A pendulum has time period `T` in air when it is made to oscillate in water it acquired a time period `T = sqrt(2)T` The density of the pendulum bob is equal to (density) of water `= 1)`

Options

A. `sqrt(2)`
B. 2
C. `2sqrt(2)`
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the time period of a pendulum
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 and \( g \) is the acceleration due to gravity.

### Step 2: Time period in water
When the pendulum is oscillating in water, the effective acceleration due to gravity \( g' \) is modified due to the buoyant force acting on the pendulum bob. The modified acceleration is given by:
\[ g' = g \left(1 - \frac{\sigma}{\rho}\right) \]
where \( \sigma \) is the density of the fluid (water in this case) and \( \rho \) is the density of the pendulum bob.

### Step 3: Write the time period in water
The time period \( T' \) of the pendulum in water can be expressed as:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} \]

### Step 4: Relate the time periods
According to the problem, the time period in water is given by:
\[ T' = \sqrt{2} T \]

### Step 5: Set up the equation
Using the expressions for \( T \) and \( T' \), we can set up the equation:
\[ \sqrt{2} T = 2\pi \sqrt{\frac{L}{g \left(1 - \frac{\sigma}{\rho}\right)}} \]
Substituting \( T \):
\[ \sqrt{2} \cdot 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{L}{g \left(1 - \frac{\sigma}{\rho}\right)}} \]

### Step 6: Simplify the equation
Cancel \( 2\pi \sqrt{L} \) from both sides:
\[ \sqrt{2} \cdot \frac{1}{\sqrt{g}} = \frac{1}{\sqrt{g \left(1 - \frac{\sigma}{\rho}\right)}} \]

### Step 7: Square both sides
Squaring both sides gives:
\[ 2 \cdot \frac{1}{g} = \frac{1}{g \left(1 - \frac{\sigma}{\rho}\right)} \]

### Step 8: Cross-multiply
Cross-multiplying leads to:
\[ 2 \left(1 - \frac{\sigma}{\rho}\right) = 1 \]

### Step 9: Solve for density ratio
This simplifies to:
\[ 2 - \frac{2\sigma}{\rho} = 1 \]
\[ \frac{2\sigma}{\rho} = 1 \]
\[ \rho = 2\sigma \]

### Step 10: Substitute the value of \( \sigma \)
Given that the density of water \( \sigma = 1 \):
\[ \rho = 2 \cdot 1 = 2 \]

### Conclusion
The density of the pendulum bob is \( \rho = 2 \).

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