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If a ball of steel ( density p = 7.8 g cm^(-3)) attains terminal velocity of 10 cm s^(-1) when falling in a water ( coefficient of viscosity eta_("water") = 8.5 xx 10^(-4) Pa.s) then its terminal velocity in glycerine (p = 1.2 g cm^(-3) , eta = 13.2 Pa.s) would be, nearly : |
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Answer» `6.25xx10^(-4)cms^(-1)` `Vg(p-p-(l))=6pietarv` `Vg(p-p_(l))=6pieta^(1)rv^(1)` `V^(1)ETA^(1)=((p-p_(l)^(1)))/((p-p_(l)))xxveta` `V^(1)=((p-p_(l)^(1)))/((p-p_(l)))xx(veta)/(eta^(1))` `=((7.8-1.2))/((7.8-1))xx(10xx8.5xx10^(-4))/(13.2)` `V^(1)=6.25xx19^(-4)cm`/s CORRECT choice is (a) |
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