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 step by step, we will analyze the situation involving the thin rod, the mass attached to it, and the effects of temperature change on the rod's length.

### Step 1: Understanding the Problem
We have a thin rod of negligible mass and a cross-sectional area \( S \) suspended vertically. The rod has an original length \( L_0 \) at temperature \( T \). A mass \( m \) is attached to the lower end of the rod. When the temperature is reduced to \( 0^\circ C \), the length of the rod remains \( L_0 \). We need to find the value of \( m \).

### Step 2: Analyzing Forces and Stresses
When the mass \( m \) is attached to the rod, it exerts a downward force due to gravity, which is \( mg \), where \( g \) is the acceleration due to gravity. This force creates a tension \( T \) in the rod, which can be expressed as:
\[ T = mg \]

### Step 3: Relating Stress and Strain
The stress \( \sigma \) in the rod due to the tension can be expressed as:
\[ \sigma = \frac{T}{S} = \frac{mg}{S} \]
The strain \( \epsilon \) in the rod is given by:
\[ \epsilon = \frac{\Delta L}{L_0} \]
where \( \Delta L \) is the change in length of the rod.

### Step 4: Using Young's Modulus
Young's modulus \( Y \) relates stress and strain:
\[ Y = \frac{\sigma}{\epsilon} \]
Substituting the expressions for stress and strain, we have:
\[ Y = \frac{\frac{mg}{S}}{\frac{\Delta L}{L_0}} \implies \Delta L = \frac{mg L_0}{Y S} \]

### Step 5: Considering Thermal Expansion
When the temperature is reduced from \( T \) to \( 0^\circ C \), the rod will contract. The decrease in length \( \Delta L' \) due to thermal contraction can be expressed as:
\[ \Delta L' = L_0 \alpha T \]
where \( \alpha \) is the coefficient of linear expansion.

### Step 6: Setting Up the Equation
Since the problem states that the length of the rod remains \( L_0 \) when the temperature is reduced, the increase in length due to the tension must equal the decrease in length due to thermal contraction:
\[ \Delta L = \Delta L' \]
Substituting the expressions we derived:
\[ \frac{mg L_0}{Y S} = L_0 \alpha T \]

### Step 7: Solving for Mass \( m \)
Now, we can solve for \( m \):
\[ mg = Y S \alpha T \]
\[ m = \frac{Y S \alpha T}{g} \]

### Final Answer
Thus, the value of \( m \) is:
\[ m = \frac{Y S \alpha T}{g} \]