A thin rod of negligible mass and area of cross section S is suspended vertically from one end. Length of the rod is `L_(0)` at `T^(@)C` . A mass m is attached to the lower end of the rod so that when temperature of the rod is reduced to `0^(@)C` its length remains `L_(0)` Y is the Young’s modulus of the rod and `alpha` is coefficient of linear expansion of rod. Value of m is :

Options

A. `(SYalphaT)/(g)`
B. `(SYalphaT)/(2g)`
C. `(SYalpha)/(2gT)`
D. `(SYalphaT)/(2gL_(0))`

Explanation

To solve the problem, we need to find the value of the mass \( m \) that is attached to the lower end of a thin rod, which is suspended vertically. The rod has a negligible mass and a cross-sectional area \( S \). The length of the rod is \( L_0 \) at temperature \( T_0 \) and remains \( L_0 \) when the temperature is reduced to \( 0^\circ C \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The rod is initially at length \( L_0 \) at temperature \( T_0 \).
- When the temperature is reduced to \( 0^\circ C \), the rod would normally contract due to thermal contraction.
- The mass \( m \) is attached to the rod to counteract this contraction, keeping the length of the rod at \( L_0 \).

2. **Calculate the Change in Length Due to Temperature**:
- The change in length \( \Delta L_1 \) when the temperature changes from \( T_0 \) to \( 0^\circ C \) can be calculated using the formula for linear expansion:
\[ \Delta L_1 = L_0 \cdot \alpha \cdot (0 - T_0) = -L_0 \cdot \alpha \cdot T_0 \]
- Here, \( \alpha \) is the coefficient of linear expansion.

3. **Extension Due to the Mass**:
- When the mass \( m \) is attached to the rod, it causes an extension \( \Delta L_2 \). The force due to the mass is \( mg \), where \( g \) is the acceleration due to gravity.
- The extension can be expressed using Young's modulus \( Y \):
\[ \Delta L_2 = \frac{mg L_0}{Y \cdot S} \]

4. **Setting Up the Equation**:
- Since the length of the rod remains the same at \( L_0 \), the total change in length must equal zero:
\[ \Delta L_1 + \Delta L_2 = 0 \]
- Substituting the expressions for \( \Delta L_1 \) and \( \Delta L_2 \):
\[ -L_0 \cdot \alpha \cdot T_0 + \frac{mg L_0}{Y \cdot S} = 0 \]

5. **Solving for the Mass \( m \)**:
- Rearranging the equation gives:
\[ \frac{mg L_0}{Y \cdot S} = L_0 \cdot \alpha \cdot T_0 \]
- Canceling \( L_0 \) from both sides (since \( L_0 \neq 0 \)):
\[ mg = Y \cdot S \cdot \alpha \cdot T_0 \]
- Finally, solving for \( m \):
\[ m = \frac{Y \cdot S \cdot \alpha \cdot T_0}{g} \]

### Final Result:
The value of the mass \( m \) is given by:
\[ m = \frac{Y \cdot S \cdot \alpha \cdot T_0}{g} \]