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JEE Main 2022 (30 Jul, Shift 2) · Elastic Behaviour, Stress & Strain

Elastic Behaviour, Stress & Strain question from JEE Main 2022 (30 Jul, Shift 2)

Q14JEE Main 2022 · 30 JUL · SHIFT 2Medium

Elastic Behaviour, Stress & Strain

A block of mass M is attached to a wire. The upper part of wire is made of steel and lower part of wire of copper ( as shown in figure ). Both parts have the same area of cross section. Neglecting the mass of wires, the ratio of longitudinal strain developed in steel and copper wires is: ( Given, Young's modulus of steel =2×1011 N/m2= 2 \times 10^{11}\text{ N/m}^2 and Young's modulus of copper =1.5×1011 N/m2= 1.5 \times 10^{11}\text{ N/m}^2)

Figure

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Answer: (d)

  1. Massless wire: both parts carry the same tension MgMg and have the same area, so the same stress.
  2. Strain =stressY∝1Y= \frac{\text{stress}}{Y} \propto \frac{1}{Y}
  3. εsεc=YcYs=1.5×10112×1011=34\frac{\varepsilon_s}{\varepsilon_c} = \frac{Y_c}{Y_s} = \frac{1.5 \times 10^{11}}{2 \times 10^{11}} = \frac{3}{4}
  4. So the answer is (d) 34\frac{3}{4}.
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