question_answer

A large cylindrical rod of length L is made by joining two identical rods of copper and steel of length \[\left( \frac{L}{2} \right)\]each. The rods are completely In sulated from the surroundings. If the free end If copper rod is maintained at \[100{}^\circ C\] and that of steel at \[0{}^\circ C\] then the temperature of junction is Thermal conductivity of copper is 9 times that if steel)     JEE Main  Online Paper (Held On 19  May  2012)

A) \[90{}^\circ C\]

B) \[50{}^\circ C\]

C) \[10{}^\circ C\]           

D) \[67{}^\circ C\]

Correct Answer: A

Solution :

Let conductivity of steel \[{{K}_{steel}}=k\]then from question Conductivity of copper\[{{K}_{copper}}=9k\] \[{{\theta }_{copper}}={{100}^{o}}C\] \[{{\theta }_{steel}}={{0}^{o}}C\] \[{{I}_{steel}}={{I}_{copper}}=\frac{L}{2}\] From formula temperature of junction; \[\theta =\frac{{{K}_{copper}}{{\theta }_{copper}}{{l}_{steel}}+{{K}_{steel}}{{\theta }_{steel}}{{l}_{copper}}}{{{K}_{copper}}{{l}_{steel}}+{{K}_{steel}}{{l}_{copper}}}\] \[=\frac{9k\times 100\times \frac{L}{2}+k\times 0\times \frac{L}{2}}{9k\times \frac{L}{2}+k\times \frac{L}{2}}\] \[=\frac{\frac{900}{2}kL}{\frac{10kL}{2}}={{90}^{o}}C\]