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. `6.25 xx 10^(-4) cm s^(-1)`
B. `6.45 xx 10^(-4) cm s^(-1)`
C. `1.5 xx 10^(-5) cm s^(-1)`
D. `1.6 xx 10^(-3) cm s^(-1)`

Answer

A

Explanation

V is the volume and v is the terminal velocity.
`mg=F_(v)+F_(b)`
`Vp_(b)g=6pi eta_(w) rv+Vp_(w)g`
`6pi eta_(w) rv=Vg(p_(b)-p_(w)) .......(i)`
Similarly, `6pieta.rv.=Vg(p_(b)-p_(g))........(ii)`
From (i) & (ii) `v.=(p_(b)-p_(g))/((p_(b)-p_(w)) xx (veta_(w))/(eta.))`
`=((7.8-1.2))/((7.8-1)) xx (10 xx 8.5 xx 10^(-4))/(13.2)`
`v.=6.25 xx 10^(-4) cm//s`