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10-2.Transmission of Heat
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In a steady state, the temperature at the end $A$ and $B$ of $20\,cm$ long rod $AB$ are $100\,^oC$ and $0\,^oC$ respectively. The temperature of a point $9\,cm$ from $A$ is....... $^oC$
A
$45$
B
$55$
C
$5$
D
$65$
Solution
$\mathrm{H}=\frac{100-0}{\mathrm{R}_{\mathrm{AH}}}=$ Heat current $=\frac{100}{\mathrm{R}_{\mathrm{AB}}}$
$100-\theta_{\mathrm{C}}=\mathrm{H} \cdot \mathrm{R}_{\mathrm{AC}}=\frac{100}{\mathrm{R}_{\mathrm{AB}}} \cdot \mathrm{R}_{\mathrm{AC}}$
$=\frac{100 \times 9}{20}=45$
$\therefore \quad \theta_{\mathrm{C}}=55^{\circ} \mathrm{C}$
Standard 11
Physics
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