1000 droplets of water having 2 mm diameter each coalesce to form a single drop . Given the surface tension of water is `0.072 Nm^(-1)` .the energy loss in the process is

Options

A. `8.146 xx 10^(-4)J`
B. `4.4 xx 10^(-4) J`
C. `2.108 xx 10^(-5) J`
D. `4.7 xx 10^(-1) J`

Answer

A

Explanation

Volume of a bigger single drop = 1000 x volume of a smaller drop
` rArr 4/3 pi R^3 = 1000 xx 4/3 pi r^3 rArr R^3 = 1000 r^3 rArr R = 10 r`
= Decrease in the surface area,
`Delta A = 1000 xx 4pi r^2 - 4pi R^2 = 1000 xx 4pi r^2 - 4pi (10 r)^2 = 900 xx 4pi r^2`
Here `r = d/2 = 1 mm = 10^(-3) m , S = 0.72 Nm^(-1)`
` Delta E = ? `
Required energy loss in the process, `Delta E = S. Delta A`
` Delta E = 0.072 xx 900 xx 4 xx 22/7 xx (10^(-3))^2 = 8.146 xx 10^(-4) J`