A spherical object of mass 1 kg and radius 1m is falling vertically downward inside a viscous liquid in a gravity free space. At a certain instant the velocity of the sphere is `2 ms^(-1)`. If the coefficient of viscosity of the liquid is `(1)/(6pi)` SI units, then velocity of ball will become `0.5 ms^(-1)` after a time.

Options

A. In 4 s
B. 2 In 4 s
C. 3 In 4 s
D. 3 In 2 s

Answer

A

Explanation

`F=(6 pi eta r v)=6pi ((1)/(6pi)) (1)(v)=v`
`therefore " Retardation" a= - (F)/(m)= - v " or " (dv)/(dt)= -v`
`implies int_(0)^(t) dt = - int_(2)^(0.5) (dv)/(v) " or " t=ln (4)s`