A simple pendulum with a metallic bob has a time preiod 10 s. The bob is now immersed in a non-viscous liquide of density 1/3 that of metal. the time period of the same pendulum becomes

Options

A. 10 s
B. `10/sqrt(2^s)`
C. `10sqrt(3/2)s`
D. `10 sqrt3 s`

Answer

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. Let's break down the solution step by step.

### Step 1: Understand the Initial Conditions
The time period \( T \) of a simple pendulum 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 the initial time period \( T = 10 \) seconds, we can express this as:
\[ 10 = 2\pi \sqrt{\frac{L}{g}} \]

### Step 2: Calculate the Effective Gravity When Immersed
When the bob is immersed in a liquid of density \( \frac{1}{3} \) that of the metal, we need to find the effective gravity \( g_{\text{effective}} \).

1. **Weight of the bob**: The weight of the bob in air is \( mg \), where \( m \) is the mass of the bob.
2. **Buoyant Force**: According to Archimedes' principle, the buoyant force \( F_b \) acting on the bob is equal to the weight of the liquid displaced:
\[ F_b = \text{Density of liquid} \times \text{Volume of bob} \times g = \left(\frac{\rho}{3}\right)Vg \]
where \( \rho \) is the density of the metal and \( V \) is the volume of the bob.

3. **Effective Weight**: The effective weight of the bob when immersed in the liquid is:
\[ F_{\text{effective}} = mg - F_b = mg - \left(\frac{\rho}{3}\right)Vg \]
Since \( m = \rho V \):
\[ F_{\text{effective}} = \rho Vg - \left(\frac{\rho}{3}\right)Vg = \rho Vg \left(1 - \frac{1}{3}\right) = \rho Vg \frac{2}{3} \]
Thus, the effective gravity \( g_{\text{effective}} \) is:
\[ g_{\text{effective}} = g \cdot \frac{2}{3} \]

### Step 3: Substitute into the Time Period Formula
Now we can substitute \( g_{\text{effective}} \) into the time period formula:
\[ T' = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} = 2\pi \sqrt{\frac{L}{\frac{2g}{3}}} = 2\pi \sqrt{\frac{3L}{2g}} \]

### Step 4: Relate to the Original Time Period
From the original time period \( T = 2\pi \sqrt{\frac{L}{g}} = 10 \) seconds, we can relate the new time period:
\[ T' = 10 \sqrt{\frac{3}{2}} \]

### Step 5: Calculate the Final Time Period
Now we can calculate \( T' \):
\[ T' = 10 \sqrt{\frac{3}{2}} \approx 10 \cdot 1.2247 \approx 12.247 \text{ seconds} \]

### Final Answer
Thus, the new time period of the pendulum when the bob is immersed in the liquid is approximately \( 12.25 \) seconds.

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