experiences a buoyant force. The density of the liquid is given as \( \frac{1}{4} \) of the density of the bob. Let:
- \( \rho \) = density of the bob
- \( V \) = volume of the bob
- The density of the liquid = \( \frac{\rho}{4} \)

The buoyant force \( F_b \) acting on the bob is:
\[ F_b = \text{density of liquid} \times \text{volume} \times g = \left(\frac{\rho}{4}\right) V g \]

The weight \( W \) of the bob is:
\[ W = \rho V g \]

### Step 3: Calculate the net force acting on the bob
The effective weight of the bob when submerged in the liquid is:
\[ W_{\text{effective}} = W - F_b = \rho V g - \left(\frac{\rho}{4} V g\right) \]
\[ W_{\text{effective}} = \rho V g \left(1 - \frac{1}{4}\right) = \rho V g \cdot \frac{3}{4} \]

### Step 4: Determine the effective gravitational acceleration
The effective gravitational acceleration \( g_{\text{effective}} \) can be derived from the net force:
\[ g_{\text{effective}} = \frac{W_{\text{effective}}}{\text{mass}} = \frac{\frac{3}{4} \rho V g}{\rho V} = \frac{3g}{4} \]

### Step 5: Calculate the new time period in the liquid
Now we can find the new time period \( T' \) using the effective gravitational acceleration:
\[ T' = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} = 2\pi \sqrt{\frac{L}{\frac{3g}{4}}} = 2\pi \sqrt{\frac{4L}{3g}} \]

### Step 6: Relate the new time period to the original time period
We know that the original time period \( T \) is:
\[ T = 2\pi \sqrt{\frac{L}{g}} = \sqrt{3} \]
From this, we can express \( \sqrt{\frac{L}{g}} \) as:
\[ \sqrt{\frac{L}{g}} = \frac{\sqrt{3}}{2\pi} \]

Substituting this into the equation for \( T' \):
\[ T' = 2\pi \sqrt{\frac{4L}{3g}} = 2\pi \cdot \frac{2}{\sqrt{3}} \cdot \sqrt{\frac{L}{g}} = 2\pi \cdot \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2\pi} = 2 \text{ seconds} \]

### Final Answer
The new time period of the simple pendulum when the bob is immersed in the liquid is \( 2 \) seconds.

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