Mechanical Properties of Solids

JEE Main level, single correct, medium

116 of 118

Question 116 of 118

A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N200 \mathrm{~N} to the lower end. The weight stretches the wire by 1 mm1 \mathrm{~mm}. Then the elastic energy stored in the wire is

Show answer and solution

Answer: (D) 0.1 J0.1 \mathrm{~J}

Elastic energy =12×F×x=\frac{1}{2} \times \mathrm{F} \times \mathrm{x}
F=200 N,x=1 mm=10−3 m∴E=12×200×1×10−3=0.1 J\mathrm{F}=200 \mathrm{~N}, \mathrm{x}=1 \mathrm{~mm}=10^{-3} \mathrm{~m} \quad \therefore \mathrm{E}=\frac{1}{2} \times 200 \times 1 \times 10^{-3}=0.1 \mathrm{~J}

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 (Work Done in Stretching a Wire), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.