One rod of length 2 m and thermal conductivity 50 unit is attached...

One rod of length 2 m and thermal conductivity 50 unit is attached to another rod of length 1 m and thermal conductivity 100 unit. Temperature of free ends are

70^{\circ} \mathrm{C}

and

50^{\circ} \mathrm{C}

respectively. Then temperature of junction point will be One rod of length 2 m and thermal conductivity 50 unit is attached to another rod of length 1 m and thermal

One rod of length 2 m and thermal conductivity 50 unit is attached to another rod of length 1 m and thermal

Options

A $60^{\circ} \mathrm{C}$
B $54^{\circ} \mathrm{C}$
C $64^{\circ} \mathrm{C}$
D $68^{\circ} \mathrm{C}$

Correct answer

B. $54^{\circ} \mathrm{C}$

Step-by-step solution

Let

\theta

be the junction temperature, then given situation is shown in the following figure. One rod of length 2 m and thermal conductivity 50 unit is attached to another rod of length 1 m and thermal

The rate of heat current due to conduction in both rod will be same, i.e,

\mathrm{H}_1=\mathrm{H}_2

\begin{aligned} \Rightarrow \quad \mathrm{K}_1 \frac{A \Delta \theta_1}{\mathrm{l}_1} & =\mathrm{K}_2 \mathrm{~A} \cdot \frac{\Delta \theta_2}{\mathrm{l}_2} \\ \mathrm{~K}_1 \frac{\Delta \theta_1}{\mathrm{l}_1} & =\mathrm{K}_2 \frac{\Delta \theta_2}{\mathrm{l}_2} \\ 50 \frac{(70-\theta)}{2} & =100 \frac{(\theta-50)}{1} \\ 5 \theta & =270 \Rightarrow \theta=54^{\circ} \mathrm{C} \end{aligned}

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