The bob of a simple pendulum performs SHM with period T in air and with period `T_(1)` in water. Relation between T and `T_1` is (neglect friction due to water, density of the material of the bob is = `9/8xx 10^3 (kg)/m^3`, density of water = `10^3(kg)/m^3`)

Options

A. `T_1=3T`
B. `T_1=2T`
C. `T_1=T`
D. `T_1=T/2`

Answer

To solve the problem, we need to find the relationship between the period of a simple pendulum in air (T) and the period in water (T₁).

### Step-by-Step Solution:

1. **Understanding the Period of a Pendulum**:
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 and \(g\) is the acceleration due to gravity.

2. **Period in Air**:
In air, the effective gravitational acceleration \(g\) is simply \(g\). Therefore, the period in air is:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]

3. **Period in Water**:
When the pendulum bob is submerged in water, it experiences a buoyant force. The effective gravitational acceleration \(g'\) in water is reduced due to this buoyant force. The apparent weight of the bob in water can be calculated as:
\[ W' = mg - \text{Buoyant Force} \]
The buoyant force \(F_b\) is given by:
\[ F_b = \rho_{water} \cdot V \cdot g \]
where \(V\) is the volume of the bob.

4. **Calculating the Buoyant Force**:
The volume \(V\) of the bob can be expressed in terms of its mass \(m\) and density \(\rho_{bob}\):
\[ V = \frac{m}{\rho_{bob}} \]
Thus, the buoyant force becomes:
\[ F_b = \rho_{water} \cdot \frac{m}{\rho_{bob}} \cdot g \]

5. **Finding the Effective Weight in Water**:
The effective weight in water is:
\[ W' = mg - \rho_{water} \cdot \frac{m}{\rho_{bob}} \cdot g \]
Simplifying this gives:
\[ W' = mg \left(1 - \frac{\rho_{water}}{\rho_{bob}}\right) \]

6. **Substituting Densities**:
Given that \(\rho_{bob} = \frac{9}{8} \times 10^3 \, \text{kg/m}^3\) and \(\rho_{water} = 10^3 \, \text{kg/m}^3\), we can substitute these values:
\[ W' = mg \left(1 - \frac{10^3}{\frac{9}{8} \times 10^3}\right) = mg \left(1 - \frac{8}{9}\right) = mg \cdot \frac{1}{9} \]

7. **Calculating Effective Gravity in Water**:
The effective gravitational acceleration in water is:
\[ g' = \frac{W'}{m} = \frac{g}{9} \]

8. **Finding the Period in Water**:
The period in water \(T_1\) can now be expressed as:
\[ T_1 = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{\frac{g}{9}}} = 2\pi \sqrt{\frac{9L}{g}} = 3 \cdot 2\pi \sqrt{\frac{L}{g}} = 3T \]

9. **Final Relation**:
Thus, the relationship between the periods is:
\[ T_1 = 3T \]

### Conclusion:
The relation between the period in water \(T_1\) and the period in air \(T\) is:
\[ T_1 = 3T \]