Stress will be same in both rods
Stress =Y1lΔl1=Y2lΔl2 … (1)
And
Δl1+Δl2=0.2×10−3m … (2)
From equation (1) and (2)
Stress ×(Y1l+Y2l)=0.2×10−3m
Stress ×(120×1091+60×1091)=0.2×10−3m
Stress =8×106N/m2
JEE Main 2019 — Physics Mechanics
In an experiment, brass and steel wires of length 1m each with areas of cross section 1mm2 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 0.2mm is,
[Given, the Young's Modulus for steel and brass are, respectively, 120×109N/m2 and 60×109N/m2 ]
Held on 10 Apr 2019 · Verified 6 Jul 2026.
8.0×106N/m2
1.2×106N/m2
0.2×106N/m2
1.8×106N/m2
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