Solution
The problem involves the thermal equilibrium of a current-carrying wire. At equilibrium, the rate of heat generation ($P_{gen}$) equals the rate of heat loss ($P_{loss}$).
1. Heat Generation:
For a wire connected to an ideal battery of voltage $V$, the power generated is:
$$ P_{gen} = \frac{V^2}{R} $$
The resistance $R$ depends on temperature:
$$ R = \rho \frac{L}{A} = \rho_0 (1 + \alpha \Delta \theta) \frac{L}{A} $$
Thus,
$$ P_{gen} = \frac{V^2 A}{\rho_0 L (1 + \alpha \Delta \theta)} $$
2. Heat Loss:
Assuming Newton’s Law of Cooling, heat loss is proportional to the surface area ($S \propto L$) and temperature excess $\Delta \theta$:
$$ P_{loss} = k \cdot L \cdot \Delta \theta $$
3. Equilibrium Equation:
$$ \frac{V^2 A}{\rho_0 L (1 + \alpha \Delta \theta)} = k L \Delta \theta $$
Rearranging to separate the length term:
$$ \frac{V^2 A}{\rho_0 k} = L^2 \Delta \theta (1 + \alpha \Delta \theta) $$
Since $V, A, \rho_0, k$ are constant, let $C$ be the constant on the left side:
$$ C = L^2 \Delta \theta (1 + \alpha \Delta \theta) $$
4. Comparison of Two Cases:
Case 1: Length $L_1$, $\Delta \theta_1 = 100^\circ \text{C}$.
Case 2: Length $L_2 = L_1/2$, $\Delta \theta_2 = ?$
Equating the constant $C$ for both cases:
$$ L_1^2 \Delta \theta_1 (1 + \alpha \Delta \theta_1) = L_2^2 \Delta \theta_2 (1 + \alpha \Delta \theta_2) $$
Substitute $L_2 = L_1/2$:
$$ L_1^2 \Delta \theta_1 (1 + \alpha \Delta \theta_1) = \frac{L_1^2}{4} \Delta \theta_2 (1 + \alpha \Delta \theta_2) $$
$$ 4 \Delta \theta_1 (1 + \alpha \Delta \theta_1) = \Delta \theta_2 (1 + \alpha \Delta \theta_2) $$
5. Numerical Solution:
Given $\Delta \theta_1 = 100$ and $\alpha = 0.0050$.
$$ \text{LHS} = 4 \times 100 \times (1 + 0.0050 \times 100) = 400 \times (1 + 0.5) = 600 $$
Now solve for $\Delta \theta_2$:
$$ 600 = \Delta \theta_2 (1 + 0.005 \Delta \theta_2) $$
$$ 600 = \Delta \theta_2 + 0.005 \Delta \theta_2^2 $$
Multiply by 200:
$$ 120000 = 200 \Delta \theta_2 + \Delta \theta_2^2 $$
$$ \Delta \theta_2^2 + 200 \Delta \theta_2 – 120000 = 0 $$
Solving the quadratic:
$$ \Delta \theta_2 = \frac{-200 + \sqrt{200^2 – 4(1)(-120000)}}{2} $$
$$ \Delta \theta_2 = \frac{-200 + \sqrt{40000 + 480000}}{2} = \frac{-200 + \sqrt{520000}}{2} $$
$$ \Delta \theta_2 \approx \frac{-200 + 721.11}{2} \approx 260.55 $$
Answer: The temperature rise is approximately $261^\circ \text{C}$.