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Properties of Matter question

2025 · 8 Apr · Shift 2 · Q59
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Properties of Matter question

2025 · 8 Apr · Shift 2 · Q59

JEE MainPhysicsProperties of MatterMCQ+4 / −1
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×1011 Nm−2P \times 10^{11} \, \text{Nm}^{-2}P×1011Nm−2, where the value of PPP is: (Take g=3π m/s2g = 3\pi \, \text{m/s}^2g=3πm/s2)
  1. A
    2.5
  2. B
    25
  3. C
    10
  4. D
    5
View written solutionFree

Correct answer: D

  1. Use Young's modulus formula

Young's modulus is

Y=stressstrain=F/AΔL/L=FLAΔLY = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}Y=strainstress​=ΔL/LF/A​=AΔLFL​

where:

  • L=3 mL = 3\,\text{m}L=3m
  • radius r=3 mm=3×10−3 mr = 3\,\text{mm} = 3 \times 10^{-3}\,\text{m}r=3mm=3×10−3m
  • extension ΔL=0.1 mm=10−4 m\Delta L = 0.1\,\text{mm} = 10^{-4}\,\text{m}ΔL=0.1mm=10−4m
  • mass m=50 kgm = 50\,\text{kg}m=50kg
  • g=3π m/s2g = 3\pi\,\text{m/s}^2g=3πm/s2
  1. Calculate the force
F=mg=50×3π=150π NF = mg = 50 \times 3\pi = 150\pi\,\text{N}F=mg=50×3π=150πN
  1. Calculate cross-sectional area
A=πr2=π(3×10−3)2=9π×10−6 m2A = \pi r^2 = \pi (3 \times 10^{-3})^2 = 9\pi \times 10^{-6}\,\text{m}^2A=πr2=π(3×10−3)2=9π×10−6m2
  1. Substitute into the formula
Y=(150π)(3)(9π×10−6)(10−4)Y = \frac{(150\pi)(3)}{(9\pi \times 10^{-6})(10^{-4})}Y=(9π×10−6)(10−4)(150π)(3)​

Cancel π\piπ:

Y=4509×10−10=50×1010=5×1011 N/m2Y = \frac{450}{9 \times 10^{-10}} = 50 \times 10^{10} = 5 \times 10^{11}\,\text{N/m}^2Y=9×10−10450​=50×1010=5×1011N/m2
  1. Find PPP

Given

Y=P×1011 N/m2Y = P \times 10^{11}\,\text{N/m}^2Y=P×1011N/m2

So,

P=5P = 5P=5
  1. Check options
  • A: 2.52.52.5 ✗
  • B: 252525 ✗
  • C: 101010 ✗
  • D: 555 ✓

Therefore, the correct option is D.

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