Current Electricity

JEE Main level, single correct, medium

56 of 380

Question 56 of 380

A wire of resistance 160Ω160Ω is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

Show answer and solution

Answer: (B) 10Ω10Ω

The formula to calculate the resistance of the original wire is given by
R=ρlA...(1)R=\rho \dfrac{l}{A}...\left(1\right)
When the new wire is formed, its resistance is given by
R′=ρl′4A...(2)R'=\rho \dfrac{l'}{4A}...\left(2\right)
Divide equation (2) by equation (1) and simplify to obtain the expression for the new resistance.
R′R=ρl′4AρlA⇒R′=Rl′4l...(3)\begin{matrix}\dfrac{R'}{R} & = & \dfrac{\rho \dfrac{l'}{4A}}{\rho \dfrac{l}{A}} \\ & \Rightarrow & R'=R\dfrac{l'}{4l}...\left(3\right)\end{matrix}
Substitute the values of the known parameters into equation (3) to calculate the required new resistance.
R′=l44l×160Ω=10Ω\begin{matrix}R' & = & \dfrac{\dfrac{l}{4}}{4l}\times 160\Omega \\ & = & 10\Omega \end{matrix}

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.