The coefficient of linear expansion `'alpha`' of a rod of length `2` m varies with the distance x from the end of the rod as ` alpha = alpha_(0) + alpha_(1) "x" ` where `alpha_(0) = 1.76 xx 10^(-5) "^(@)C^(-1)"and" alpha_(1) = 1.2 xx 10^(-6)m^(-1) "^(@)C^(-1)`. The increase in the length of the rod. When heated through `100^(@)C` is :-

Options

A. `2`cm
B. `3.76`mm
C. `1.2`mm
D. None of these

Explanation

To solve the problem of finding the increase in the length of a rod when heated, we will follow these steps:

### Step 1: Understand the Coefficient of Linear Expansion
The coefficient of linear expansion \( \alpha \) varies with the distance \( x \) from the end of the rod as:
\[ \alpha = \alpha_0 + \alpha_1 x \]
where:
- \( \alpha_0 = 1.76 \times 10^{-5} \, ^\circ C^{-1} \)
- \( \alpha_1 = 1.2 \times 10^{-6} \, m^{-1} \, ^\circ C^{-1} \)

### Step 2: Set Up the Expression for Length Change
The increase in length \( dL \) of an infinitesimal element \( dx \) of the rod when the temperature changes by \( \Delta T \) is given by:
\[ dL = \alpha \cdot dx \cdot \Delta T \]
Substituting the expression for \( \alpha \):
\[ dL = (\alpha_0 + \alpha_1 x) \cdot dx \cdot \Delta T \]

### Step 3: Integrate Over the Length of the Rod
To find the total increase in length \( L \) when the entire rod is heated, we need to integrate from \( x = 0 \) to \( x = L \):
\[ \Delta L = \int_0^L (\alpha_0 + \alpha_1 x) \, dx \cdot \Delta T \]
where \( L = 2 \, m \) and \( \Delta T = 100 \, ^\circ C \).

### Step 4: Calculate the Integral
The integral can be split into two parts:
\[ \Delta L = \Delta T \left( \int_0^L \alpha_0 \, dx + \int_0^L \alpha_1 x \, dx \right) \]
Calculating the first integral:
\[ \int_0^L \alpha_0 \, dx = \alpha_0 \cdot L \]
Calculating the second integral:
\[ \int_0^L \alpha_1 x \, dx = \alpha_1 \cdot \frac{L^2}{2} \]
Thus, we have:
\[ \Delta L = \Delta T \left( \alpha_0 L + \alpha_1 \frac{L^2}{2} \right) \]

### Step 5: Substitute the Values
Substituting the known values:
\[ \Delta L = 100 \left( 1.76 \times 10^{-5} \cdot 2 + 1.2 \times 10^{-6} \cdot \frac{2^2}{2} \right) \]
Calculating each term:
1. \( 1.76 \times 10^{-5} \cdot 2 = 3.52 \times 10^{-5} \)
2. \( 1.2 \times 10^{-6} \cdot \frac{4}{2} = 2.4 \times 10^{-6} \)

Now, adding these:
\[ \Delta L = 100 \left( 3.52 \times 10^{-5} + 2.4 \times 10^{-6} \right) \]
Calculating the sum:
\[ 3.52 \times 10^{-5} + 2.4 \times 10^{-6} = 3.76 \times 10^{-5} \]

### Step 6: Final Calculation
Now, multiplying by 100:
\[ \Delta L = 100 \cdot 3.76 \times 10^{-5} = 3.76 \times 10^{-3} \, m \]
Converting to millimeters:
\[ \Delta L = 3.76 \, mm \]

### Conclusion
The increase in the length of the rod when heated through \( 100^\circ C \) is \( 3.76 \, mm \).

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