Two rods P an...

Question

Two rods $P$ and $Q$ have equal lengths. Their thermal conductivities are $K_{1}$ and $K_{2}$ and cross-sectional areas are $A_{1}$ and $A_{2}$ . When the temperature at ends of each rod are $T_{1}$ and $T_{2}$ respectively, the rate of flow of heat through $P$ and $Q$ will be equal, if

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

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

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

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\frac{A_{1}}{A_{2}} = {(\frac{K_{2}}{K_{1}})}^{2}

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Solution

The correct option is A $\frac{A_{1}}{A_{2}} = \frac{K_{2}}{K_{1}}$
As, ${(\frac{Δ Q}{Δ t})}_{P} = {(\frac{Δ Q}{Δ t})}_{Q}$
$\Rightarrow K_{1} A_{1} \frac{(T_{1} - T_{2})}{l} = K_{2} A_{2} \frac{(T_{1} - T_{2})}{l}$
or $K_{1} A_{1} = K_{2} A_{2}$
or $\frac{A_{1}}{A_{2}} = \frac{K_{2}}{K_{1}}$

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As, ${(\frac{Δ Q}{Δ t})}_{P} = {(\frac{Δ Q}{Δ t})}_{Q}$
$\Rightarrow K_{1} A_{1} \frac{(T_{1} - T_{2})}{l} = K_{2} A_{2} \frac{(T_{1} - T_{2})}{l}$
or $K_{1} A_{1} = K_{2} A_{2}$
or $\frac{A_{1}}{A_{2}} = \frac{K_{2}}{K_{1}}$