Correct Answer - B
When the sphere of radius R is falling in a liquid of density `sigma` and coefficient of viscosity `eta` it attains a terminal velocity v, under two forces
(i) Effective force acting downward
`=V(rho-sigma)g=(4)/(3)piR^3(rho-sigma)g`
(ii) Viscous forces acting upwards `=6pietaRv` Since the sphere is moving with a constant velocity v, There is no acceleration in it, the net force acting on it must be zero. That is
`6pietaRv=(4)/(3)piR^3(rho-sigma)g`
`impliesv=(2)/(9)(R^2(rho-sigma)g)/(eta)`
`impliesvpropR^2`
Thus terminal velocity is proportional to the square of its radius.