A simple pendulam with a solid metal bob has a period T. The metal bob is now immersed in a liquid having density one-teeth that of the metal of the bob. The liquid is non-viscuous. Now the period of the same pendulum with its bob remaining all the time in the liquid will be

Options

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

Explanation

To solve the problem step by step, we will analyze the situation of a simple pendulum and how its period changes when the bob is immersed in a liquid.

### Step 1: Understand the Period of a Simple Pendulum
The 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.

### Step 2: Determine the Effect of Immersion in a Liquid
When the pendulum bob is immersed in a liquid, the effective gravitational acceleration changes due to the buoyant force acting on the bob. The new period \( T' \) can be expressed as:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} \]
where \( g' \) is the effective gravitational acceleration when the bob is in the liquid.

### Step 3: Calculate the Effective Gravitational Acceleration
The effective gravitational acceleration \( g' \) can be calculated using the formula:
\[ g' = g \left(1 - \frac{1}{\sigma}\right) \]
where \( \sigma \) is the ratio of the density of the bob \( \rho_b \) to the density of the liquid \( \rho_l \):
\[ \sigma = \frac{\rho_b}{\rho_l} \]

Given that the density of the liquid is one-tenth that of the metal bob, we have:
\[ \sigma = \frac{\rho_b}{\frac{1}{10} \rho_b} = 10 \]

### Step 4: Substitute the Value of \( \sigma \) into the Equation for \( g' \)
Now substituting \( \sigma \) into the equation for \( g' \):
\[ g' = g \left(1 - \frac{1}{10}\right) = g \left(\frac{9}{10}\right) = \frac{9g}{10} \]

### Step 5: Substitute \( g' \) into the Period Formula
Now substitute \( g' \) back into the formula for the period \( T' \):
\[ T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{\frac{9g}{10}}} = 2\pi \sqrt{\frac{10L}{9g}} \]

### Step 6: Relate \( T' \) to the Original Period \( T \)
Since \( T = 2\pi \sqrt{\frac{L}{g}} \), we can express \( T' \) in terms of \( T \):
\[ T' = T \sqrt{\frac{10}{9}} = T \cdot \frac{\sqrt{10}}{3} \]

### Final Result
Thus, the period of the pendulum with its bob remaining all the time in the liquid is:
\[ T' = T \sqrt{\frac{10}{9}} \]