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Standard XII

Physics

Thermal Resistance

Heat flows th...

Question

Heat flows through two rods having same temperature difference at the ends. One rod is of length $l_{1}$ , radius $r_{1}$ and thermal conductivity $K_{1}$ and the other rod of $l_{1}, r_{2}$ and $K_{2}$ . The heat flow rate through the two rods will be equal, if :

K_{1} l_{2} = K_{2} l_{1}

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K_{1} r_{1}^{2} = K_{2} r_{2}^{2}

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k_{1} l_{2} r_{1} = k_{2} l_{1} r_{2}

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K_{1} l_{2} r_{1}^{2} = K_{2} l_{1} r_{2}^{2}

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Solution

The correct option is D $K_{1} l_{2} r_{1}^{2} = K_{2} l_{1} r_{2}^{2}$
Heat flow
$= \frac{Q}{t} = \frac{K_{1} A_{1} △ T}{l_{1}}$
$= \frac{K_{2} A_{2} △ T}{l_{2}}$
$\Rightarrow \frac{K_{1} π r_{1}^{2}}{l_{1}} = \frac{K_{2} π r_{2}^{2}}{l_{2}}$
$K_{1} r_{1}^{2} l_{2} = K_{2} r_{2}^{2} l_{1}$ .

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The correct option is D $K_{1} l_{2} r_{1}^{2} = K_{2} l_{1} r_{2}^{2}$
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$= \frac{Q}{t} = \frac{K_{1} A_{1} △ T}{l_{1}}$
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$\Rightarrow \frac{K_{1} π r_{1}^{2}}{l_{1}} = \frac{K_{2} π r_{2}^{2}}{l_{2}}$
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