Two rods of same length and area of cross-section $A_{1}$ and $A_{2}$ have their ends at the same temperature. If $K_{1}$ and $K_{2}$ are their thermal conductivities. $c_{1}$ and $c_{2}$ are their specific heats and $d_{1}$ and $d_{2}$ are their density, then the rate of flow of heat is the same in both the rods if

\frac{A_{1}}{A_{2}} = \frac{- K_{1}}{K_{2}}

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\frac{A_{1}}{A_{2}} = \frac{K_{1} c_{1} d_{1}}{K_{2} c_{2} d_{2}}

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\frac{A_{1}}{A_{2}} = \frac{K_{2} c_{1} d_{1}}{c_{2} d_{2} K_{1}}

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\frac{A_{1}}{A_{2}} = \frac{K_{2}}{K_{1}}

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Solution

The correct option is C $\frac{A_{1}}{A_{2}} = \frac{K_{2}}{K_{1}}$
Rate of flow of heat through a rod with temperature gradient $\frac{d Q}{d t} = K A \frac{d θ}{d x}$
Since both the rods are kept at same temperature difference and have same length, the temperature gradient for both the rods is same.
It is given that the rate of heat flow is same.
Thus $K_{1} A_{1} \frac{d θ}{d x} = K_{2} A_{2} \frac{d θ}{d x}$
$⟹ \frac{A_{1}}{A_{2}} = \frac{K_{2}}{K_{1}}$

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