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Question

$\frac{Q}{t}=\frac{k A\left(T_{1}-T_{2}ight)}{\ell}=\frac{k A\left(T_{1}-T_{2}ight)}{\ell} 	imes \frac{A}{A}$

$=\mathrm{A}^{2}\left[\frac{\mat

Vedclass · Accepted Answer

$\frac{Q}{16}$

Vedclass · Answer

$\frac{Q}{t}=\frac{k A\left(T_{1}-T_{2}ight)}{\ell}=\frac{k A\left(T_{1}-T_{2}ight)}{\ell} 	imes \frac{A}{A}$

$=\mathrm{A}^{2}\left[\frac{\mathrm{k}\left(\mathrm{T}_{1}-\mathrm{T}_{2}ight)}{\mathrm{V}}ight]$

$	herefore Q \propto A^{2}$

$\Rightarrow \frac{Q_{2}}{Q_{1}}=\left(\frac{A_{2}}{A_{1}}ight)^{2}=\left(\frac{r_{2}}{r_{1}}ight)^{4}=\frac{1}{16}$

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