Question

Two conductors A and B are made of the same material and have the same length. Conductor A is a solid wire of diameter \(d_1\), and conductor B is a hollow tube of outer diameter 2.0 mm and inner diameter 1.0 mm. The resistances of the conductors A and B are \(R_1\) and \(R_2\) respectively. If \(\frac{R_1}{R_2} = 3\), then \(d_1\) is:

• A. 1.0 mm
• B. 1.5 mm
• C. 0.75 mm
• D. 2.0 mm

Hint:

For comparing resistances of solid and hollow wires of same length and material, use \(R = \rho L / A\) and calculate the effective cross-sectional area.

Answer 1

The Correct Option is A

Step 1: Resistance formula for a wire.
For a conductor of length \(L\), cross-sectional area \(A\) and resistivity \(\rho\): \[ R = \rho \frac{L}{A} \]

Step 2: Cross-sectional areas of solid and hollow wire.
Solid wire of diameter \(d_1\): \[ A_1 = \pi \left(\frac{d_1}{2}\right)^2 \] Hollow tube with outer diameter 2 mm and inner diameter 1 mm: \[ A_2 = \pi \left(\frac{2^2 - 1^2}{4}\right) \text{ mm}^2 = \pi \frac{4-1}{4} = \frac{3\pi}{4} \text{ mm}^2 \]

Step 3: Express resistance ratio.
\[ \frac{R_1}{R_2} = \frac{\rho L / A_1}{\rho L / A_2} = \frac{A_2}{A_1} = 3 \]

Step 4: Substitute areas and solve for \(d_1\).
\[ \frac{3\pi/4}{\pi (d_1/2)^2} = 3 \] \[ \frac{3/4}{(d_1^2/4)} = 3 \] \[ \frac{3/4}{d_1^2/4} = \frac{3}{4} \cdot \frac{4}{d_1^2} = \frac{3}{d_1^2} = 3 \]

Step 5: Solve for \(d_1\).
\[ d_1^2 = 1 \] \[ d_1 = 1.0 \text{ mm} \]

Step 6: Final conclusion.
Hence, diameter of solid wire is: \[ \boxed{1.0 \, \text{mm}} \]