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(B) The rate of heat emission (power) from a body is given by the Stefan-Boltzmann law: $E = \sigma e A T^4$.Since the emitters are identical in natur

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$1:1$

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(B) The rate of heat emission (power) from a body is given by the Stefan-Boltzmann law: $E = \sigma e A T^4$.Since the emitters are identical in nature,emissivity $e$ and Stefan-Boltzmann constant $\sigma$ are the same.The surface area of a cylinder is $A = 2\pi r L + 2\pi r^2$. Given that length $L$ is proportional to radius $r$ $(L \propto r)$,the area $A$ is proportional to $r^2$.Thus,$E \propto r^2 T^4$.The ratio of heat emitted is $\frac{E_1}{E_2} = \left( \frac{r_1}{r_2} ight)^2 \left( \frac{T_1}{T_2} ight)^4$.Substituting the given values: $\frac{E_1}{E_2} = \left( \frac{1}{4} ight)^2 	imes \left( \frac{2}{1} ight)^4 = \frac{1}{16} 	imes 16 = 1$.Therefore,the ratio is $1:1$.

For two identical cylindrical emitters,the ratio of their radii is $1:4$ and the ratio of their temperatures is $2:1$. The ratio of the amount of heat emitted by them is ....... (Assume length is proportional to radius for the cylinders).

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