A) 1 : 1
B) \[\frac{4\pi }{3}:1\]
C) \[{{\left( \frac{\pi }{6} \right)}^{1/3}}:1\]
D) \[\frac{1}{2}{{\left( \frac{4\pi }{3} \right)}^{2/3}}:1\]
Correct Answer: C
Solution :
[c] \[\operatorname{Q}=\sigma At\left( {{T}^{4}}-T_{0}^{4} \right)\] If \[\operatorname{T}, {{T}_{0}}, \sigma \]and t are same for both bodies, then \[\frac{{{Q}_{sphere}}}{{{Q}_{cube}}} \frac{{{A}_{sphere}}}{{{A}_{cube}}}=\frac{4\pi {{r}^{2}}}{6{{a}^{2}}}\] But according to problem, Volume of sphere = Volume of cube \[\Rightarrow \frac{4}{3}\pi {{r}^{3}}={{a}^{3}}\Rightarrow a=r{{\left( \frac{4}{3}\pi \right)}^{1/3}}\] Substituting the value of a in equation (i), we get \[\frac{{{Q}_{sphere}}}{{{Q}_{cube}}}=\frac{4\pi {{r}^{2}}}{{{\left\{ 6{{\left( \frac{4}{3}\pi \right)}^{1/3}}r \right\}}^{2}}}={{\left( \frac{\pi }{6} \right)}^{1/3}}:1\]
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