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`.

Options

A. `7 xx 10^(-12) J`
B. `7.5 xx 10^(-12) J`
C. `7.9 xx 10^(-12) J`
D. `8.5 xx 10^(-12) J`

Answer

C

Explanation

Let r be the radius of smaller drops and R of bigger one. Equating the initial and final volumes, we have
`(4)/(3) piR^(3) = (1000) ((4)/(3) pi r^(3))`
`rArr R = 10 r = (10) (10^(-7)) m rArr R = 10^(-6) m`
Further, the water drops have only one free surface. Therefore,
`Delta A = 4 piR^(2) - (1000) (4 pir^(2))`
`= 4 pi [(10^(-6))^(2) - (10^(3)) (10^(-7))^(2)] = - 36 pi (1)^(-12) m^(2)`
Here, negative sign implies that surface area is decreasing. Hence, energy released in the process,
`U = S |DeltaA| = (7 xx 10^(-2)) (36 pi xx 10^(-12)) J`
`= 7.9 xx 10^(-12) J`