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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Physics
Chapter
Thermal Properties of Matter

BITSAT 2013 · Physics · Thermal Properties of Matter

Figure shows a copper rod joined to a steel rod. The rods have...

Figure shows a copper rod joined to a steel rod. The rods have equal length and equal crosssectional area. The free end of the copper rod is kept at 0∘C0^{{}^{\circ}} \mathrm{C} and that of steel rod is kept at 100∘C100^{{}^{\circ}} \mathrm{C}. Find the temperature of the junction of the rod. Conductivity of copper =390W/\mathrmm∘C=390 \mathrm{ W} / \mathrmm^{{}^{\circ}} \mathrm{C} Conductivity of steel =46W/\mathrmm∘C=46 \mathrm{ W} / \mathrmm^{{}^{\circ}} \mathrm{C} Figure shows a copper rod joined to a steel rod. The rods have equal length and equal crosssectional area. The free end

Options

  1. A18.01∘C18.01^{{}^{\circ}} \mathrm{C}
  2. B26∘C26^{{}^{\circ}} \mathrm{C}
  3. C10.6∘C10.6^{{}^{\circ}} \mathrm{C}
  4. D20∘C20^{{}^{\circ}} \mathrm{C}

Correct answer

C. 10.6∘C10.6^{{}^{\circ}} \mathrm{C}

Step-by-step solution

Heat current in first rod (copper) =390×A(0−θ)ℓ\frac{390 \times \mathrm{A}(0-\theta)}{\ell} Here θ\theta is temperature of the junction and A  and ℓ\mathrm{A}\ \text{ and } \ell are area and length of copper rod. Heat current in second rod (steel) =46×A(θ−100)ℓ\frac{46 \times \mathrm{A}(\theta-100)}{\ell} In series combination, heat current remains same. So, 390×A(0−θ)ℓ=46×A(θ−100)ℓ−390θ==46θ−4600436θ=4600⇒θ=10.6∘C\begin{array}{l}\frac{390 \times \mathrm{A}(0-\theta)}{\ell}=\frac{46 \times \mathrm{A}(\theta-100)}{\ell}\\ -390 \theta==46 \theta-4600 \\ 436 \theta=4600 \Rightarrow \theta=10.6^{{}^{\circ}} \mathrm{C} \end{array}
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