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(C) The rate of heat flow in the steady state is given by the formula $\frac{Q}{t} = \frac{KA(	heta_1 - 	heta_2)}{l}$.Since the material is the same

Vedclass · Accepted Answer

$8:9$

Vedclass · Answer

(C) The rate of heat flow in the steady state is given by the formula $\frac{Q}{t} = \frac{KA(	heta_1 - 	heta_2)}{l}$.Since the material is the same,the thermal conductivity $K$ is constant. Given that the temperature difference $(	heta_1 - 	heta_2)$ is also constant,we have $\frac{Q}{t} \propto \frac{A}{l}$.Since the cross-sectional area $A = \pi r^2$,we get $\frac{Q}{t} \propto \frac{r^2}{l}$.Given ratios: $\frac{l_1}{l_2} = \frac{1}{2}$ and $\frac{r_1}{r_2} = \frac{2}{3}$.Therefore,the ratio of heat flow is $\frac{(Q/t)_1}{(Q/t)_2} = \left(\frac{r_1}{r_2}ight)^2 	imes \left(\frac{l_2}{l_1}ight) = \left(\frac{2}{3}ight)^2 	imes \left(\frac{2}{1}ight) = \frac{4}{9} 	imes 2 = \frac{8}{9}$.

The lengths and radii of two rods made of the same material are in the ratios $1:2$ and $2:3$ respectively. If the temperature difference between the ends for the two rods is the same,then in the steady state,the amount of heat flowing per second through them will be in the ratio:

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