JEE MAIN - Physics (2021 - 27th August Morning Shift - No. 15)
In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0 $$\times$$ 10$$-$$5 m and density 1.2 $$\times$$ 103 kgm$$-$$3 ? Take viscosity of liquid = 1.8 $$\times$$ 10$$-$$5 Nsm$$-$$2. (Neglect buoyancy due to air).
3.8 $$\times$$ 10$$-$$11 N
3.9 $$\times$$ 10$$-$$10 N
1.8 $$\times$$ 10$$-$$10 N
5.8 $$\times$$ 10$$-$$10 N
Explanation
Viscous force = Weight
$$ = \rho \times \left( {{4 \over 3}\pi {r^3}} \right)g$$
= 3.9 $$\times$$ 10$$-$$10
$$ = \rho \times \left( {{4 \over 3}\pi {r^3}} \right)g$$
= 3.9 $$\times$$ 10$$-$$10
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