A mercury drop of radius 1.0 cm is sprayed into `10^(6)` droplets of equal sizes. The energy expended in this process is (Given, surface tension of mercury is `32 xx 10^(-2) Nm^(-1)`)

Answer

3.98

Explanation

Volume of `10^6` droplets = Volume of larger drop `10^6 xx 4/3 pi r^3 =4/3 piR^3`
`r = 10^(-2) R =10^(-2) xx 1.0 =10^(-2) cm =10^(-4) m`
Surface area of larger drop
`= 4pi R^2 = 4 pi xx(10^(-2))^(2) = 4pi xx10^(-4)m^2`
Surface area of `10^6` droplets
`= 4pir^2 xx10^6 = 4pi xx (10^(-4))^2xx10^6`
`= 4pi xx10^(-2) m^2`
`:.` Increase in surface area
`= 4pi xx 10^(-4)(100 – 1) = 4pi xx99xx10^(-4)m^2`
`:.` Work done in spraying a spherical drop of mercury
= Surface tension x increase in surface area
`= 32xx10^(-2) xx4pi xx99xx10^(-4) = 3.98xx10^(-2)J`