The ratio of the diameters of two metallic ro | vedclass

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Question

(d) $\frac{Q}{t} = \frac{{kA(\Delta 	heta )}}{l}$

==>$\frac{Q}{t} \propto \frac{A}{l} \propto \frac{{{r^2}}}{l}$

==> $\frac{{{{\left( {Q/

Vedclass · Accepted Answer

$16:1$

Vedclass · Answer

(d) $\frac{Q}{t} = \frac{{kA(\Delta 	heta )}}{l}$

==>$\frac{Q}{t} \propto \frac{A}{l} \propto \frac{{{r^2}}}{l}$

==> $\frac{{{{\left( {Q/t} ight)}_1}}}{{{{(Q/t)}_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} ight)^2} 	imes \frac{{{l_2}}}{{{l_1}}}$$ = {\left( {\frac{2}{1}} ight)^2} 	imes \left( {\frac{4}{1}} ight) = \frac{{16}}{1}$

The ratio of the diameters of two metallic rods of the same material is $2 : 1$ and their lengths are in the ratio $1 : 4$ . If the temperature difference between their ends are equal, the rate of flow of heat in them will be in the ratio

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