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BITSAT 2013 Paper

Section: Physics

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Question : 16 of 150

Marks: +1, -0

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 0ºC and that of steel rod is kept at 100ºC. Find the temperature of the junction of the rod. Conductivity of copper = 390 W/mºC. Conductivity of steel = 46 W/m ºC

18.01 °C
26 °C
10.6 °C
20 °C

Solution:

Heat current in first rod (copper)

=

\frac{390 \times A (0-\theta)}{l}

Here θ is temperature of the junction and A & l are area and length of copper rod.

Heat current in second rod (steel)

=

\frac{46 \times A (\theta-100)}{l}

In series combination, heat current remainssame. So,

\frac{390 \times A (0-\theta)}{l}

=

\frac{46 \times A (\theta-100)}{l}

- 390 θ = 46 θ - 4600

436 θ = 4600 ⇒ θ = 10.6 °C

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