Question

If the diameter of a wire of $ 1 \, \Omega $ is tripled, then its resistance will become:

• A. $ 9 \, \Omega $
• B. $ \frac{1}{9} \, \Omega $
• C. $ \frac{1}{3} \, \Omega $
• D. $ 3 \, \Omega $

Hint:

When the diameter of a wire is changed, the resistance changes inversely with the square of the diameter because the cross-sectional area depends on the square of the diameter. Always use the formula $ R \propto \frac{1}{A} $ to analyze such problems.

Answer 1

The Correct Option is B

Step 1: Understand Resistance Formula.
The resistance ($ R $) of a wire is given by: \[ R = \rho \frac{L}{A}, \] where:
$ \rho $ is the resistivity of the material,
$ L $ is the length of the wire,
$ A $ is the cross-sectional area of the wire. The cross-sectional area ($ A $) of a wire is related to its diameter ($ d $) by: \[ A = \pi \left( \frac{d}{2} \right)^2 = \frac{\pi d^2}{4}. \] Step 2: Analyze the Effect of Tripling the Diameter.
If the diameter is tripled, new diameter = $ 3d $. Then, \[ A_{\text{new}} = \pi \left( \frac{3d}{2} \right)^2 = \frac{\pi (3d)^2}{4} = \frac{9\pi d^2}{4} = 9A. \] Since resistance is inversely proportional to cross-sectional area: \[ R \propto \frac{1}{A}. \] So if $ A $ becomes 9 times, then: \[ R_{\text{new}} = \frac{R_{\text{original}}}{9}. \] Given $ R_{\text{original}} = 1 \, \Omega $, we have: \[ R_{\text{new}} = \frac{1}{9} \, \Omega. \] Step 3: Analyze the Options.
Option (1): $ 9 \, \Omega $ — Incorrect
Option (2): $ \frac{1}{9} \, \Omega $ — Correct
Option (3): $ \frac{1}{3} \, \Omega $ — Incorrect
Option (4): $ 3 \, \Omega $ — Incorrect Step 4: Final Answer.
\[ \boxed{(2) \quad \frac{1}{9} \, \Omega} \]