(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 identical conducting rods are first connected independently in parallel to two vessels,one containing water at $100^oC$ and the other containing ice at $0^oC$. In the second case,the rods are joined end to end (in series) and connected to the same vessels. Let $q_1$ and $q_2$ $g/s$ be the rate of melting of ice in the two cases respectively. The ratio of $q_1/q_2$ is

DifficultIIT 2004

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Two rods of equal length and diameter have thermal conductivities $3$ and $4$ units respectively. If they are joined in series,the thermal conductivity of the combination would be

MediumWBJEE 2009

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Two rods of same length and material transfer a given amount of heat in $12 \, s$ when they are joined end to end. But when they are joined lengthwise (parallel),they will transfer the same amount of heat under the same conditions in .......... $s$.

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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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Consider a compound slab consisting of two different materials having equal thickness and thermal conductivities $K$ and $2K$ respectively. The equivalent thermal conductivity of the slab is

MediumAIPMT 2003

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Two rods of equal length and diameter have thermal conductivities $3$ and $4$ units respectively. If they are joined in series,the thermal conductivity of the combination would be

Two rods of same length and material transfer a given amount of heat in $12 \, s$ when they are joined end to end. But when they are joined lengthwise (parallel),they will transfer the same amount of heat under the same conditions in .......... $s$.

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: $\frac{2}{K} = \frac{1}{K_1} + \frac{1}{K_2}$

Consider a compound slab consisting of two different materials having equal thickness and thermal conductivities $K$ and $2K$ respectively. The equivalent thermal conductivity of the slab is

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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: $\frac{2}{K} = \frac{1}{K_1} + \frac{1}{K_2}$