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A certain number of spherical drops of a liquid of radius `r` coalesce to form a single drop of radius `R` and volume `V`. If `T` is the surface tension of the liquid, then
A. energy `= 4VT((1)/(r)-(1)/(R))` is released
B. energy `= 3VT((1)/(r)+(1)/(R))` is absorbed
C. energy `= 3VT((1)/(r)-(1)/(R))` is released
D. energy is neither released nor absorbed

1 Answer

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Best answer
Correct Answer - c
As energy released `= (A_(f)-A_(i))T`
where, `A_(i) = 4pi R^(2) = (3)/(3) xx 4pi (R^(3))/(R) = (3V)/(R)`
and `A_(f) = 4pi r^(2) = (V)/((4)/(3)pi r^(2))4pir^(2) = (3V)/(r)`
`:.` Energy released `= 3VT[(1)/(r)-(1)/(R)]`

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