JEE Main 2019PhysicsMechanical Properties of SolidsHardMCQ

JEE Main 2019Mechanical Properties of Solids Question with Solution

JEE Main 2019 (10 Apr Shift 2)

Question

In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mm2 are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is,
[Given, the Young's Modulus for steel and brass are, respectively, 120×109 N/m2 and 60×109 N/m2 ]

Choose an option

Show full solutionCorrect option: A
Correct answer
A8.0×106 N/m2

Step-by-step explanation

Stress will be same in both rods
Stress =Y1Δl1l=Y2Δl2l … (1)
And
Δl1+Δl2=0.2×10-3 m … (2)
From equation (1) and (2)
Stress ×lY1+lY2=0.2×10-3 m
Stress ×1120×109+160×109=0.2×10-3 m
Stress =8×106 N/m2

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About this question

This is a previous-year question from JEE Main 2019, covering the Mechanical Properties of Solids chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.