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10-2.Transmission of Heat
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The heat is flowing through two cylindrical rods of same material. The diameters of the rods are in the ratio $1 : 2$ and their lengths are in the ratio $2 : 1$ . If the temperature difference between their ends is the same, the ratio of rate of flow of heat through them will be
A
$1:1$
B
$2:1$
C
$1:4$
D
$1:8$
(AIPMT-1995)
Solution
(d) $\frac{Q}{t} = \frac{{KA\,\Delta \theta }}{l}$
==> $\frac{Q}{t} \propto \frac{A}{l} \propto \frac{{{d^2}}}{l}$ ( $d =$ Diameter of rod)
==> $\frac{{{{(Q/t)}_1}}}{{{{(Q/t)}_2}}} = {\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^2} \times \frac{{{l_2}}}{{{l_1}}} = {\left( {\frac{1}{2}} \right)^2} \times \left( {\frac{1}{2}} \right) = \frac{1}{8}$
Standard 11
Physics
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