Mechanical Properties of Solids

JEE Main level, numerical answer, medium

27 of 118

Question 27 of 118

Each of three blocks P,QP,Q and RR shown in figure has a mass of 3kg3kg. Each of the wire AA and BB has cross-sectional area 0.005cm20.005{cm}^{2} and Young's modulus 2×1011Nm−22\times {10}^{11}N{m}^{-2}. Neglecting friction, the longitudinal strain on wire BB is _____ ×10−4\times {10}^{-4}. (Take g=10ms−2g=10m{s}^{-2})

Show answer and solution

Answer: 2


Acceleration of the system,
a=3g9=103ms−2a=\dfrac{3g}{9}=\dfrac{10}{3}m{s}^{-2}
Now, for
RR we can write
30−T1=3×a30-{T}_{1}=3\times a
⇒T1=20N\Rightarrow {T}_{1}=20N
Therefore, strain developed in wire
BB,
strain=stressY=T1AY=200.005×10−42×1011\text{strain}=\dfrac{\text{stress}}{Y} =\dfrac{\dfrac{{T}_{1}}{A}}{Y}=\dfrac{\dfrac{20}{0.005\times {10}^{-4}}}{2\times {10}^{11}}
=2×10−4=2\times {10}^{-4}

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.