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Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of crosssection of each rod \[=4c{{m}^{2}}.\]End of copper rod is maintained at \[100{}^\circ C\] where as ends of brass and steel are kept at \[0{}^\circ C\]. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is :   JEE Main  Solved  Paper-2014

A) 4.8 cal/s               

B) 6.0 cal/s

C) 1.2 cal/s                               

D) 2.4 cal/s

Correct Answer: A

Solution :

\[H={{H}_{1}}+{{H}_{2}}\] \[\frac{100-T}{\frac{{{\ell }_{1}}}{{{k}_{1}}A}}=\frac{T-0}{\frac{{{\ell }_{2}}}{{{k}_{2}}A}}+\frac{T-0}{\frac{{{\ell }_{3}}}{{{k}_{3}}A}}\] \[\frac{{{k}_{1}}(100-T)}{{{\ell }_{1}}}=\frac{{{k}_{2}}T}{{{\ell }_{2}}}+\frac{{{k}_{3}}T}{{{\ell }_{3}}}\] \[\frac{0.92}{46}(100-T)=\frac{0.26}{13}T+\frac{0.12}{12}T\] \[2\times {{10}^{-2}}(100-T)=2\times {{10}^{-2}}T+1\times {{10}^{-2}}T\] \[2(100-T)=2T+T\] \[2(100-T)=3T\] \[200=5T\] \[T={{40}^{o}}C\] \[H=\frac{{{k}_{1}}(100-T)A}{{{\ell }_{1}}}=\frac{0.92}{46}(100-10)\times 4\] \[=2\times {{10}^{-2}}\times 60\times 4=120\times {{10}^{-2}}\times 4=4.8\]cal/s