Transmission Of Heat Question 129
Question: A sphere and a cube of same material and same volume are heated upto same temperature and allowed to cool in the same surroundings. The ratio of the amounts of radiations emitted will be
Options:
A) 1 : 1
B) $ \frac{4\pi }{3},,:,,1 $
C) $ {{( \frac{\pi }{6} )}^{1/3}}:,,1 $
D) $ \frac{1}{2},{{( \frac{4\pi }{3} )}^{2/3}}:,,1 $
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Answer:
Correct Answer: C
Solution:
Q = s A t (T4 ? T04) If T, T0, s and t are same for both bodies then $ \frac{Q _{sphere}}{Q _{cube}}=\frac{A _{sphere}}{A _{cube}}=\frac{4\pi r^{2}}{6a^{2}} $ ?..(i) But according to problem, volume of sphere = Volume of cube
Therefore $ \frac{4}{3}\pi r^{3}=a^{3} $
Therefore $ a={{( \frac{4}{3}\pi )}^{1/3}}r $ Substituting the value of a in equation (i) we get $ \frac{Q _{sphere}}{Q _{cube}}=\frac{4\pi r^{2}}{6a^{2}}=\frac{4\pi r^{2}}{6{{{ {{( \frac{4}{3}\pi )}^{1/3}}r }}^{2}}} $
$ =\frac{4\pi r^{2}}{6,{{( \frac{4}{3}\pi )}^{2/3}}r^{2}}={{( \frac{\pi }{6} )}^{1/3}}:1 $
BETA
