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10-2.Transmission of Heat
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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
A
$2:1$
B
$4:1$
C
$8:1$
D
$16:1$
Solution
(d) $\frac{Q}{t} = \frac{{kA(\Delta \theta )}}{l}$
==>$\frac{Q}{t} \propto \frac{A}{l} \propto \frac{{{r^2}}}{l}$
==> $\frac{{{{\left( {Q/t} \right)}_1}}}{{{{(Q/t)}_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} \times \frac{{{l_2}}}{{{l_1}}}$$ = {\left( {\frac{2}{1}} \right)^2} \times \left( {\frac{4}{1}} \right) = \frac{{16}}{1}$
Standard 11
Physics
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