A metal ball `B_(1)` (density `3.2g//"cc")` is dropped in water, while another metal ball `B_(2)` (density `6.0g//"cc")` is dropped in a liquid of density `1.6g//"cc"`. If both the balls have the same diameter and attain the same terminal velocity, the ratio of viscosity of water to that of the liquid is

Options

A. `2.0`
B. `0.5`
C. `4.0`
D. indeterminate due to insufficient data

Answer

B

Explanation

The terminal velocity of the body of radius `r`, density `rho` falling through a medium of density `sigma` is given by
`v=(2)/(9)(r^2(rho-sigma)g)/(eta)`
Where `eta` is the coefficient of viscosity of the medium
`v_(B1)=(2)/(9)(f_(B1)^2)/(eta_(water))(rho_(B1)-sigma_(water))g` ..(i)
and `v_(B2)=(2)/(9)(f_(B2)^2)/(eta_(liquid))(rho_(B2)-sigma_(liquid))g` ..(ii)
Where the subscripts `B_1` and `B_2` represent metal ball `B_1` and metal ball `B_2` respectively.
`r_(B1)=r_(B2)` and `v_(B1)=r_(B2)` .(Given)
Sustituting the given values in (i) and (ii) we get
`(eta_(water))/(eta_(liquid))=((rho_(B1)-sigma_(water)))/((rho_(B2)-sigma_(liquid)))`
Sustituting the given values we get
`(eta_(water))/(eta_(liquid))=((3.2-1))/((6.0-1.6))`
`=(2.2)/(4.4)=0.5`