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JEE Main 2023 (6 Apr, Shift 2) · Elastic Behaviour, Stress & Strain

Elastic Behaviour, Stress & Strain question from JEE Main 2023 (6 Apr, Shift 2)

Q11JEE Main 2023 · 6 APR · SHIFT 2Easy

Elastic Behaviour, Stress & Strain

A metal block of mass mm is suspended from a rigid support through a metal wire of diameter 14 mm14\text{ mm}. The tensile stress developed in the wire under equilibrium state is 7×105 N m−27 \times 10^{5}\text{ N m}^{-2}. The value of mass mm is _________ kg\text{kg}. (Take, g=9.8 m s−2g = 9.8\text{ m s}^{-2} and π=227\pi = \frac{22}{7})

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

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Answer: 11

  1. Stress σ=mgA\sigma = \frac{mg}{A}, so m=σAgm = \frac{\sigma A}{g}.
  2. r=7 mmr = 7\text{ mm}: A=227×(7×10−3)2=1.54×10−4 m2A = \frac{22}{7} \times (7 \times 10^{-3})^2 = 1.54 \times 10^{-4}\ \text{m}^2
  3. m=7×105×1.54×10−49.8=107.89.8m = \frac{7 \times 10^{5} \times 1.54 \times 10^{-4}}{9.8} = \frac{107.8}{9.8}
  4. m=11 kgm = 11\ \text{kg}
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