The excess pressure inside a spherical drop of water is four times that of another drop. Then, their respective mass ratio is

Options

A. `1: 16`
B. `8:1`
C. `1:4`
D. `1: 64`

Answer

D

Explanation

Excess pressure inside the liqui drop, `P= (2S)/(R )`, where S is the surface tension and R its radius.
`:. P_(1) = (2S)/(R_(1)) and P_(2) = (2S)/(R_(2))`
But `P_(1) = 4P_(2)` (Given)
`:. (2S)/(R_(1)) = 4 ((2S)/(R_(2))) or (R_(1))/(R_(2)) = (1)/(4)`
If `rho` is the density of water, then
`m_(1) = (4)/(3) pi R_(1)^(3) rho and m_(2) = (4)/(3) pi R_(2)^(3) rho :. (m_(1))/(m_(2)) = ((R_(1))/(R_(2)))^(3) = ((1)/(4))^(3) = (1)/(64)`