Mechanical Properties of Solids

JEE Main level, single correct, medium

68 of 118

Question 68 of 118

The elastic limit of brass is 379MPa379MPa. The minimum diameter of a brass rod if it is to support a 400N400N load without exceeding its elastic limit will be

Show answer and solution

Answer: (C) 1.16mm1.16mm

Stress =FA=\dfrac{F}{A}
Elastic limit is 379MPa379 MPa
⇒379×106=400A=400πd2/4\Rightarrow 379\times {10}^{6}=\dfrac{400}{A}=\dfrac{400}{\pi {d}^{2}/4}
d=400×4π×379×106d=\sqrt{\dfrac{400\times 4}{\pi \times 379\times {10}^{6}}}
=1.16×10−3m=1.16\times {10}^{-3} m
=1.16mm=1.16 mm

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.