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Physics · Mechanical Properties of Matter

JEE Main 2025 — 8 April, Evening Shift — Question 64

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×1011Nm−2P \times 10^{11} \mathrm{Nm}^{-2}, where the value of PP is: (Take g=3π m/s2g=3 \pi \mathrm{~m} / \mathrm{s}^{2} )

  1. Option A:

    2.5

  2. Option B:

    25

  3. Option C:

    5

    Correct
  4. Option D:

    10

Answer: C

Step-by-step solution

Y=F/AΔL/L=FLAN=(500N)(3)π×9×10−6×1×10−4Y=\frac{F / A}{\Delta L / L}=\frac{F L}{A N}=\frac{(500 N)(3)}{\pi \times 9 \times 10^{-6} \times 1 \times 10^{-4}} =5×1011 N/m2=5 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
8 April, Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Matter
Topic
Longitudinal Strain and Elastic Potential Energy

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