Question

# The energy required to split a liquid drop having surface tension $$T$$ and radius $$R$$ into $$n$$ identical droplets is:

A
8πR2(n1/31)T
B
4πR2(n1/31)T
C
8πR2(n2/31)T
D
4πR2(n2/31)T

Solution

## The correct option is C $$4\pi R^{2}(n^{1/3}-1)T$$Initial energy $$= 4\pi R^{2}T$$initial volume = final volume$$\dfrac { 4 }{ 3 } \pi { R }^{ 3 }=n\dfrac { 4 }{ 3 } \pi { r }^{ 3 }\\ \Rightarrow r=R{ (n) }^{ -1/3 }$$New $$SE = (n4\pi r^{2})T$$$$=4\pi T R^{2} n^{1/3}$$Energy required $$= 4\pi T R^{2} n^{1/3}-4\pi R^{2}T$$$$= 4\pi R^{2}(n^{1/3-1})T$$Hence,Option B is correct.Physics

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