if a ball of steel (density `rho=7.8 g//cm^(3)`) attains a terminal velocity of `10 cm//s` when falling in a tank of water (coefficient of viscosity, `eta_(water) =8.5xx10^(-4)`Ps s), then its terminal velocity in glycerine `(rho =1.2 g//cm^(2), eta =13.2 Pa s)` would be nearly

Options

A. `6.45xx10^(-4)cm//s`
B. `1.5xx10^(-5)cm//s`
C. `1.6xx10^(-5)cm//s`
D. `6.25xx10^(-4)cm//s`

Answer

D

Explanation

Here, `rho_(s)=7.8 g//c^(3), v_(1) =10 cm//s`,
`eta_(water) =8.5xx10^(-4)Pa S`
`rho_(w)=1, rho_(g) =1.2 g//cm^(3), eta_(g) =13.2 Pa S, v_(2)=?`
As terminal velocity `v=(2r^(2)(rho-rho_(0))g)/(9eta)`
`:. v prop((rho-rho_(0)))/(eta)`
`(v_(2))/(v_(1))=((rho-rho_(g)))/(eta_(g))xx(eta_(water))/((rho-rho_(w)))`
`=((7.8-1.2))/(13.2)xx(8.5xx10^(-4))/((7.8-1)) = (6.6xx8.5xx10^(-4))/(13.2xx6.8)`
`(v_(2))/(v_(1)) =0.625xx10^(4)` or `v_(2) =0.625xx10^(-4) xxv_(1)`
`v_(2)=0.625xx10^(-4)xx10= 6.25xx10^(-4) cm//s`