Question

A cylindrical shape resistance is connected to a battery with emf $5V$. 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 $20W$, 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
$\frac{dR}{dx}=\rho {\left(\frac{x}{L}\right)}^{\alpha }\Rightarrow dR=\rho {\left(\frac{x}{L}\right)}^{\alpha }\cdot dx$
So, total resistance of cylinder,
$<br/>R={\int }_0^{L}dR={\int }_0^L\frac{\rho }{L^{\alpha }}\cdot x^{\alpha }dx=\frac{\rho }{L^{\alpha }}\cdot {\left(\frac{x^{\alpha +1}}{\alpha +1}\right)}_0^L=\frac{\rho L^{\alpha +1}}{L^{\alpha }(\alpha +1)}=\frac{\rho L}{\alpha +1}<br/>$
Power generated,
$<br/>\begin{array}{l}<br/>P=\frac{V^2}{R}=20\Rightarrow \frac{25}{\left(\frac{\rho L}{\alpha +1}\right)}=20\\ <br/>\Rightarrow \frac{\rho L}{\alpha +1}=\frac{5}{4}\Rightarrow \frac{10}{\alpha +1}=\frac{5}{4}\Rightarrow \alpha =7<br/>\end{array}<br/>$