(C) Let $A_1$ and $A_2$ be the cross-sectional areas of the inner cylinder and the outer shell respectively.$A_1 = \pi R^2$$A_2 = \pi (2R)^2 - \pi R^2

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(C) Let $A_1$ and $A_2$ be the cross-sectional areas of the inner cylinder and the outer shell respectively.$A_1 = \pi R^2$$A_2 = \pi (2R)^2 - \pi R^2 = 3\pi R^2$Total area $A = A_1 + A_2 = 4\pi R^2$.Since the two ends are at different temperatures and the system is in a steady state,the heat flow through both parts occurs in parallel.The equivalent thermal resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.Using $R = \frac{l}{kA}$,we have $\frac{kA}{l} = \frac{k_1 A_1}{l} + \frac{k_2 A_2}{l}$.Substituting the values:$k(4\pi R^2) = k_1(\pi R^2) + k_2(3\pi R^2)$$4k = k_1 + 3k_2$$k = \frac{k_1 + 3k_2}{4}$

Class 11 Physics 10-2.Heat Transfer Thermal Resistance and it's Combination

Difficult

$A$ cylinder of radius $R$ made of a material of thermal conductivity $k_1$ is surrounded by a cylindrical shell of inner radius $R$ and outer radius $2R$ made of a material of thermal conductivity $k_2$. The two ends of the combined system are maintained at different temperatures. There is no loss of heat from the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is

A
$k_1 + k_2$
B
$\frac{k_1 k_2}{k_1 + k_2}$
C
$\frac{1}{4}(k_1 + 3k_2)$
D
$\frac{1}{4}(3k_1 + k_2)$

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10-2.Heat Transfer

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Thermal Resistance and it's Combination

Two diagonally opposite corners of a square made of four thin rods of the same material and same dimensions are at temperatures $40^{\circ} C$ and $10^{\circ} C$. If only heat conduction takes place,then the temperature difference between the other two corners will be .......... $^{\circ} C$.

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The temperatures of the two outer surfaces of a composite slab,consisting of two materials having coefficients of thermal conductivity $K$ and $2K$ and thicknesses $x$ and $4x$ respectively,are $T_2$ and $T_1$ $(T_2 > T_1)$. The rate of heat transfer through the slab in a steady state is $\left( \frac{A(T_2 - T_1)K}{x} \right)f$,where $f$ is equal to:

DifficultAIEEE 2004

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$A$ slab consists of two parallel layers of copper and brass of equal thickness. The ratio of their thermal conductivities is $1:4$. If the temperature of the free side of the brass is $100^{\circ}C$ and that of the copper is $0^{\circ}C$,find the temperature of the interface in $^{\circ}C$.

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The rate of flow of heat through a copper rod with a temperature difference of $28^{\circ} C$ is $1400 \ cal s^{-1}$. The thermal resistance of the copper rod will be:

EasyMHT CET 2021

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Assertion: The equivalent thermal conductivity of two plates of the same thickness in contact is less than the smaller value of thermal conductivity.
Reason: For two plates of equal thickness in contact,the equivalent thermal conductivity is given by: $\frac{2}{K} = \frac{1}{K_1} + \frac{1}{K_2}$

MediumAIIMS 1997

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Two diagonally opposite corners of a square made of four thin rods of the same material and same dimensions are at temperatures $40^{\circ} C$ and $10^{\circ} C$. If only heat conduction takes place,then the temperature difference between the other two corners will be .......... $^{\circ} C$.

$A$ slab consists of two parallel layers of copper and brass of equal thickness. The ratio of their thermal conductivities is $1:4$. If the temperature of the free side of the brass is $100^{\circ}C$ and that of the copper is $0^{\circ}C$,find the temperature of the interface in $^{\circ}C$.

The rate of flow of heat through a copper rod with a temperature difference of $28^{\circ} C$ is $1400 \ cal s^{-1}$. The thermal resistance of the copper rod will be:

Assertion: The equivalent thermal conductivity of two plates of the same thickness in contact is less than the smaller value of thermal conductivity.Reason: For two plates of equal thickness in contact,the equivalent thermal conductivity is given by: $\frac{2}{K} = \frac{1}{K_1} + \frac{1}{K_2}$

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Assertion: The equivalent thermal conductivity of two plates of the same thickness in contact is less than the smaller value of thermal conductivity.
Reason: For two plates of equal thickness in contact,the equivalent thermal conductivity is given by: $\frac{2}{K} = \frac{1}{K_1} + \frac{1}{K_2}$