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Question

A metal sphere of radius 1 mm and mass 50 mg falls vertically in glycerine. Find (a) the viscous force exerted by the glycerine on the sphere when the speed of the sphere is 1 cm s1, (b) the hydrostatic force exerted by the glycerine on the sphere and (c) the terminal velocity with which the sphere will move down without acceleration. Density of glycerine = 1260 kg m−3 and its coefficient of viscosity at room temperature = 8.0 poise.

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Solution

Given:
Radius of metallic sphere r = 1 mm = 10−3 m
Speed of the sphere v = 10−2 m/s
Coefficient of viscosity η = 8 poise = 0.8 decapoise
Mass m = 50 mg = 50 × 10−3 kg
Density of glycerin σ = 1260 kg/m3
(a)
Viscous force exerted by glycerine on the sphere F = 6πηrv
⇒ F= 6 × (3.14) × (0.8) × 10−3 × (10−2)
= 1.50 × 10−4 N

(b)
Let V be the volume of the sphere.
Hydrostatic force exerted by glycerin on the sphere F'=Vσg
F'=43πr2σg
=43×3.14×10-6×1260×10=5.275×10-5 N

(c)
Let the terminal velocity of the sphere be v'.
The forces acting on the drops are
(i) The weight mg acting downwards
(ii) The force of buoyance, i.e., 43πr3σg acting upwards
(iii) The force of viscosity, i.e., 6πηrv' acting upwards
From the free body diagram:



6πηrv'+43πr3σg=mgv=mg-43πr2σg6πηr =50×10-3-43×3.14×10-6×1260×106×3.14×0.8×10-3 =500-43×3.14×10-3×1260×106×3.14×0.8 =2.3 cm/s

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