A heavy brass sphere is hung from a weightless inelastic string and used as a simple pendulum. Its time period of osciallation is T. When the sphere is immersed in a non-viscous liquid of density `(1)/(10)` that of brass. It acts as a simple pendulum of period.

Options

A. T
B. `(10)/(9)T`
C. ` sqrt((9)/(10))T`
D. `sqrt((10)/(9))T`

Answer

To solve the problem, we need to find the new time period of the pendulum when the brass sphere is immersed in a liquid of density \( \frac{1}{10} \) that of brass. Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Understanding the Time Period of a Simple Pendulum**:
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]
where \( L \) is the length of the pendulum and \( g_{\text{effective}} \) is the effective gravitational acceleration.

2. **Time Period in Air**:
When the brass sphere is hanging in air, the effective gravitational acceleration is simply \( g \) (acceleration due to gravity). Thus, the time period in air is:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]

3. **Buoyant Force When Immersed in Liquid**:
When the sphere is immersed in a liquid, it experiences a buoyant force \( B \) given by Archimedes' principle:
\[ B = V \cdot \rho_L \cdot g \]
where \( V \) is the volume of the sphere and \( \rho_L \) is the density of the liquid.

4. **Net Effective Gravitational Force**:
The effective weight of the sphere when immersed in the liquid is:
\[ g_{\text{effective}} = g - \frac{B}{m} \]
where \( m \) is the mass of the sphere.

5. **Substituting the Values**:
The mass \( m \) of the sphere can be expressed as:
\[ m = V \cdot \rho_B \]
where \( \rho_B \) is the density of brass. Thus, we can express the effective gravitational acceleration as:
\[ g_{\text{effective}} = g - \frac{V \cdot \rho_L \cdot g}{V \cdot \rho_B} = g - \frac{\rho_L}{\rho_B} g \]

6. **Using Given Density**:
Given that \( \rho_L = \frac{1}{10} \rho_B \), we substitute this into the equation:
\[ g_{\text{effective}} = g - \frac{1}{10} g = g \left(1 - \frac{1}{10}\right) = \frac{9}{10} g \]

7. **Finding the New Time Period**:
Now substituting \( g_{\text{effective}} \) back into the time period formula:
\[ T' = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} = 2\pi \sqrt{\frac{L}{\frac{9}{10}g}} = 2\pi \sqrt{\frac{10L}{9g}} \]
We can express this in terms of the original time period \( T \):
\[ T' = T \sqrt{\frac{10}{9}} \]

### Final Answer:
Thus, the new time period \( T' \) when the sphere is immersed in the liquid is:
\[ T' = T \sqrt{\frac{10}{9}} \]