The ends of a rod of length $ℓ$ are maintained at temperature $Q_{1} and Q_{2} (< Q_{1}) .$ The thermal conductivity of rod at a point is given as $K = \frac{α}{Q}$ when $Q$ = temperature at that point of the rod. $Q$ as a function of x is :

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$Q_{1} {(\frac{Q_{2}}{Q_{1}})}^{x / ℓ}$

${(\frac{x}{ℓ})}^{2} \frac{Q_{1}^{3}}{Q_{2}^{2}}$

$\frac{Q_{1} x}{ℓ} e^{- \frac{Q_{1}}{Q_{2}}}$

$\frac{x}{ℓ} \frac{Q_{1}^{2}}{Q_{2}}$

answer is D.

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Detailed Solution

Current in the element, $i = - KA \frac{dQ}{dx}$
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$\begin{aligned} i \int_{0}^{ℓ} d x = - A \propto \int_{Q_{2}}^{Q_{1}} \frac{dQ}{Q} \\ i ℓ - A \propto \ln (\frac{Q_{1}}{Q_{2}}) \end{aligned}$
$A l s o, i \int_{0}^{x} dx = - A \propto \int_{Q_{1}}^{Q} \frac{dQ}{Q} ix = A \propto \ln (\frac{Q_{1}}{Q}) U \sin g (i), {(\frac{Q_{1}}{Q_{2}})}^{x / ℓ} = \frac{Q_{1}}{Q} Q = Q_{1} {(\frac{Q_{2}}{Q_{1}})}^{x / ℓ}$