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10-2.Transmission of Heat
medium
The lengths and radii of two rods made of same material are in the ratios $1 : 2$ and $2 : 3$ respectively. If the temperature difference between the ends for the two rods be the same, then in the steady state, the amount of heat flowing per second through them will be in the ratio
A
$1: 3$
B
$4:3$
C
$8:9$
D
$3:2$
Solution
(c) $\frac{Q}{t} = \frac{{KA({\theta _1} – {\theta _2})}}{l}$ ==> $\frac{Q}{t} \propto \frac{A}{l} \propto \frac{{{r^2}}}{l}$
[As $({\theta _1} – {\theta _2})$ and K are constants]
$\Rightarrow \frac{{{{\left( {\frac{Q}{t}} \right)}_1}}}{{{{\left( {\frac{Q}{t}} \right)}_2}}} = \frac{{r_1^2}}{{r_2^2}} \times \frac{{{l_2}}}{{{l_1}}} = \frac{4}{9} \times \frac{2}{1} = \frac{8}{9}$
Standard 11
Physics
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