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(C) According to Stefan-Boltzmann law,the rate of loss of heat $dQ/dt$ is given by $dQ/dt = e \sigma A (T^4 - T_0^4)$.Since both objects are black,mad

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$(\frac{\pi}{6})^{\frac{1}{3}}$

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(C) According to Stefan-Boltzmann law,the rate of loss of heat $dQ/dt$ is given by $dQ/dt = e \sigma A (T^4 - T_0^4)$.Since both objects are black,made of the same material,and cooling in the same atmosphere at the same temperature,the rate of loss of heat is directly proportional to the surface area $A$.Let $V$ be the volume of both objects. For a sphere of radius $r$,$V = \frac{4}{3} \pi r^3$. For a cube of side $a$,$V = a^3$.Since $V$ is equal,$a^3 = \frac{4}{3} \pi r^3$,which implies $a = (\frac{4}{3} \pi r^3)^{1/3}$.The surface area of the sphere is $A_s = 4 \pi r^2$.The surface area of the cube is $A_c = 6 a^2 = 6 (\frac{4}{3} \pi r^3)^{2/3}$.The ratio of the rate of loss of heat is $\frac{A_s}{A_c} = \frac{4 \pi r^2}{6 (\frac{4}{3} \pi r^3)^{2/3}}$.Simplifying this,$\frac{A_s}{A_c} = \frac{4 \pi r^2}{6 (\frac{4}{3})^{2/3} \pi^{2/3} r^2} = \frac{4^{1/3} \pi^{1/3}}{6 (\frac{1}{3})^{2/3}} = \frac{4^{1/3} \pi^{1/3}}{6 \cdot 3^{-2/3}} = \frac{4^{1/3} \pi^{1/3} \cdot 3^{2/3}}{6} = \frac{(4 \cdot 9)^{1/3} \pi^{1/3}}{6} = \frac{(36)^{1/3} \pi^{1/3}}{6} = (\frac{36 \pi}{216})^{1/3} = (\frac{\pi}{6})^{1/3}$.

$A$ sphere and a cube,both made of copper,have equal volumes and are black. They are allowed to cool at the same temperature and in the same atmosphere. The ratio of their rate of loss of heat is:

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