Two drops of same radius are falling through air with steady speed `v`. If the two drops coalesce, what would be the terminal speed ?

Options

A. `v(R )/(r )`
B. `v(R^(2))/(r^(2))`
C. v
D. 2v

Answer

B

Explanation

Terminal valocity, `v=2/9(r^(2)(rho-sigma)g)/eta`
`therefore" "vpropr^(2)rArrv/V=r^(2)/R^(2)rArrV=(vR^(2))/(r^(2))`