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A cylindrical shape resistance is connected to a battery with emf $5 \mathrm{~V}$. 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 $20 \mathrm{~W}$, then the value of $\alpha$ is
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Verified Answer
The correct answer is:
7
Resistance of cylinder per unit Iength 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,
$$
\begin{aligned}
R=\int_0^L d R & =\int_0^L \frac{\rho}{L^\alpha} \cdot x^\alpha d x=\frac{\rho}{L^\alpha} \cdot\left(\frac{x^{\alpha+1}}{\alpha+1}\right)_0^L \\
& =\frac{\rho L^{\alpha+1}}{L^\alpha(\alpha+1)}=\frac{\rho L}{\alpha+1}
\end{aligned}
$$
Power generated,
$$
\begin{array}{r}
P=\frac{V^2}{R}=20 \Rightarrow \frac{25}{\left(\frac{\rho L}{\alpha+1}\right)}=20 \\
\Rightarrow \quad \frac{\rho L}{\alpha+1}=\frac{5}{4} \Rightarrow \frac{10}{\alpha+1}=\frac{5}{4} \Rightarrow \alpha=7
\end{array}
$$
$$
\frac{d R}{d x}=\rho\left(\frac{x}{L}\right)^\alpha \Rightarrow d R=\rho\left(\frac{X}{L}\right)^\alpha \cdot d x
$$
So, total resistance of cylinder,
$$
\begin{aligned}
R=\int_0^L d R & =\int_0^L \frac{\rho}{L^\alpha} \cdot x^\alpha d x=\frac{\rho}{L^\alpha} \cdot\left(\frac{x^{\alpha+1}}{\alpha+1}\right)_0^L \\
& =\frac{\rho L^{\alpha+1}}{L^\alpha(\alpha+1)}=\frac{\rho L}{\alpha+1}
\end{aligned}
$$
Power generated,
$$
\begin{array}{r}
P=\frac{V^2}{R}=20 \Rightarrow \frac{25}{\left(\frac{\rho L}{\alpha+1}\right)}=20 \\
\Rightarrow \quad \frac{\rho L}{\alpha+1}=\frac{5}{4} \Rightarrow \frac{10}{\alpha+1}=\frac{5}{4} \Rightarrow \alpha=7
\end{array}
$$
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