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

Options

A. energy = 3VT `(1/r-1/R)` is released
B. energy is neither released nor absorbed
C. energy = 4VT`(1/r=1/R)` is released
D. energy = 3VT `(1/r-1/R)` is absorbed

Answer

A

Explanation

Volume remains constant
` :. n xx 4/3 pir^(3) =4/3piR^(3) " " :. R^(3) =nr^(3)`
` :. (R/3)^(3) = n :. R/r = n^(1//3)`
Initial energy of n drop `(U_(1)) = n . 4 pir^(2)T`
Final energy of one drop `(U_(f)) =4piR^(2)T`
`Delta U = U_(i) -U_(f)=4piT (nr^(2)-R^(2))`
` =4piTR^(2) ((nr^(2))/(R^(2)) -1)`
` =4piTR^(2)((R^(3))/(r^(3))xx(r^(2))/(R^(2)) -1)`
` = 4piTR^(2) (R/r-1)`
` :. Delta U = 4piTR^(3)(1/r-1/R) =4/3 piR^(3)(3T) (1/r-1/R)`
` = 3VT(1/r-1/R)`
` :' R gt r " " :. 1/r gt 1/R ` is positive
` :. DeltaU ` is + ve ` " " :. " Energy " (Delta U) ` is related