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(B) The rate of cooling $R$ of a body is given by $R = \frac{dQ}{dt} \cdot \frac{1}{mc} = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{mc}$.Since the mat

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The cube cools down faster than the sphere

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(B) The rate of cooling $R$ of a body is given by $R = \frac{dQ}{dt} \cdot \frac{1}{mc} = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{mc}$.Since the material is the same,the emissivity $\varepsilon$,specific heat $c$,and density $\rho$ are the same.Thus,$R \propto \frac{A}{m} \propto \frac{A}{\rho V} \propto \frac{A}{V}$,where $A$ is the surface area and $V$ is the volume.For a given surface area $A$,the rate of cooling $R \propto \frac{1}{V}$.For a cube of side $a$,$A = 6a^2$ and $V = a^3 = (\sqrt{A/6})^3$.For a sphere of radius $r$,$A = 4\pi r^2$ and $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (\sqrt{A/4\pi})^3$.Comparing volumes for the same surface area,the volume of the cube is less than the volume of the sphere $(V_{cube} Therefore,the rate of cooling of the cube is greater than that of the sphere $(R_{cube} > R_{sphere})$.

$A$ solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature $120^{\circ}C$,then

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