Q76 / 90 · JEE Main 2019 (10 Apr Shift 2) · Physics

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 ]

  • Correct answer
    A
    8.0×106 N/m2
  • B
    1.2×106 N/m2
  • C
    0.2×106 N/m2
  • D
    1.8×106 N/m2

Solution

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

Hence, the correct answer is option: 'A'.

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