Coefficient of thermal expansion of a wire varies with temperature T as `alpha = 1.6 xx 10^(-6) T`, where T is temperature in degree Celsius. If the length of wire is 10 m at` 0^@C`, what is the change in length of wire (in mm), when its temperature is raised to `25^@C`?

Explanation

To find the change in length of the wire when its temperature is raised from \(0^\circ C\) to \(25^\circ C\), we will use the given coefficient of thermal expansion, which varies with temperature \(T\) as:

\[ \alpha = 1.6 \times 10^{-6} T \]

### Step 1: Determine the change in temperature
The initial temperature \(T_1\) is \(0^\circ C\) and the final temperature \(T_2\) is \(25^\circ C\). Therefore, the change in temperature \(\Delta T\) is:

\[ \Delta T = T_2 - T_1 = 25^\circ C - 0^\circ C = 25^\circ C \]

**Hint:** Remember that the change in temperature is simply the final temperature minus the initial temperature.

### Step 2: Set up the expression for change in length
The change in length \(\Delta L\) of the wire can be expressed as:

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

where \(L_0\) is the initial length of the wire, which is \(10 \, m\).

**Hint:** The integral of the coefficient of thermal expansion over the temperature range gives the total change in length.

### Step 3: Substitute \(\alpha(T)\) into the integral
Substituting the expression for \(\alpha\):

\[ \Delta L = L_0 \int_{0}^{25} (1.6 \times 10^{-6} T) \, dT \]

### Step 4: Evaluate the integral
Calculating the integral:

\[ \Delta L = 10 \int_{0}^{25} (1.6 \times 10^{-6} T) \, dT \]

The integral of \(T\) is:

\[ \int T \, dT = \frac{T^2}{2} \]

Thus,

\[ \Delta L = 10 \times 1.6 \times 10^{-6} \left[ \frac{T^2}{2} \right]_{0}^{25} \]

### Step 5: Calculate the definite integral
Calculating the limits:

\[ \Delta L = 10 \times 1.6 \times 10^{-6} \left( \frac{25^2}{2} - \frac{0^2}{2} \right) \]

Calculating \(25^2\):

\[ 25^2 = 625 \]

So,

\[ \Delta L = 10 \times 1.6 \times 10^{-6} \times \frac{625}{2} \]

Calculating further:

\[ \Delta L = 10 \times 1.6 \times 10^{-6} \times 312.5 = 5.0 \times 10^{-4} \, m \]

### Step 6: Convert to mm
To convert meters to millimeters, multiply by \(1000\):

\[ \Delta L = 5.0 \times 10^{-4} \, m \times 1000 = 0.5 \, mm \]

### Final Answer
The change in length of the wire when its temperature is raised to \(25^\circ C\) is:

\[ \Delta L = 0.5 \, mm \]

---