One end of a cylindrical rod is kept in stem chamber and the other end in meting $I C E$ . Now $0.5 g m$ of ice melts in $1 s e c$ . If the rod is replaced by another rod of same length , half the diameter and double the conductivity of the first rod, then rate of melting of ice will be

0.25 g m / s e c

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0.5 g m / s e c

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1 g m / s e c

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2 g m / s e c

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Solution

The correct option is A $0.25 g m / s e c$
We know
$Q = \frac{K A Δ Q}{L}$
$\frac{Q_{2}}{Q_{1}} = \frac{K_{2}}{K_{1}} \times \frac{A_{2}}{A_{1}}$
When L & $Δ Q$ are constant
$\frac{Q_{2}}{Q_{1}} = \frac{K_{2}}{K_{1}} \times \frac{D_{2}^{2}}{D_{1}^{2}}$
$∴ Q_{2} = 0.5 \times 2 \times {(\frac{1}{2})}^{2} = 0.25 g / s$
$Q_{2} = 0.25 g / s$

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