Properties Of Solids And Liquids Question 444
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 $
Show Answer
Answer:
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}}{6a^{2}} $
But according to problem, Volume of sphere = Volume of cube $ \Rightarrow \frac{4}{3}\pi r^{3}=a^{3}\Rightarrow a=r{{( \frac{4}{3}\pi )}^{1/3}} $
Substituting the value of a in equation (i), we get $ \frac{Q _{sphere}}{Q _{cube}}=\frac{4\pi r^{2}}{{{{ 6{{( \frac{4}{3}\pi )}^{1/3}}r }}^{2}}}={{( \frac{\pi }{6} )}^{1/3}}:1 $
BETA
