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Properties of Matter question

2021 · 31 Aug · Shift 1 · Q55
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  5. /2021 · 31 Aug · Shift 1 · Q55

Properties of Matter question

2021 · 31 Aug · Shift 1 · Q55

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A uniform heavy rod of weight 10 kg ms −-− 2, cross-sectional area 100 cm2 and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 ×\times× 1011 Nm −-− 2. Neglecting the lateral contraction, find the elongation of rod due to its own weight.
  1. A
    2 ×\times× 10 −-− 9 m
  2. B
    5 ×\times× 10 −-− 8 m
  3. C
    4 ×\times× 10 −-− 8 m
  4. D
    5 ×\times× 10 −-− 10 m
View written solutionFree

Correct answer: D

  1. Given data
  • Weight of rod: W=10 kg m s−2=10 NW = 10\ \text{kg m s}^{-2} = 10\ \text{N}W=10 kg m s−2=10 N
  • Cross-sectional area: A=100 cm2=100×10−4=10−2 m2A = 100\ \text{cm}^2 = 100 \times 10^{-4} = 10^{-2}\ \text{m}^2A=100 cm2=100×10−4=10−2 m2
  • Length: L=20 cm=0.2 mL = 20\ \text{cm} = 0.2\ \text{m}L=20 cm=0.2 m
  • Young's modulus: Y=2×1011 N m−2Y = 2 \times 10^{11}\ \text{N m}^{-2}Y=2×1011 N m−2
  1. Formula for elongation due to self-weight

For a uniform rod hanging vertically, the tension varies from top to bottom. The elongation due to its own weight is:

ΔL=WL2AY\Delta L = \frac{W L}{2 A Y}ΔL=2AYWL​

This is because the average tension in the rod is W/2W/2W/2.

  1. Substitute values
ΔL=10×0.22×10−2×2×1011\Delta L = \frac{10 \times 0.2}{2 \times 10^{-2} \times 2 \times 10^{11}}ΔL=2×10−2×2×101110×0.2​

First, numerator:

10×0.2=210 \times 0.2 = 210×0.2=2

Denominator:

2×10−2×2×1011=4×1092 \times 10^{-2} \times 2 \times 10^{11} = 4 \times 10^92×10−2×2×1011=4×109

So,

ΔL=24×109=12×10−9=5×10−10 m\Delta L = \frac{2}{4 \times 10^9} = \frac{1}{2} \times 10^{-9} = 5 \times 10^{-10}\ \text{m}ΔL=4×1092​=21​×10−9=5×10−10 m
  1. Match with options
ΔL=5×10−10 m\boxed{\Delta L = 5 \times 10^{-10}\ \text{m}}ΔL=5×10−10 m​

So the correct option is:

D

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