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(B) The rate of heat flow is given by the formula $\frac{dQ}{dt} = \frac{KA\Delta T}{L}$.For the first rod (connected to $100\,^{\circ}C$): $\left( \f

Vedclass · Accepted Answer

$70$

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(B) The rate of heat flow is given by the formula $\frac{dQ}{dt} = \frac{KA\Delta T}{L}$.For the first rod (connected to $100\,^{\circ}C$): $\left( \frac{dQ}{dt} ight)_1 = \frac{3KA}{L}(100 - 	heta)$.For the second rod (connected to $50\,^{\circ}C$): $\left( \frac{dQ}{dt} ight)_2 = \frac{2KA}{L}(	heta - 50)$.For the third rod (connected to $20\,^{\circ}C$): $\left( \frac{dQ}{dt} ight)_3 = \frac{KA}{L}(	heta - 20)$.At the junction,by the principle of conservation of energy (steady state),the heat flowing in must equal the heat flowing out:$\left( \frac{dQ}{dt} ight)_1 = \left( \frac{dQ}{dt} ight)_2 + \left( \frac{dQ}{dt} ight)_3$.Substituting the expressions:$\frac{3KA}{L}(100 - 	heta) = \frac{2KA}{L}(	heta - 50) + \frac{KA}{L}(	heta - 20)$.Canceling $\frac{KA}{L}$ from both sides:$3(100 - 	heta) = 2(	heta - 50) + (	heta - 20)$.$300 - 3	heta = 2	heta - 100 + 	heta - 20$.$300 - 3	heta = 3	heta - 120$.$6	heta = 420$.$	heta = 70\,^{\circ}C$.

Three rods of the same dimensions have thermal conductivities $3K, 2K$ and $K$. They are arranged as shown in the figure with their ends at $100\,^{\circ}C, 50\,^{\circ}C$ and $20\,^{\circ}C$. The temperature of their junction is....... $^{\circ}C$

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