Mechanical Properties of Solids

JEE Main level, single correct, medium

42 of 118

Question 42 of 118

A 100m100m long wire having cross-sectional area 6.25×10−4m26.25\times {10}^{-4}{m}^{2} and Young's modulus is 1010Nm−2{10}^{10}N{m}^{-2} is subjected to a load of 250N250N, then the elongation in the wire will be :

Show answer and solution

Answer: (D) 4×10−3m4\times {10}^{-3}m

As we know, elongation in wire is given by,
∆l=FlAY∆l=\dfrac{Fl}{AY}
⇒∆l=250×1006.25×10−4×1010\Rightarrow ∆l=\dfrac{250\times 100}{6.25\times {10}^{-4}\times {10}^{10}}
⇒∆l=4×10−3m\Rightarrow ∆l=4\times {10}^{-3}m

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: Elasticity: stress, strain and Young’s modulus simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Mechanical Properties of Solids (Young's Modulus and Breaking Stres), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.