Mechanical Properties of Solids

JEE Main level, single correct, medium

53 of 118

Question 53 of 118

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50×103kg.50\times {10}^{3}kg. The inner and outer radii of each column are 50cm50cm and 100cm100cm respectively. Assuming uniform local distribution, calculate the compression strain of each column.
[use Y=2.0×1011Pa,g=9.8ms−2Y=2.0\times {10}^{11}Pa,g=9.8m{s}^{-2}]

Show answer and solution

Answer: (B) 2.60×10−72.60\times {10}^{-7}

ΔLL=mg4AY\dfrac{\Delta L}{L}=\dfrac{mg}{4AY}
=50×103×9.84π(1002−502)×10−4×2×1011=5×1054×π×150×50×10−4×2×1011=\dfrac{50\times {10}^{3}\times 9.8}{4\pi \left({100}^{2}-{50}^{2}\right)\times {10}^{-4}\times 2\times {10}^{11}}=\dfrac{5\times {10}^{5}}{4\times \pi \times 150\times 50\times {10}^{-4}\times 2\times {10}^{11}}
ΔLL=112π×10−5=2.60×10−7\dfrac{\Delta L}{L}=\dfrac{1}{12\pi }\times {10}^{-5}=2.60\times {10}^{-7}

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.