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A tiny sphere of mass m and density x is dropped in a tall jar of glycerine of density y. When the sphere acquires terminal velocity, the magnitude of the viscous force acting on it is

Option: 1

\mathrm{\frac{m g x}{y}}


Option: 2

\mathrm{\frac{m g y}{x}}


Option: 3

\mathrm{m g\left(1-\frac{y}{x}\right)}


Option: 4

\mathrm{m g\left(1+\frac{x}{y}\right)}


Answers (1)

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When the sphere acquires terminal velocity, the upward viscous force equals the apparent weight in glycerine. Therefore, the magnitude of the viscous force is (here r is the radius of the sphere)

                        \mathrm{ F=\frac{4}{3} \pi r^3(x-y) g=\frac{m}{x}(x-y) g=m g\left(1-\frac{y}{x}\right) }

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