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Q.

A Sphere and a cube both made of copper have equal volumes and are blackened. These are heated to same temperature and are allowed to cool under same surroundings. The ratio of their rates of loss of heat is

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a

${(\frac{π}{6})}^{\frac{1}{3}}$

b

${(\frac{6}{π})}^{\frac{1}{3}}$

c

${(\frac{π}{6})}^{\frac{1}{2}}$

d

$1 : 1$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{d Q}{d t} \propto A T^{4} \\ V_{s p} = V_{c u b e} \\ \frac{4}{3} Π r^{3} = a^{3} \\ a = (\frac{4}{3} Π) \frac{1}{3 V} \\ \frac{{(\frac{d Q}{d t})}_{S p h e r e}}{{(\frac{d Q}{d t})}_{C u b e}} = \frac{4 Π r^{2}}{6 a^{2}} \\ = {(\frac{Π}{6})}^{\frac{1}{3}} \end{array}$

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