If R is the radius of big drop, then
`(4)/(3) pi R^(3) =8xx(4)/(3) pi r^(3)`
or `R=2 r=2xx0.1=0.2cm`
As terminal velocity, `v=(2)/(9) (r^(2)(rho-sigma)g)/(eta)`
or `v alpha r^(2)`
`:. (v_(2))/(v_(1))=((R)/(r))^(2)`
or `v_(2)=v_(1)((R)/(r))^(2) =5((2r)/(r))^(2) =20 cm//s`