Solution
We start with a cylindrical conductor connected to a battery. The wire is then stretched uniformly to a new length, while the volume remains constant.
1. Geometric Transformations
The wire is stretched to $\eta$ times its original length. Let the initial length be $L$ and radius be $r$. The new length is $L’ = \eta L$.
Since the volume of the material remains constant ($V = \pi r^2 L = \text{constant}$):
$$ \pi r^2 L = \pi (r’)^2 (\eta L) $$ $$ r^2 = \eta (r’)^2 \implies r’ = \frac{r}{\sqrt{\eta}} $$The new surface area $S’$ (curved surface area $2\pi r L$) becomes:
$$ S’ = 2\pi r’ L’ = 2\pi \left( \frac{r}{\sqrt{\eta}} \right) (\eta L) = \sqrt{\eta} (2\pi r L) = \sqrt{\eta} S $$2. Resistance and Power
The new resistance $R’$ is given by:
$$ R’ = \frac{\rho L’}{A’} = \frac{\rho (\eta L)}{(\pi (r’)^2)} = \frac{\rho \eta L}{\pi (r^2 / \eta)} = \eta^2 \frac{\rho L}{\pi r^2} = \eta^2 R $$Since the conductor remains connected to the same battery, the voltage $V$ is constant. The power generated is:
$$ P = \frac{V^2}{R} \quad \text{and} \quad P’ = \frac{V^2}{R’} = \frac{V^2}{\eta^2 R} = \frac{P}{\eta^2} $$3. Steady State Temperature
At steady state, Power Generated = Heat Dissipated. Using Newton’s Law of Cooling ($H = h S \Delta \theta$):
$$ P = h S \Delta \theta $$ $$ P’ = h S’ \Delta \theta’ $$Taking the ratio of the two states:
$$ \frac{P’}{P} = \frac{h S’ \Delta \theta’}{h S \Delta \theta} $$Substituting the values derived above:
$$ \frac{1}{\eta^2} = \frac{\sqrt{\eta} S}{S} \cdot \frac{\Delta \theta’}{\Delta \theta} $$ $$ \frac{1}{\eta^2} = \sqrt{\eta} \cdot \frac{\Delta \theta’}{\Delta \theta} $$ $$ \Delta \theta’ = \Delta \theta \cdot \frac{1}{\eta^2 \sqrt{\eta}} = \Delta \theta \left( \frac{1}{\eta} \right)^{5/2} $$4. Numerical Calculation
Given values:
- Room temperature $\theta_0 = 0^\circ \text{C}$
- Initial steady temperature $\theta = 20\sqrt{2}^\circ \text{C}$
- Stretching factor $\eta = 2$
Initial temperature excess $\Delta \theta = \theta – \theta_0 = 20\sqrt{2}^\circ \text{C}$.
Calculating the new temperature excess $\Delta \theta’$:
$$ \Delta \theta’ = (20\sqrt{2}) \times \left( \frac{1}{2} \right)^{5/2} $$ $$ \Delta \theta’ = 20\sqrt{2} \times \frac{1}{2^2 \cdot 2^{1/2}} = 20\sqrt{2} \times \frac{1}{4\sqrt{2}} $$ $$ \Delta \theta’ = \frac{20}{4} = 5^\circ \text{C} $$The final temperature of the conductor is:
$$ \theta_{\text{final}} = \theta_0 + \Delta \theta’ = 0 + 5 = 5.0^\circ \text{C} $$