A sphere of density `rho` falls vertically downward through a fluid of density `sigma`. At a certain instant its velocity is u. The terminal velocity of the sphere is `u_0`. Assuming that stokes’s law for viscous drag is applicable, the instantaneous acceleration of the sphere is found to be `beta(1-(sigma)/(rho))(1-u/(u_(0)))`. here `beta` is in an integer. find `beta`.

Explanation

`F_(d)=6 pi mu r u`
`F_(B)=4/3 pi r^(3) sigma g, mg=4/3 pir^(3) rhog`
`mg-F_(d)-F_(B)=ma, u_(0)=(2r^(2))/9 g(rho-sigma)/(mu)`
`:. A=(1-(sigma)/(rho))(1-u/(u_(0)))g`