Question

A cylindrical shape resistance is connected to a battery with emf $5\, V$. The resistance per unit length varies as $\rho(x)=\rho_{0}\left(\frac{x}{L}\right)^{\alpha}$, where $\rho_{0}$ and $\alpha$ are constants and $x$ is the distance from one end of the resistor. The magnitude or product $\rho_{0} L$ is $10\, \Omega$, where $L$ is the length of the resistor. If the thermal power generated by the resistor is $20\, W$, then the value of $\alpha$ is

• A. 3
• B. 5
• C. 7
• D. 9

Answer 1

Answer: (C)

Resistance of cylinder per unit length is
$\frac{d R}{d x}=\rho\left(\frac{x}{L}\right)^{\alpha} \Rightarrow d R=\rho\left(\frac{x}{L}\right)^{\alpha} \cdot d x$
So, total resistance of cylinder,
$
R=\int_{0}^{L} d R=\int_{0}^{L} \frac{\rho}{L^{\alpha}} \cdot x^{\alpha} d x=\frac{\rho}{L^{\alpha}} \cdot\left(\frac{x^{\alpha+1}}{\alpha+1}\right)_{0}^{L}=\frac{\rho L^{\alpha+1}}{L^{\alpha}(\alpha+1)}=\frac{\rho L}{\alpha+1}
$
Power generated,
$
\begin{array}{l}
P=\frac{V^{2}}{R}=20 \Rightarrow \frac{25}{\left(\frac{\rho L}{\alpha+1}\right)}=20 \\
\Rightarrow \frac{\rho L}{\alpha+1}=\frac{5}{4} \Rightarrow \frac{10}{\alpha+1}=\frac{5}{4} \Rightarrow \alpha=7
\end{array}
$