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Properties of Matter question

2023 · 31 Jan · Shift 1 · Q71
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  5. /2023 · 31 Jan · Shift 1 · Q71

Properties of Matter question

2023 · 31 Jan · Shift 1 · Q71

JEE MainPhysicsProperties of MatterNumerical+4 / −1
A thin rod having a length of 1 m1 \mathrm{~m}1 m and area of cross-section 3×10−6 m23 \times 10^{-6} \mathrm{~m}^{2}3×10−6 m2 is suspended vertically from one end. The rod is cooled from 210∘C210^{\circ} \mathrm{C}210∘C to 160∘C160^{\circ} \mathrm{C}160∘C. After cooling, a mass M\mathrm{M}M is attached at the lower end of the rod such that the length of rod again becomes 1 m1 \mathrm{~m}1 m. Young's modulus and coefficient of linear expansion of the rod are 2×1011 N m−22 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}2×1011 N m−2 and 2×10−5 K−12 \times 10^{-5} \mathrm{~K}^{-1}2×10−5 K−1, respectively. The value of M\mathrm{M}M is ‾\underline{\hspace{2cm}}​kg\mathrm{kg}kg. (Take g=10 m s−2\mathrm{g=10~m~s^{-2}}g=10 m s−2)
Numerical answer
View written solutionFree

Correct answer: 60

  1. Given data
  • Original length: L=1 mL = 1\,\text{m}L=1m
  • Cross-sectional area: A=3×10−6 m2A = 3 \times 10^{-6}\,\text{m}^2A=3×10−6m2
  • Temperature changes from 210∘C210^\circ\text{C}210∘C to 160∘C160^\circ\text{C}160∘C
  • So, temperature drop: ΔT=50 K\Delta T = 50\,\text{K}ΔT=50K
  • Young's modulus: Y=2×1011 N m−2Y = 2 \times 10^{11}\,\text{N m}^{-2}Y=2×1011N m−2
  • Coefficient of linear expansion: α=2×10−5 K−1\alpha = 2 \times 10^{-5}\,\text{K}^{-1}α=2×10−5K−1
  • g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2
  1. Contraction due to cooling

The rod shortens by

ΔLthermal=αLΔT\Delta L_{\text{thermal}} = \alpha L \Delta TΔLthermal​=αLΔT

Substitute values:

ΔLthermal=(2×10−5)(1)(50)=10−3 m\Delta L_{\text{thermal}} = (2 \times 10^{-5})(1)(50) = 10^{-3}\,\text{m}ΔLthermal​=(2×10−5)(1)(50)=10−3m

So after cooling, its length becomes

1−10−3=0.999 m1 - 10^{-3} = 0.999\,\text{m}1−10−3=0.999m
  1. Extension required to bring length back to 1 m1\,\text{m}1m

To restore the rod to length 1 m1\,\text{m}1m, required extension is exactly

ΔL=10−3 m\Delta L = 10^{-3}\,\text{m}ΔL=10−3m
  1. Use Young's modulus relation

For extension under load,

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

So,

ΔL=FLAY\Delta L = \frac{FL}{AY}ΔL=AYFL​

Here F=MgF = MgF=Mg. Thus,

10−3=Mg⋅1(3×10−6)(2×1011)10^{-3} = \frac{Mg \cdot 1}{(3 \times 10^{-6})(2 \times 10^{11})}10−3=(3×10−6)(2×1011)Mg⋅1​
  1. Simplify denominator
(3×10−6)(2×1011)=6×105(3 \times 10^{-6})(2 \times 10^{11}) = 6 \times 10^5(3×10−6)(2×1011)=6×105

Thus,

10−3=Mg6×10510^{-3} = \frac{Mg}{6 \times 10^5}10−3=6×105Mg​

So,

Mg=10−3×6×105=6×102=600 NMg = 10^{-3} \times 6 \times 10^5 = 6 \times 10^2 = 600\,\text{N}Mg=10−3×6×105=6×102=600N
  1. Find mass MMM
M=60010=60 kgM = \frac{600}{10} = 60\,\text{kg}M=10600​=60kg
  1. Final answer
60\boxed{60}60​
  1. Comparison with stored answer

Stored correct answer = 606060.

Our derived answer matches the stored answer.

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