A cylindrical rod having temperature $T_{1}$ and $T_{2}$ at its end. The rate of flow of heat is $Q_{1}$ cal/sec. If all the linear dimensions are doubled keeping temperature constant, then the rate of flow of heat $Q_{2}$ will be:

4 Q_{1}

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2 Q_{1}

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Q_{1} / 4

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4 Q_{1} / 2

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Solution

The correct option is B $2 Q_{1}$
Heat flow rate $\frac{d Q}{d t} = \frac{K A (T_{1} - T_{2})}{L}$
Given that all linear dimensions are doubled
So $r$ will become $2 r$ and $L$ will become $2 L$
Since $A$ is directly proportional to $r^{2}$ , the value of $A$ will become $4 A$
Therefore the heat flow $\frac{d Q}{d t} = Q_{2} = 2 Q_{1}$
Therefore option $B$ is correct.

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