Three rods of Copper, Brass and Steel are welded together to form a Y shaped structure. Area of cross - section of each rod = 4 cm². End of copper rod is maintained at 100 where as ends of brass and steel are kept at 0. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings excepts 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: (a) 1.2 cal/s (b) 2.4 cal/s (c) 4.8 cal/s (d) 6.0 cal/s

Answer: (c). Let junction temperature be T. Heat in through copper = heat out through brass + steel (same area). (0.92(100-T))/(46) = (0.26T)/(13) + (0.12T)/(12) ⇒ 0.02(100-T) = 0.03T T = 40° C HCu = (0.92×4×60)/(46) = 4.8 cal/s So the answer is (c) 4.8 cal/s.

Three rods of Copper, Brass and Steel are welded together to form a Y… | JEE Main 2014

Q50JEE Main 2014Difficult

Heat Conduction

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

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Answer: (c)

Let junction temperature be $T$ . Heat in through copper = heat out through brass + steel (same area).
$\dfrac{0.92(100-T)}{46} = \dfrac{0.26T}{13} + \dfrac{0.12T}{12} \Rightarrow 0.02(100-T) = 0.03T$
$T = 40^\circ$ C
$H_{Cu} = \dfrac{0.92\times4\times60}{46} = 4.8$ cal/s
So the answer is (c) 4.8 cal/s.

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