A pendulum made of a uniform wire of cross sectional area A has time T. When an additional mass M is added to its bob, the time period changes to `T_M`. If the Young's modulus of the material of the wire is Y then `1/Y` is equal to:

Options

A. `[(T_(M)/T)^(2) - 1](Mg)/(A)`
B. `[1 - (T_(M)/T)^(2)](A)/(Mg)`
C. `[1 - (T/T_(M))^(2)](A)/(Mg)`
D. `[(T_(M)/T)^(2) - 1](A)/(Mg)`

Answer

To solve the problem, we need to analyze the effect of adding an additional mass \( M \) to the pendulum's bob on its time period and relate it to the Young's modulus \( Y \) of the wire.

### Step-by-Step Solution:

1. **Initial Time Period of the 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. **Adding an Additional Mass**:
When an additional mass \( M \) is added to the bob, the new time period \( T_M \) becomes:
\[ T_M = 2\pi \sqrt{\frac{L'}{g}} \]
where \( L' \) is the new effective length of the pendulum after the wire stretches due to the added mass.

3. **Stress and Strain in the Wire**:
The stress \( \sigma \) in the wire due to the additional mass \( M \) is given by:
\[ \sigma = \frac{Mg}{A} \]
where \( A \) is the cross-sectional area of the wire.

The strain \( \epsilon \) experienced by the wire is defined as:
\[ \epsilon = \frac{\Delta L}{L} \]
where \( \Delta L \) is the change in length of the wire.

4. **Relating Stress to Young's Modulus**:
Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\sigma}{\epsilon} \]
Rearranging gives:
\[ \epsilon = \frac{\sigma}{Y} \]

5. **Change in Length**:
The change in length \( \Delta L \) can be expressed as:
\[ \Delta L = \epsilon L = \frac{\sigma L}{Y} = \frac{MgL}{AY} \]

6. **New Length of the Pendulum**:
The new length \( L' \) after the extension is:
\[ L' = L + \Delta L = L + \frac{MgL}{AY} = L \left(1 + \frac{Mg}{AY}\right) \]

7. **New Time Period**:
Substituting \( L' \) into the expression for \( T_M \):
\[ T_M = 2\pi \sqrt{\frac{L(1 + \frac{Mg}{AY})}{g}} = 2\pi \sqrt{\frac{L}{g}} \sqrt{1 + \frac{Mg}{AY}} = T \sqrt{1 + \frac{Mg}{AY}} \]

8. **Relating the Time Periods**:
We can express the ratio of the new time period to the original time period:
\[ \frac{T_M}{T} = \sqrt{1 + \frac{Mg}{AY}} \]

9. **Squaring Both Sides**:
Squaring both sides gives:
\[ \left(\frac{T_M}{T}\right)^2 = 1 + \frac{Mg}{AY} \]

10. **Rearranging for Young's Modulus**:
Rearranging this equation to find \( \frac{1}{Y} \):
\[ \frac{Mg}{AY} = \left(\frac{T_M}{T}\right)^2 - 1 \]
Therefore,
\[ \frac{1}{Y} = \frac{A}{Mg} \left(\left(\frac{T_M}{T}\right)^2 - 1\right) \]

### Final Result:
Thus, the expression for \( \frac{1}{Y} \) is:
\[ \frac{1}{Y} = \frac{A}{Mg} \left(\left(\frac{T_M}{T}\right)^2 - 1\right) \]