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JEE Main 2024 (1 Feb, Shift 2) · Elastic Behaviour, Stress & Strain

Elastic Behaviour, Stress & Strain question from JEE Main 2024 (1 Feb, Shift 2)

Q6JEE Main 2024 · 1 FEB · SHIFT 2Medium

Elastic Behaviour, Stress & Strain

One end of a metal wire is fixed to a ceiling and a load of 2 kg2\text{ kg} hangs from the other end. A similar wire is attached to the bottom of the load and another load of 1 kg1\text{ kg} hangs from this lower wire. Then the ratio of longitudinal strain of upper wire to that of the lower wire will be ________. [Area of cross section of wire =0.005 cm2,Y=2×1011 N m−2= 0.005\text{ cm}^2, Y = 2 \times 10^{11}\text{ N m}^{-2} and g=10 m s−2g = 10\text{ m s}^{-2} ]

Numerical answer type. Work it out, then open the answer.

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Answer: 3

  1. Upper wire holds both loads: T1=(2+1)×10=30 NT_1 = (2 + 1) \times 10 = 30\ \text{N}.
  2. Lower wire holds only the 1 kg load: T2=1×10=10 NT_2 = 1 \times 10 = 10\ \text{N}.
  3. Same AA and YY, so strain ∝T\propto T: ε1ε2=3010\frac{\varepsilon_1}{\varepsilon_2} = \frac{30}{10}
  4. Ratio =3= 3
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