(C) In a steady state,the rate of heat flow through both rods must be equal.Let the junction temperature be $\phi$.The rate of heat flow is given by $

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(C) In a steady state,the rate of heat flow through both rods must be equal.Let the junction temperature be $\phi$.The rate of heat flow is given by $H = \frac{KA(\Delta T)}{L}$.For the first rod: $H_1 = \frac{K A (\phi - 0)}{1}$.For the second rod: $H_2 = \frac{(3K) A (100 - \phi)}{2}$.Since $H_1 = H_2$:$\frac{KA\phi}{1} = \frac{3KA(100 - \phi)}{2}$$\phi = \frac{3}{2}(100 - \phi)$$2\phi = 300 - 3\phi$$5\phi = 300$$\phi = 60^{\circ}C$.

Class 11 Physics 10-2.Heat Transfer Heat Conduction and Thermal Conductivity

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Two rods have thermal conductivities $K$ and $3K$ and lengths $1 \ cm$ and $2 \ cm$ respectively. Their cross-sectional areas are equal. They are joined in series as shown in the figure. If the temperatures of the outer ends of this composite rod are $0^{\circ}C$ and $100^{\circ}C$ respectively,find the junction temperature $\phi$ in $^{\circ}C$.

A
$50$
B
$\frac{100}{3}$
C
$60$
D
$\frac{200}{3}$

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10-2.Heat Transfer

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Heat Conduction and Thermal Conductivity

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Two closed containers of same dimensions made of different materials are completely filled with ice. The ice in the first container takes $20 \ min$ and that in the second container takes $10 \ min$,respectively,for complete melting. The ratio of the thermal conductivities of the materials of the two containers is . . . . . . .

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$A$ cylindrical rod has temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is $Q$. All the linear dimensions of the rod are doubled while keeping the temperatures constant. The new rate of flow of heat is

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