A rain drop of radius r is falling through air startin from rest the work done by all the forces on the drop when it attains terminal velocity is proportional to

Options

A. `r^(3)`
B. `r^(7)`
C. `r^(5)`
D. `r^(4)`

Answer

B

Explanation

Given radius of rain drop = r
Since rain drop starts falling from rest hence its initial speed u= 0
final velocity of rain drop is equal to the terminal velocity v which is given by
`v= (2g r^(2) (p -Sigma))/( 9etc )`
Where `p to ` density of the rain drop
`Sigma to ` density of air
`etc to` coefficient of viscosity
`:.` According to the work - energy theorem
= change in its kinetic energy
`w= (1)/(2) mv^(2) - (1)/(2) mu^(2) = (1)/(2) mv^(2)`
`= (1)/(2) xx (4)/(3) pir^(3) . p ((2gr^(2) (p - Sigma ))/(9 etc ))^(2) `
`k = (8 pi p g^(2))(p- sigma )^(2) r^(7))/( 234 etc^(2)) rArr 2= kr^(7) " or " W prop r^(7)`