(A) The thermal resistance $R$ of a rod is given by the formula $R = \frac{l}{k \cdot A}$,where $l$ is the length,$k$ is the thermal conductivity,and

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$\frac{R_A}{R_B} = \frac{1}{3}$

(A) The thermal resistance $R$ of a rod is given by the formula $R = \frac{l}{k \cdot A}$,where $l$ is the length,$k$ is the thermal conductivity,and $A$ is the cross-sectional area.For rod $A$ with thermal conductivity $3K$,length $l$,and area $A$:$R_A = \frac{l}{(3K) \cdot A} = \frac{l}{3KA}$For rod $B$ with thermal conductivity $K$,length $l$,and area $A$:$R_B = \frac{l}{K \cdot A} = \frac{l}{KA}$Now,calculating the ratio of the thermal resistances:$\frac{R_A}{R_B} = \frac{\frac{l}{3KA}}{\frac{l}{KA}} = \frac{1}{3}$Therefore,the ratio $\frac{R_A}{R_B} = \frac{1}{3}$.

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

Medium

Two rods $A$ and $B$ of the same cross-sectional area $A$ and length $l$ are connected in series between a source $(T_1 = 100^{\circ}C)$ and a sink $(T_2 = 0^{\circ}C)$ as shown in the figure. The rods are laterally insulated. The thermal conductivities of rods $A$ and $B$ are $3K$ and $K$ respectively. The ratio of the thermal resistance of rod $A$ to that of rod $B$ is:

A
$\frac{R_A}{R_B} = \frac{1}{3}$
B
$\frac{R_A}{R_B} = 3$
C
$\frac{R_A}{R_B} = \frac{3}{4}$
D
$\frac{R_A}{R_B} = \frac{4}{3}$

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10-2.Heat Transfer

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Thermal Resistance and it's Combination

Two identical metal rods are welded end-to-end as shown in figure $(1)$. It takes $4 \ min$ for $20 \ cal$ of heat to flow through them. If these rods are now welded side-by-side as shown in figure $(2)$,the time taken for the same amount of heat to flow through them is .......... $\min$.

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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:

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Two identical metal rods are welded end-to-end as shown in figure $(1)$. It takes $4 \ min$ for $20 \ cal$ of heat to flow through them. If these rods are now welded side-by-side as shown in figure $(2)$,the time taken for the same amount of heat to flow through them is .......... $\min$.

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