Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of `6cms^-1` . If they coalesce to form one big drop , what will be the terminal speed of bigger drop ? ( Neglect the buoyancy of the air)

Options

A. `1.5cms^-1`
B. `6cms^-1`
C. `24cms^-1`
D. `32cms^-1`

Answer

C

Explanation

`4/3pir^(3).n= 4/3 piR^(3)`
R=2r
Terminal velocity,
`V=2/9 xx r^(2)/eta (rho-sigma)g`
After combining to form a bigger drop
`V^(1)=2/9xx R^(2)/eta (rho-sigma)g`
`V/V_(1) = 6/V^(1)= (2/9 xx (rho-sigma)g)/(2/9 R^(2)/eta(rho-sigma)g)`
`therefore 6/V^(1)=r^(2)/(2r^(2))`
`V^(1) = 24cm//s`