Question

Consider a cylindrical conductor of length $L$ and area of cross section $A$. The specific conductivity varies as $\sigma(x)=\sigma_0\frac{L}{\sqrt{x}}$ where $x$ is the distance along the axis of the cylinder from one of its ends. The resistance of the system along the cylindrical axis is

• A. \( \frac{2\sqrt{L}}{3A\sigma_0} \)
• B. \( \frac{3\sqrt{L}}{2A\sigma_0} \)
• C. \( \frac{\sqrt{L}}{3A\sigma_0} \)
• D. \( \frac{2\sqrt{L}}{A\sigma_0} \)
• E. \( \frac{4\sqrt{L}}{3A\sigma_0} \)

Hint:

Whenever a physical property (like conductivity, density, or mass) varies along a dimension, always set up an integral for a small element \(dx\). Ensure your limits of integration cover the physical boundaries of the object (in this case, 0 to L).

Answer 1

The Correct Option is A

Concept: The resistance \(R\) of a conductor with uniform cross-sectional area \(A\) and length \(L\) is given by \(R = \rho \frac{L}{A}\). Since conductivity \(\sigma\) is the reciprocal of resistivity (\(\rho = 1/\sigma\)), the formula becomes \(R = \frac{L}{\sigma A}\). When the conductivity \(\sigma(x)\) varies along the length, we must consider an infinitesimal element of thickness \(dx\) at a distance \(x\) from the end. The resistance of this small element \(dR\) is: \[ dR = \frac{dx}{\sigma(x) A} \] The total resistance is found by integrating this expression over the entire length from \(x = 0\) to \(x = L\).

Step 1: Express the elemental resistance \(dR\).
Given the specific conductivity function: \[ \sigma(x) = \sigma_0 \frac{L}{\sqrt{x}} \] Substitute \(\sigma(x)\) into the elemental resistance formula: \[ dR = \frac{dx}{\left( \sigma_0 \frac{L}{\sqrt{x}} \right) A} = \frac{\sqrt{x} \, dx}{\sigma_0 L A} \]

Step 2: Integrate to find the total resistance \(R\).
To find the total resistance, integrate \(dR\) from \(x = 0\) to \(x = L\): \[ R = \int_0^L \frac{\sqrt{x}}{\sigma_0 L A} \, dx \] Take the constants out of the integral: \[ R = \frac{1}{\sigma_0 L A} \int_0^L x^{1/2} \, dx \]

Step 3: Perform the integration and simplify.
Applying the power rule for integration \(\int x^n dx = \frac{x^{n+1}}{n+1}\): \[ R = \frac{1}{\sigma_0 L A} \left[ \frac{x^{3/2}}{3/2} \right]_0^L \] \[ R = \frac{1}{\sigma_0 L A} \cdot \frac{2}{3} \left[ L^{3/2} - 0 \right] \] \[ R = \frac{2 L^{3/2}}{3 \sigma_0 L A} \] Using the law of exponents \(\frac{L^{3/2}}{L^1} = L^{3/2 - 1} = L^{1/2} = \sqrt{L}\): \[ R = \frac{2 \sqrt{L}}{3 A \sigma_0} \]