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(C) The resistance per unit length is given by $\frac{dR}{dx} = \rho_0 \left(\frac{x}{L}\right)^\alpha$. Integrating this from $x=0$ to $x=L$ gives th

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$7$

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(C) The resistance per unit length is given by $\frac{dR}{dx} = \rho_0 \left(\frac{x}{L}\right)^\alpha$. Integrating this from $x=0$ to $x=L$ gives the total resistance $R$: $R = \int_0^L \frac{\rho_0}{L^\alpha} x^\alpha dx = \frac{\rho_0}{L^\alpha} \left[ \frac{x^{\alpha+1}}{\alpha+1} \right]_0^L = \frac{\rho_0 L^{\alpha+1}}{L^\alpha(\alpha+1)} = \frac{\rho_0 L}{\alpha+1}$. Given the power $P = 20 \ W$ and voltage $V = 5 \ V$,we use $P = \frac{V^2}{R}$: $20 = \frac{5^2}{R} \Rightarrow R = \frac{25}{20} = 1.25 \ \Omega$. Substituting $R = \frac{\rho_0 L}{\alpha+1}$ and $\rho_0 L = 10 \ \Omega$: $1.25 = \frac{10}{\alpha+1} \Rightarrow \alpha+1 = \frac{10}{1.25} = 8 \Rightarrow \alpha = 7$.

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