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A steel ball of density 8.0 xx 10^(3) kg//m^(3)and radius 2 mm is observed to fall with a terminal velocity 1.0 xx 10^(-2) m//sin a liquid of density 1.8 xx 10^(3) kg//m^(3) . Calculate the viscosity of the liquid. |
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Answer» SOLUTION :We have, `ETA = (2R^(2)(d-d_(0))g)/(9u)` `=(2(2 xx 10^(-3))^(2) (8.0 xx 10^(3) - 1.8 xx 10^(3)) xx 9.8)/(9 xx 1.0 xx 10^(-2))` `=5.4 Nm^(-2) s` |
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