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Heat and Thermodynamics question

2023 · 24 Jan · Shift 1 · Q66
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  5. /2023 · 24 Jan · Shift 1 · Q66

Heat and Thermodynamics question

2023 · 24 Jan · Shift 1 · Q66

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
A hole is drilled in a metal sheet. At 27∘C\mathrm{27^\circ C}27∘C, the diameter of hole is 5 cm. When the sheet is heated to 177∘C\mathrm{177^\circ C}177∘C, the change in the diameter of hole is d×10−3\mathrm{d\times10^{-3}}d×10−3 cm. The value of d will be ‾\underline{\hspace{2cm}}​ if coefficient of linear expansion of the metal is 1.6×10−5/∘1.6\times10^{-5}/^\circ1.6×10−5/∘ C.
Numerical answer
View written solutionFree

Correct answer: 12

  1. Concept used

A hole in a metal sheet expands exactly as if it were made of the same metal. So its diameter increases according to linear expansion:

ΔD=αD0ΔT\Delta D = \alpha D_0 \Delta TΔD=αD0​ΔT

where:

  • α=1.6×10−5 /∘C\alpha = 1.6\times 10^{-5}\,/^\circ\text{C}α=1.6×10−5/∘C
  • D0=5 cmD_0 = 5\,\text{cm}D0​=5cm
  • ΔT=177−27=150∘C\Delta T = 177 - 27 = 150^\circ\text{C}ΔT=177−27=150∘C
  1. Calculate change in diameter
ΔD=(1.6×10−5)(5)(150)\Delta D = (1.6\times 10^{-5})(5)(150)ΔD=(1.6×10−5)(5)(150)

First compute the numerical part:

1.6×5=81.6 \times 5 = 81.6×5=8 8×150=12008 \times 150 = 12008×150=1200

So,

ΔD=1200×10−5 cm\Delta D = 1200 \times 10^{-5}\,\text{cm}ΔD=1200×10−5cm ΔD=1.2×10−2 cm\Delta D = 1.2\times 10^{-2}\,\text{cm}ΔD=1.2×10−2cm

Now write it in the form d×10−3d\times 10^{-3}d×10−3 cm:

1.2×10−2=12×10−31.2\times 10^{-2} = 12\times 10^{-3}1.2×10−2=12×10−3

Hence,

d=12d = 12d=12
  1. Comparison with stored answer

Stored correct answer = 121212

Our derived answer also gives:

d=12d = 12d=12

So the answer matches.

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