Physics Current Electricity MDCAT 2019
PMDC Verified Question 87 of 112
A copper wire has length L and cross-sectional area A. Its resistance is R. If we halved the length and halved the diameter of wire, then what will be the resistance of this wire?
A
R
B
2R
C
3R
D
4R
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: 2R
Concept:

Resistance depends on length and the square of the diameter (since Area \( \propto d^2 \)).

Formula:

$$ R = \rho \frac{L}{A} \propto \frac{L}{d^2} $$

Solution:

  • Original Resistance \( R \propto \frac{L}{d^2} \).


  • New length \( L' = \frac{L}{2} \).


  • New diameter \( d' = \frac{d}{2} \), so new Area \( A' \propto \left(\frac{d}{2}\right)^2 = \frac{d^2}{4} \). This means area is 1/4th of the original.


  • New Resistance \( R' \propto \frac{L/2}{d^2/4} = \frac{1/2}{1/4} \left( \frac{L}{d^2} \right) = 2 \left( \frac{L}{d^2} \right) = 2R \).


Why other options are incorrect:

  • Opt A: Assumes linear relation with diameter rather than area.


  • Opt D: Happens if one forgets to half the length, only scaling the area.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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