The corfficient of linear expansion for a certain metal varies with temperature as `alpha(T)`. If `L_0` is the initial elgnth of the metal and the temperature of metal is changed from `T_0` to `T(T_0gtT)`, then

Options

A. `L=L_0int_(T_(0))^(T) alpha(T)dT`
B. `L=L_0[1+int_(T0)^(T)alpha(T)dT]`
C. `L=L_0[1-int_(T_(0))^(T)alpha(T)dT]`
D. `LgtL_0`

Explanation

To solve the problem, we need to derive the expression for the change in length of a metal rod when its temperature changes from \( T_0 \) to \( T \) (where \( T_0 > T \)) and the coefficient of linear expansion \( \alpha(T) \) varies with temperature.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Let \( L_0 \) be the initial length of the metal at temperature \( T_0 \).
- Let \( L \) be the final length of the metal at temperature \( T \).

2. **Understand the Coefficient of Linear Expansion:**
- The coefficient of linear expansion \( \alpha(T) \) is defined as:
\[ \frac{dL}{L_0} = \alpha(T) dT \]
- This equation relates the change in length \( dL \) to the change in temperature \( dT \).

3. **Set Up the Integral:**
- Rearranging the equation gives:
\[ dL = L_0 \alpha(T) dT \]
- To find the total change in length as the temperature changes from \( T_0 \) to \( T \), we integrate both sides:
\[ \int_{L_0}^{L} dL = L_0 \int_{T}^{T_0} \alpha(T) dT \]

4. **Evaluate the Integral:**
- The left side evaluates to:
\[ L - L_0 \]
- Thus, we have:
\[ L - L_0 = L_0 \int_{T}^{T_0} \alpha(T) dT \]

5. **Rearranging for Final Length:**
- Adding \( L_0 \) to both sides gives:
\[ L = L_0 + L_0 \int_{T}^{T_0} \alpha(T) dT \]
- This can be factored as:
\[ L = L_0 \left( 1 + \int_{T}^{T_0} \alpha(T) dT \right) \]

6. **Consider the Direction of Temperature Change:**
- Since \( T_0 > T \), the metal is cooling down, which means it will contract. Therefore, \( L < L_0 \).

### Final Expression:
The final expression for the length of the metal at temperature \( T \) is:
\[ L = L_0 \left( 1 + \int_{T}^{T_0} \alpha(T) dT \right) \]

### Conclusion:
The correct option is option B, which corresponds to this derived expression.