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(D) Let the junction temperature be $T^{\circ} C$.In the steady state,the rate of heat flow through the copper rod must be equal to the rate of heat f

Vedclass · Accepted Answer

$88$

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(D) Let the junction temperature be $T^{\circ} C$.In the steady state,the rate of heat flow through the copper rod must be equal to the rate of heat flow through the steel rod,assuming no heat loss from the sides.The rate of heat flow $H$ is given by $H = \frac{KA(T_1 - T_2)}{l}$.Equating the heat flow for both rods:$\frac{K_{	ext{copper}} A (100 - T)}{l} = \frac{K_{	ext{steel}} A (T - 0)}{l}$Since the length $l$ and cross-sectional area $A$ are the same for both rods,they cancel out:$K_{	ext{copper}} (100 - T) = K_{	ext{steel}} T$Substituting the given values $K_{	ext{copper}} = 385$ and $K_{	ext{steel}} = 50$:$385(100 - T) = 50T$$38500 - 385T = 50T$$38500 = 435T$$T = \frac{38500}{435} \approx 88.5^{\circ} C$.Rounding to the nearest integer,the junction temperature is $88^{\circ} C$.

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