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