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If a ball of steel (density p = 7.8 g cm...

If a ball of steel (density `p = 7.8 g cm^(-3)`) attains a terminal velocity of `10cm s^(-1)` when falling in a tank of water (coefficient of viscosity) water `= 8.5 xx 10^(-4)` pa s. What will the terminal velocity in glycerine `(P = 1.2 g cm^(-3), eta = 13.2 rho a. s)` be ?

Text Solution

Verified by Experts

Here `rho = 7.8 g//cm^(3), r_(w) = 10 cm s^(-1)`
`eta_(w) = 8.5 xx 10^(-4) rho a.s`
`Pg = 1.2g//cm^(3) eta_(g) = 13.2 rho a.s , V_(g) = ?`
Terminal velocity `V = (2r^(2)(rho - rho_(0))g)/(9 eta)`
`V prop ((rho - rho_(0)))/(eta)`
When ball falls in water, then
`V_(w) = prop ((rho - rho_(w)))/(eta_(w))`
When ball falls in glyercine
`V_(g) prop ((rho - rho_(g)))/(eta_(g))`
`(V_(g))/(V_(w)) = ((rho - rho_(g))/(rho - rho_(w))) (eta_(w))/(eta_(g))`
`V_(g) = V_(w) = ((rho - rho_(g))/(rho - rho_(w)))(eta_(w))/(eta_(g))`
`= 10((7.8 - 1.2)/(7.8 - 1)) xx (8.5 xx 10^(-4))/(13.2)`
`= 6.25 xx 10^(-4) cm s^(-1)`
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