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Two small drops of mercury, each of radi...

Two small drops of mercury, each of radius `R`, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

A

`1:2^(1//3)`

B

`2^(1//3) :1`

C

`2:1`

D

`1:2`

Text Solution

Verified by Experts

The correct Answer is:
B
`W_S/W_B=(2xx4piTr^2)/(4piTR^2)=(2r^2)/((n^(1//3)r)^2)=(2r^2)/(2^(1//3)r^2)=1/(2^(-1//3))`
`implies (2^(1//3))/(1)=2^(1//3):1`
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