A simple pendulum has time period T = 2s in air. If the whole arrangement is placed in a non viscous liquid whose density is 1/2 times the density of bob. The time period in the liquid will be

Options

A. `2/sqrt(2)`s
B. 4 s
C. `2sqrt(2)`s
D. `4sqrt(2)`s

Answer

To solve the problem of finding the time period of a simple pendulum when it is immersed in a non-viscous liquid, we can follow these steps:

### Step 1: Understand the Time Period in Air
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.

Given that \( T = 2 \) seconds in air, we can express this as:
\[ 2 = 2\pi \sqrt{\frac{L}{g}} \]

### Step 2: Calculate the Effective Gravity in the Liquid
When the pendulum bob is placed in a liquid, the effective gravitational force acting on it changes due to buoyancy. The buoyant force \( F_b \) acting on the bob is given by:
\[ F_b = \rho_l V g \]
where:
- \( \rho_l \) is the density of the liquid,
- \( V \) is the volume of the bob.

The weight of the bob \( F_g \) is:
\[ F_g = \rho_s V g \]
where \( \rho_s \) is the density of the bob.

Given that the density of the liquid \( \rho_l \) is \( \frac{1}{2} \) times the density of the bob \( \rho_s \):
\[ \rho_l = \frac{\rho_s}{2} \]

### Step 3: Calculate the Net Force
The net force acting on the bob when submerged in the liquid is:
\[ F_{net} = F_g - F_b = \rho_s V g - \rho_l V g \]
Substituting for \( \rho_l \):
\[ F_{net} = \rho_s V g - \left(\frac{\rho_s}{2}\right) V g = \rho_s V g - \frac{\rho_s V g}{2} = \frac{\rho_s V g}{2} \]

### Step 4: Determine the Effective Gravity
The effective gravitational force \( g_{effective} \) can be expressed as:
\[ m \cdot g_{effective} = \frac{mg}{2} \]
Thus, we have:
\[ g_{effective} = \frac{g}{2} \]

### Step 5: Calculate the New Time Period in the Liquid
Now, we can find the new time period \( T' \) in the liquid:
\[ T' = 2\pi \sqrt{\frac{L}{g_{effective}}} = 2\pi \sqrt{\frac{L}{\frac{g}{2}}} = 2\pi \sqrt{\frac{2L}{g}} = \sqrt{2} \cdot 2\pi \sqrt{\frac{L}{g}} \]
Since we know \( 2\pi \sqrt{\frac{L}{g}} = 2 \) seconds (the time period in air), we substitute this value:
\[ T' = \sqrt{2} \cdot 2 = 2\sqrt{2} \text{ seconds} \]

### Final Answer
The time period of the pendulum in the liquid is:
\[ T' = 2\sqrt{2} \text{ seconds} \]