n' droplets of equal of radius r coalesce to form a bigger drop of radius R. The energy liberated is equal to ( T = Surface tension of water )

Options

A. `(2S)/(rJ)`
B. `(3S)/(J) ((1)/(r ) - (1)/(R ))`
C. `(3S)/(rJ)`
D. `(3S)/(J) ((1)/(r ) + (1)/(R ))`

Answer

B

Explanation

Let n drops coalesce to form big drop. Then `n xx (4)/(3) pi r^(3) = (4)/(3) pi R^(3) or n = (R^(3))/(r^(3))`
As heat produced `= ("Word done")/(J)`
`:. Ms Delta T= ("surface tension" xx "decrease in area")/(J)`
or `ms Delta T = (S)/(J) xx [n xx 4pi r^(2)- 4pi R^(2)]`
or `(4)/(3) pi R^(3) xx 1 xx 1 xx Delta T = (S)/(J) xx [n xx 4pi r^(2) - 4pi R^(2)]`
or `Delta T = (3S xx 4pi)/(J 4pi R^(3)) [nr^(2) - R^(2)] = (3S)/(JR^(3)) [(R^(3))/(r^(3)) r^(2) - R^(2)]`
`= (3SR^(3))/(JR^((3)) [(1)/(r ) - (1)/(R )] = (3S)/(J) [(1)/(r ) - (1)/( R)]`