Three rods of same dimensions have thermal co | vedclass

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(A) The amount of heat $Q$ transmitted through a rod in time $t$ is given by the formula:$Q = \frac{k A t (T_1 - T_2)}{d}$where $k$ is the thermal con

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$\frac{200}{3}{ }^{\circ} C$

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(A) The amount of heat $Q$ transmitted through a rod in time $t$ is given by the formula:$Q = \frac{k A t (T_1 - T_2)}{d}$where $k$ is the thermal conductivity,$A$ is the cross-sectional area,$d$ is the length of the rod,and $(T_1 - T_2)$ is the temperature difference.Since the rods have the same dimensions,$A$ and $d$ are constant for all rods.At the junction $P$,let the temperature be $T$. According to the principle of conservation of energy in the steady state,the heat flowing into the junction must equal the heat flowing out of the junction.Heat flowing from the $100^{\circ} C$ end to $P$ equals the sum of heat flowing from $P$ to the $50^{\circ} C$ end and from $P$ to the $0^{\circ} C$ end:$\frac{3K A (100 - T) t}{d} = \frac{2K A (T - 50) t}{d} + \frac{K A (T - 0) t}{d}$Canceling the common terms $\frac{K A t}{d}$ from both sides:$3(100 - T) = 2(T - 50) + (T - 0)$$300 - 3T = 2T - 100 + T$$300 - 3T = 3T - 100$$400 = 6T$$T = \frac{400}{6} = \frac{200}{3}{ }^{\circ} C$

Three rods of same dimensions have thermal conductivities $3K$,$2K$,and $K$. They are arranged as shown in the figure. In the steady state,the temperature of the junction $P$ is

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