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(B) The rate of cooling of a body is given by the formula:$R = \frac{d	heta}{dt} = \frac{A \epsilon \sigma (T^4 - T_0^4)}{mc}$Since the material is t

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The cube cools down faster than the sphere

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(B) The rate of cooling of a body is given by the formula:$R = \frac{d	heta}{dt} = \frac{A \epsilon \sigma (T^4 - T_0^4)}{mc}$Since the material is the same,the emissivity $\epsilon$,specific heat $c$,and density $ho$ are constant.$R \propto \frac{A}{m} = \frac{A}{ho V} \propto \frac{A}{V}$Given that the surface areas $A$ are equal,the rate of cooling $R$ is inversely proportional to the volume $V$ $(R \propto \frac{1}{V})$.For a given surface area,the sphere has the maximum volume among all geometric shapes.Therefore,$V_{	ext{sphere}} > V_{	ext{cube}}$.Since $R \propto \frac{1}{V}$,it follows that $R_{	ext{cube}} > R_{	ext{sphere}}$.Thus,the cube cools down faster than the sphere.

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

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