A) \[energy=4VT\text{ }\left( \frac{1}{r}-\frac{1}{R} \right)\text{ }is\text{ }released\]
B) \[energy\text{ }=3VT\text{ }\left( \frac{1}{r}+\frac{1}{R} \right)is\text{ }absorbed\]
C) \[energy\text{ }=\text{ }3VT\left( \frac{1}{r}-\frac{1}{R} \right)is\text{ }released\]
D) Energy is neither released nor absorbed
Correct Answer: C
Solution :
[c] As surface area decreases so energy is released. Energy released\[=4\pi {{R}^{2}}T[{{n}^{1/3}}-1]\] (where\[R={{n}^{1/3}}r\]) \[=4\pi {{R}^{3}}T\left[ \frac{1}{r}-\frac{1}{R} \right]=3VT\left[ \frac{1}{r}-\frac{1}{R} \right]\]
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