Calculate the energy released when 1000 small water drops each of same radius `10^(-7)m` coalesce to form one large drop. The surface tension of water is `7.0xx10^(-2)N//m`.

Explanation

Let r be the radius of smaller drops and R of bigger one. Equating the initial and final volumes, we have
`( 4)/( 3) pi R^(3) = ( 1000) ((4)/( 3) pi r^(3))`
or `R = 10 r = ( 10 ) ( 10^(-7)) m ` or `R = 10^(-6) m`
Further, the water drops have only one free surface. Therefore,
`Delta A = 4pi R^(2) - ( 1000) ( 4pi r^(2))`
`= 4 pi [ ( 10^(-6))^(2) - ( 10^(3)) ( 10^(-7)) ^(2) ]` Here, negative sign implies that surface area is deceasing . Hence, energy released in the process.
`U= T | Delta A | = ( 7 xx 10^(-2)) ( 36 pi xx 10^(-12))`
`= 7.9 xx 10^(-12)J`