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, we need to find the linear expansion of the rod when heated from temperature \( T_1 \) to \( T_2 = 2T_1 \). The coefficient of linear expansion \( \alpha \) is given by the equation:

\[ \alpha = aT - bT^2 \]

### Step 1: Write the expression for linear expansion
The linear expansion \( \Delta L \) of the rod can be expressed as:

\[ \Delta L = L_0 \int_{T_1}^{T_2} \alpha \, dT \]

### Step 2: Substitute the expression for \( \alpha \)
Substituting the expression for \( \alpha \) into the integral, we have:

\[ \Delta L = L_0 \int_{T_1}^{T_2} (aT - bT^2) \, dT \]

### Step 3: Evaluate the integral
Now, we can evaluate the integral:

\[ \Delta L = L_0 \left[ \frac{aT^2}{2} - \frac{bT^3}{3} \right]_{T_1}^{T_2} \]

### Step 4: Substitute the limits of integration
Substituting the limits \( T_1 \) and \( T_2 = 2T_1 \):

\[ \Delta L = L_0 \left( \left( \frac{a(2T_1)^2}{2} - \frac{b(2T_1)^3}{3} \right) - \left( \frac{aT_1^2}{2} - \frac{bT_1^3}{3} \right) \right) \]

### Step 5: Simplify the expression
Calculating the terms:

1. For \( T_2 = 2T_1 \):
\[ \frac{a(2T_1)^2}{2} = \frac{a \cdot 4T_1^2}{2} = 2aT_1^2 \]
\[ \frac{b(2T_1)^3}{3} = \frac{b \cdot 8T_1^3}{3} = \frac{8bT_1^3}{3} \]

2. For \( T_1 \):
\[ \frac{aT_1^2}{2} = \frac{aT_1^2}{2} \]
\[ \frac{bT_1^3}{3} = \frac{bT_1^3}{3} \]

Now substituting these back into the expression for \( \Delta L \):

\[ \Delta L = L_0 \left( \left( 2aT_1^2 - \frac{8bT_1^3}{3} \right) - \left( \frac{aT_1^2}{2} - \frac{bT_1^3}{3} \right) \right) \]

### Step 6: Combine the terms
Combining the terms gives:

\[ \Delta L = L_0 \left( 2aT_1^2 - \frac{aT_1^2}{2} - \frac{8bT_1^3}{3} + \frac{bT_1^3}{3} \right) \]

This simplifies to:

\[ \Delta L = L_0 \left( \frac{4aT_1^2}{2} - \frac{aT_1^2}{2} - \frac{7bT_1^3}{3} \right) \]
\[ \Delta L = L_0 \left( \frac{3aT_1^2}{2} - \frac{7bT_1^3}{3} \right) \]

### Final Result
Thus, the linear expansion of the rod when heated from \( T_1 \) to \( T_2 = 2T_1 \) is:

\[ \Delta L = L_0 \left( \frac{3aT_1^2}{2} - \frac{7bT_1^3}{3} \right) \]