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