Current Electricity

JEE Main level, single correct, medium

91 of 380

Question 91 of 380

The length of a given cylindrical wire is increased by 100%100 \%. Due to the consequent decrease in diameter the change in the resistance of the wire will be

Show answer and solution

Answer: (D) 300%300 \%

L1=2l\mathrm{L}_1=2 \mathrm{l} or (πr2l)=(πr22)(2l)\left(\pi \mathrm{r}^2 \mathrm{l}\right)=\left(\pi \mathrm{r}_2^2\right)(2 \mathrm{l})
⇒r2=r2\Rightarrow \mathrm{r}_2=\frac{\mathrm{r}}{\sqrt{2}};; R=ρ1π2\mathrm{R}=\rho \frac{1}{\pi^2}
Rnew =(ρ)21(π)(r/2)2=(ρ)4l(π)r2=4×R\mathrm{R}_{\text {new }}=(\rho) \frac{21}{(\pi)(\mathrm{r} / \sqrt{2})^2}=\frac{(\rho) 4 \mathrm{l}}{(\pi) \mathrm{r}^2}=4 \times \mathrm{R}
∴ΔR=4R−R=3R\therefore \Delta \mathrm{R}=4 \mathrm{R}-\mathrm{R}=3 \mathrm{R}
ΔRR%=3RR×100=300%\frac{\Delta \mathrm{R}}{\mathrm{R}} \%=\frac{3 \mathrm{R}}{\mathrm{R}} \times 100=300 \%

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Circuit builder simulationDrag the sliders and watch the numbers in this question change.Revise Current Electricity firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Current Electricity (Resistance and Resistivity), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.