An aluminium rod of length $1 \mathrm{~m}$ and a steel rod of length $2 \mathrm{~m}$ both having same cross-sectional area, are soldered together end-to-end. The thermal conductivity of aluminium rod and steel rod is $200 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ and $50 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ respectively. The temperatures of the free ends are maintained at $300 \mathrm{~K}$ and $500 \mathrm{~K}$. What is the temperature of the junction?

• Correct answer
  A

  $322 \mathrm{~K}$
• B

  $350 \mathrm{~K}$
• C

  $367 \mathrm{~K}$
• D

  $400 \mathrm{~K}$

Solution

Let junction temperature is $T K$, then in steady state heat flow rate is same through both rods.
So, using, $\frac{Q}{t}=\frac{k A\left(T_2-T_1\right)}{l}$, we have, $$ \frac{Q}{t}=\frac{k_s A_s(500-T)}{2}=\frac{k_{\mathrm{Al}} A_{\mathrm{Al}}(T-300)}{1} $$ As areas of both rods is same, so $$ \begin{array}{rlrl} & & \frac{k_s(500-T)}{2} & =\frac{k_{\mathrm{Al}}(T-300)}{1} \\ \Rightarrow & & \frac{50(500-T)}{2} & =\frac{200(T-300)}{1} \\ \Rightarrow & 25 \times 500-25 T & =200 T-200 \times 300 \\ \Rightarrow & & 225 T & =12500+60000 \\ \Rightarrow & & T & =\frac{72500}{225}=322.2 \approx 322 \mathrm{~K} \end{array} $$

Hence, the correct answer is option: 'A'.