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JEE Main 2025 (8 Apr, Shift 2) · Young's Modulus & Elongation

Young's Modulus & Elongation question from JEE Main 2025 (8 Apr, Shift 2)

Q4JEE Main 2025 · 8 APR · SHIFT 2Easy

Young's Modulus & Elongation

A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is P×1011P \times 10^{11} Nm−2{}^{-2}, where the value of PP is: (Take g=3πg = 3\pi m/s2{}^{2})

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Answer: (d)

  1. Y=MgLπr2 ΔLY = \frac{MgL}{\pi r^2\, \Delta L}
  2. =50×3π×3π×(3×10−3)2×10−4=4509×10−10= \frac{50 \times 3\pi \times 3}{\pi \times (3 \times 10^{-3})^2 \times 10^{-4}} = \frac{450}{9 \times 10^{-10}}
  3. =5×1011 N m−2= 5 \times 10^{11}\ \text{N m}^{-2}
  4. So P=5P = 5, answer (d).
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