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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Physics
Chapter
Thermal Properties of Matter

JIPMER 2018 · Physics · Thermal Properties of Matter

How much should the temperature of a brass rod be increased so as...

How much should the temperature of a brass rod be increased so as to increase its length by 1%1 \% ? (Given, α\alpha for brass is 0.00002∘C−10.00002^{\circ} \mathrm{C}^{-1} ).

Options

  1. A300∘C300^{\circ} \mathrm{C}
  2. B400∘C400^{\circ} \mathrm{C}
  3. C500∘C500^{\circ} \mathrm{C}
  4. D550∘C550^{\circ} \mathrm{C}

Correct answer

C. 500∘C500^{\circ} \mathrm{C}

Step-by-step solution

Here, ΔT=\Delta \mathrm{T}= ?, ΔLL=1100\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{1}{100} α=0.00002∘C−1\alpha=0.00002^{\circ} \mathrm{C}^{-1} As ΔL=αLΔT\Delta \mathrm{L}=\alpha \mathrm{L} \Delta \mathrm{T} ∴αΔT=ΔLL\therefore \quad \alpha \Delta \mathrm{T}=\frac{\Delta \mathrm{L}}{\mathrm{L}} or ΔT=ΔLLα=1100×0.00002\Delta \mathrm{T}=\frac{\Delta \mathrm{L}}{\mathrm{L} \alpha}=\frac{1}{100 \times 0.00002} ΔT=1052×102=500∘C\Delta \mathrm{T}=\frac{10^5}{2 \times 10^2}=500^{\circ} \mathrm{C}
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