(B) For an uncharged oil drop falling at terminal velocity,the viscous force $(F_v)$ is equal to the weight of the drop $(W)$ when buoyancy is neglect

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(B) For an uncharged oil drop falling at terminal velocity,the viscous force $(F_v)$ is equal to the weight of the drop $(W)$ when buoyancy is neglected.$W = m \cdot g = \rho \cdot V \cdot g = \rho \cdot (\frac{4}{3} \pi r^3) \cdot g$Given:$\rho = 1.2 \times 10^3 \, kg/m^3$$r = 2.0 \times 10^{-5} \, m$$g = 9.8 \, m/s^2$$F_v = (1.2 \times 10^3) \times \frac{4}{3} \times 3.14 \times (2.0 \times 10^{-5})^3 \times 9.8$$F_v = 1.2 \times 10^3 \times 4.1867 \times 8.0 \times 10^{-15} \times 9.8$$F_v \approx 3.9 \times 10^{-10} \, N$

Class 11 Physics Fluid Mechanics and Surface Tension Viscosity and Stoke's Law and Terminal Velocity

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In Millikan's oil drop experiment,what is the viscous force acting on an uncharged drop of radius $2.0 \times 10^{-5} \, m$ and density $1.2 \times 10^{3} \, kg/m^3$? Take the viscosity of air $= 1.8 \times 10^{-5} \, Nsm^{-2}$. (Neglect buoyancy due to air).

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A
$3.8 \times 10^{-11} \, N$
B
$3.9 \times 10^{-10} \, N$
C
$1.8 \times 10^{-10} \, N$
D
$5.8 \times 10^{-10} \, N$

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Fluid Mechanics and Surface Tension

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Viscosity and Stoke's Law and Terminal Velocity

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In Millikan's oil drop experiment,what is the viscous force acting on an uncharged drop of radius $2.0 \times 10^{-5} \, m$ and density $1.2 \times 10^{3} \, kg/m^3$? Take the viscosity of air $= 1.8 \times 10^{-5} \, Nsm^{-2}$. (Neglect buoyancy due to air).

Similar Questions

$A$ lead sphere of mass $m$ falls with a terminal velocity $V$ in a viscous liquid. With what terminal velocity will another lead sphere of mass $8m$ fall in the same liquid?

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$A$ heavy spherical ball is dropped near the surface in a long column of viscous liquid. Which of the following graphs represent the variation of:
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