A solid cylinder of length L and radius r is | vedclass

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(A) Since both objects have the same volume,we have $a^3 = \pi r^2 L$  --- $(1)$.The amount of radiation emitted $Q$ is proportional to the surface ar

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$\frac{a}{3r}$

Vedclass · Answer

(A) Since both objects have the same volume,we have $a^3 = \pi r^2 L$  --- $(1)$.The amount of radiation emitted $Q$ is proportional to the surface area $A$ of the object,given that the temperature and material are the same for both $(Q = \sigma A e T^4 t)$.For the cylinder,the curved surface area is $A_{	ext{cyl}} = 2\pi r L$ (neglecting flat ends).For the cube,the total surface area is $A_{	ext{cube}} = 6a^2$.The ratio of radiation is $\frac{R_{	ext{cyl}}}{R_{	ext{cube}}} = \frac{2\pi r L}{6a^2}$.From equation $(1)$,we know $\pi r^2 L = a^3$,so $L = \frac{a^3}{\pi r^2}$.Substituting $L$ into the ratio: $\frac{R_{	ext{cyl}}}{R_{	ext{cube}}} = \frac{2\pi r (a^3 / \pi r^2)}{6a^2} = \frac{2a^3 / r}{6a^2} = \frac{2a}{6r} = \frac{a}{3r}$.

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