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)`
when `eta` is the coefficient of viscosity of the medium.
`:. V_(B_(1)) = (2)/(9) (r_(B_(1)^(2)))/(eta_("water")) (rho_(B_(1)) - sigma_("water"))g`...(i)
and `v_(B_(2)) = (2)/(9) (r_(B_(1)^(2)))/(eta_("liquid")) (rho_(B_(2)) - sigma_("liquid"))g` ...(ii)
where the subscripts `B_(1) and B_(2)` represent metal ball `B_(1)` and metal ball `B_(2)` respectively.
`:. (v_(B_(1)))/(v_(B_(2))) = (r_(B_(1)^(2)))/(r_(B_(2)^(2))) (eta_("liquid"))/(eta_("water")) ((rho_(B_(1)) - sigma_("water")))/((rho_(B_(2)) - sigma_("liquid")))`
But `r_(B_(1)) = r_(B_(2)) and v_(B_(1)) = v_(B_(2))` (Given)
`:. (eta_("water"))/(eta_("liquid")) = ((rho_(B_(1)) - sigma_("water")))/((rho_(B_(2)) - sigma_("liquid"))) = ((3.2-1))/((6.0- 1.6)) = (2.2)/(4.4) = 0.5` ( `:.` Density of water = `1 g cm^(-3)`)