The coefficient of linear expansion `'alpha`' of the material of a rod of length `l_(0)` varies with absolute temperature as `alpha = aT -bT^(2)` where a & b are constant. The linear expansion of the rod when heated from `T_(1)` to `T_(2) = 2T_(1)` is :-

Options

A. `((3)/(2)aT_(1)^(2) - (7b)/(3)T_(1)^(3))L_(0)`
B. `(4a -(7b)/(3))T_(1)L_(0)`
C. `(2aT_(1)^(2) - (7b)/(3)T_(1)^(3))L_(0)`
D. None of these

Explanation

To solve the problem of finding the linear expansion of a rod when heated from temperature \( T_1 \) to \( T_2 = 2T_1 \), given that the coefficient of linear expansion \( \alpha \) varies with temperature as \( \alpha = aT - bT^2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Coefficient of Linear Expansion**:
The coefficient of linear expansion is given by:
\[ \alpha(T) = aT - bT^2 \]
This means that \( \alpha \) depends on the temperature \( T \).

2. **Set Up the Formula for Linear Expansion**:
The change in length \( \Delta L \) of the rod when heated can be expressed as:
\[ \Delta L = L_0 \int_{T_1}^{T_2} \alpha(T) \, dT \]
where \( L_0 \) is the original length of the rod.

3. **Substitute \( T_2 \)**:
Since \( T_2 = 2T_1 \), we can rewrite the integral:
\[ \Delta L = L_0 \int_{T_1}^{2T_1} (aT - bT^2) \, dT \]

4. **Evaluate the Integral**:
We can break the integral into two parts:
\[ \Delta L = L_0 \left( \int_{T_1}^{2T_1} aT \, dT - \int_{T_1}^{2T_1} bT^2 \, dT \right) \]

- For the first integral:
\[ \int aT \, dT = \frac{aT^2}{2} \]
Evaluating from \( T_1 \) to \( 2T_1 \):
\[ \left[ \frac{a(2T_1)^2}{2} - \frac{aT_1^2}{2} \right] = \frac{a(4T_1^2)}{2} - \frac{aT_1^2}{2} = \frac{3aT_1^2}{2} \]

- For the second integral:
\[ \int bT^2 \, dT = \frac{bT^3}{3} \]
Evaluating from \( T_1 \) to \( 2T_1 \):
\[ \left[ \frac{b(2T_1)^3}{3} - \frac{bT_1^3}{3} \right] = \frac{b(8T_1^3)}{3} - \frac{bT_1^3}{3} = \frac{7bT_1^3}{3} \]

5. **Combine the Results**:
Now substituting back into the expression for \( \Delta L \):
\[ \Delta L = L_0 \left( \frac{3aT_1^2}{2} - \frac{7bT_1^3}{3} \right) \]

6. **Final Expression**:
Thus, the linear expansion of the rod when heated from \( T_1 \) to \( 2T_1 \) is:
\[ \Delta L = L_0 \left( \frac{3a}{2} T_1^2 - \frac{7b}{3} T_1^3 \right) \]