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A cylindrical metallic rod, in thermal contact with two reservoirs of heat at its two ends, conducts an amount of heat $Q$ in time $t.$ The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. Amount of heat conducted by the new rod, when placed in thermal contact with same two reservoirs in time $t$ , is
$\frac{Q}{2}$
$\frac{Q}{4}$
$\frac{Q}{16}$
$2Q$
Solution
$\frac{Q}{t}=\frac{k A\left(T_{1}-T_{2}\right)}{\ell}=\frac{k A\left(T_{1}-T_{2}\right)}{\ell} \times \frac{A}{A}$
$=\mathrm{A}^{2}\left[\frac{\mathrm{k}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{V}}\right]$
$\therefore Q \propto A^{2}$
$\Rightarrow \frac{Q_{2}}{Q_{1}}=\left(\frac{A_{2}}{A_{1}}\right)^{2}=\left(\frac{r_{2}}{r_{1}}\right)^{4}=\frac{1}{16}$
