Question

A cylinder of length \(L\) is made up of an inner core of steel of radius \(r\) and an outer sheath of copper of thickness \(r\). The resistivities of steel and copper are \(\rho_1\) and \(\rho_2\) respectively. The total resistance of the cylinder is:

• A. \(\dfrac{\rho_1 \rho_2 L}{\pi r^2 (3\rho_1 + \rho_2)}\)
• B. \(\dfrac{(3\rho_1 + \rho_2) L}{\pi r^2 (\rho_1 \rho_2)}\)
• C. \(\dfrac{(3\rho_1 + \rho_2) L}{\pi r^2}\)
• D. \(\dfrac{(\rho_1 + \rho_2) L}{\pi r^2}\)

Hint:

Treat core (area \(\pi r^2\)) and sheath (annulus area \(3\pi r^2\)) as two resistors in parallel over the same length.

Answer 1

The Correct Option is A

Step 1: The steel core and the copper sheath share the same length \(L\) and the same end faces, so they carry current in parallel. Compute each resistance separately.

Step 2: The steel core has radius \(r\), so its cross-sectional area is \(A_1 = \pi r^2\).
\[R_1 = \frac{\rho_1 L}{\pi r^2}\]
Step 3: The copper sheath is an annulus of inner radius \(r\) and outer radius \(2r\) (thickness \(r\)). Its area is
\[A_2 = \pi\big((2r)^2 - r^2\big) = \pi(4r^2 - r^2) = 3\pi r^2\]
\[R_2 = \frac{\rho_2 L}{3\pi r^2}\]
Step 4: Combine in parallel:
\[R = \frac{R_1 R_2}{R_1 + R_2}\]
Numerator: \(R_1 R_2 = \dfrac{\rho_1 \rho_2 L^2}{3\pi^2 r^4}\).
Denominator: \(R_1 + R_2 = \dfrac{L}{\pi r^2}\left(\rho_1 + \dfrac{\rho_2}{3}\right) = \dfrac{L(3\rho_1 + \rho_2)}{3\pi r^2}\).
Step 5: Divide:
\[R = \frac{\rho_1 \rho_2 L^2}{3\pi^2 r^4} \times \frac{3\pi r^2}{L(3\rho_1 + \rho_2)} = \frac{\rho_1 \rho_2 L}{\pi r^2 (3\rho_1 + \rho_2)}\]
This matches option (A).
\[\boxed{R = \dfrac{\rho_1 \rho_2 L}{\pi r^2 (3\rho_1 + \rho_2)}}\]