Question

Two rods of the same length and area of cross-section ${A_1}$and ${A_2}$have their ends at the same temperature. If ${{\rm K}_1}$and ${{\rm K}_2}$ are the conductivities, ${C_1}$ and ${C_2}$their specific heats and ${\rho _1},{\rho _2}$ their densities, then the condition that the rate of flow of heat is same in both the rods is?

(A) $\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{{\rm K}_1}{\rho _1}}}{{{{\rm K}_2}{\rho _2}}}$
(B) $\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{{\rm K}_1}{C_1}{\rho _1}}}{{{{\rm K}_2}{C_2}{\rho _2}}}$
(C) $\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\dfrac{{{{\rm K}_1}}}{{{C_1}{\rho _1}}}}}{{\dfrac{{{{\rm K}_2}}}{{{C_2}{\rho _2}}}}}$
(D) $\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{{\rm K}_2}}}{{{{\rm K}_1}}}$

Answer 1

Hint It is given that the temperature of the ends of both rods is the same, and the rate of heat flow has to be kept the same for both rods. Their ratios must be equal. Thus the conduction equation can be rearranged and can be equated for both rods. This gives the necessary condition for the same rate of heat flow.
Formula used:
$H = \dfrac{{{\rm K}A\left( {{T_h} - {T_l}} \right)}}{l}$

Complete Step by step solution
In the question, it is given that-
Length of rods 1 and 2 is the same, let us assume the length to be $L$.
The area of cross-sections of the rods is ${A_1}$ and ${A_2}$.
The conductivities of the rods are ${{\rm K}_1}$and ${{\rm K}_2}$.
The specific heats of the rods are ${C_1}$and ${C_2}$.
The densities of the rods are ${\rho _1}$and${\rho _2}$.
From the conduction equation, we know that-
$H = \dfrac{{{\rm K}A\left( {{T_h} - {T_l}} \right)}}{l}$
Here, $H$is the heat flow per unit time.
${\rm K}$ is the conductivity of the rod.
${T_h}$ is the hotter temperature.
${T_l}$ is the lower or colder temperature.
$l$ is the length of the rod.
$A$ is the cross-section area of the rod.
${T_h} - {T_l}$can be written as $\Delta T$.
It is known that the temperature of the ends of both rods is similar, therefore,
$\Delta {T_1} = \Delta {T_2}$
According to the question, the rate of heat flow in both rods should be the same.
Therefore,
$\dfrac{{{H_1}}}{{\Delta {T_1}}} = \dfrac{{{H_2}}}{{\Delta {T_2}}}$
For rod $1$,
$\dfrac{H}{{\Delta T}} = \dfrac{{{{\rm K}_1}{A_1}}}{L}$
For rod $2$,
$\dfrac{H}{{\Delta T}} = \dfrac{{{{\rm K}_2}{A_2}}}{L}$
Equating both of these expressions, we have-
$\dfrac{{{{\rm K}_1}{A_1}}}{L} = \dfrac{{{{\rm K}_2}{A_2}}}{L}$
${{\rm K}_1}{A_1} = {{\rm K}_2}{A_2}$
On rearranging this,
$\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{{\rm K}_2}}}{{{{\rm K}_1}}}$

Therefore option (D) is correct.

Note It is to be noted that for a solid rod, the rate of heat transfer in conductivity does not depend on its specific heat and density. It only depends on the temperature difference between ends of the rod, cross-section area, thermal conductivity, and the length of the rod.