Rod of 40, cm in length and temperature difference of {80^o}C at its two ends. A nother rod B of len

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10-2.Transmission of Heat

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rod of $40\, cm$ in length and temperature difference of ${80^o}C$ at its two ends. $A$ nother rod $B$ of length $60\, cm$ and of temperature difference ${90^o}C$, having the same area of cross-section. If the rate of flow of heat is the same, then the ratio of their thermal conductivities will be

A

$3:4$

B

$4:3$

C

$1:2$

D

$2:1$

Solution

(a) $\frac{{dQ}}{{dt}} = \frac{{KA({\theta _1} – {\theta _2})}}{d}$

==>$\frac{{{K_1}\Delta {\theta _1}}}{{{l_1}}} = \frac{{{K_2}\Delta {\theta _2}}}{{{l_2}}}$ ($\because$ $\frac{{dQ}}{{dt}}$ and $A$ are same)

==>$\frac{{{K_1} \times 80}}{{40}} = \frac{{{K_2} \times 90}}{{60}} \Rightarrow \frac{{{K_1}}}{{{K_2}}} = \frac{3}{4}$

Standard 11

Physics

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