Two cylinders P and Q have the same length an | vedclass

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(B) Since the cylinders are connected in series,the rate of heat flow through both cylinders must be the same in the steady state.Let $K_P = 2K$ and $

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$40$

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(B) Since the cylinders are connected in series,the rate of heat flow through both cylinders must be the same in the steady state.Let $K_P = 2K$ and $K_Q = 3K$ be the thermal conductivities,and $L$ be the length and $A$ be the cross-sectional area of each cylinder.Let $	heta$ be the temperature at the interface.The rate of heat flow $H$ is given by $H = \frac{KA(\Delta T)}{L}$.For cylinder $P$: $H_P = \frac{2KA(100 - 	heta)}{L}$.For cylinder $Q$: $H_Q = \frac{3KA(	heta - 0)}{L}$.Since $H_P = H_Q$,we have:$\frac{2KA(100 - 	heta)}{L} = \frac{3KA(	heta - 0)}{L}$.$2(100 - 	heta) = 3	heta$.$200 - 2	heta = 3	heta$.$5	heta = 200$.$	heta = 40^{\circ}C$.

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