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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 rates of flow of heat through them will be
A
$1 : 1$
B
$2 : 1$
C
$1 : 4$
D
$1 : 8$
Solution
$\frac{\left(\frac{\mathrm{Q}}{\mathrm{t}}\right)_{1}}{\left(\frac{\mathrm{Q}}{\mathrm{t}}\right)_{2}}=\frac{\mathrm{A}_{1} \mathrm{L}_{2}}{\mathrm{A}_{2} \mathrm{L}_{1}}=\frac{1 \times 1}{4 \times 2}=\frac{1}{8}$
Standard 11
Physics
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