(D) At terminal velocity,the viscous force is equal to the weight of the water drop (since buoyancy is neglected).$6 \pi \eta r v_t = \frac{4}{3} \pi

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(D) At terminal velocity,the viscous force is equal to the weight of the water drop (since buoyancy is neglected).$6 \pi \eta r v_t = \frac{4}{3} \pi r^3 \rho g$Rearranging for terminal velocity $v_t$:$v_t = \frac{2 r^2 \rho g}{9 \eta}$Given values:$r = 1\,\mu m = 10^{-6}\,m$$\eta = 1.8 \times 10^{-5}\,Nsm^{-2}$$\rho = 10^3\,kgm^{-3}$$g = 10\,ms^{-2}$Substituting the values:$v_t = \frac{2 \times (10^{-6})^2 \times 10^3 \times 10}{9 \times 1.8 \times 10^{-5}}$$v_t = \frac{2 \times 10^{-12} \times 10^4}{16.2 \times 10^{-5}}$$v_t = \frac{2 \times 10^{-8}}{16.2 \times 10^{-5}} = \frac{2}{16.2} \times 10^{-3} \approx 0.1234 \times 10^{-3} = 123.4 \times 10^{-6}\,ms^{-1}$

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

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$A$ water drop of radius $1\,\mu m$ falls in a situation where the effect of buoyant force is negligible. The coefficient of viscosity of air is $1.8 \times 10^{-5}\,Nsm^{-2}$ and its density is negligible as compared to that of water $10^{3}\,kgm^{-3}$. The terminal velocity of the water drop is . . . . . . $\times 10^{-6}\,ms^{-1}$. (Take acceleration due to gravity $= 10\,ms^{-2}$)

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A
$145.4$
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$118.0$
C
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D
$123.4$

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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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The terminal velocity of a copper ball of radius $2.0 \; mm$ falling through a tank of oil at $20 \; ^{\circ}C$ is $6.5 \; cm \; s^{-1}$. Compute the viscosity of the oil at $20 \; ^{\circ}C$. Density of oil is $1.5 \times 10^{3} \; kg \; m^{-3}$,density of copper is $8.9 \times 10^{3} \; kg \; m^{-3}$.

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Consider two solid spheres $P$ and $Q$ each of density $8 \ g \ cm^{-3}$ and diameters $1 \ cm$ and $0.5 \ cm$,respectively. Sphere $P$ is dropped into a liquid of density $0.8 \ g \ cm^{-3}$ and viscosity $\eta = 3 \ \text{poiseuille}$. Sphere $Q$ is dropped into a liquid of density $1.6 \ g \ cm^{-3}$ and viscosity $\eta = 2 \ \text{poiseuille}$. The ratio of the terminal velocities of $P$ and $Q$ is:

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The terminal velocity of a small-sized spherical body of radius $r$ falling vertically in a viscous liquid is given by the proportionality:

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$A$ solid ball of volume $V$ is dropped in a viscous liquid. It experiences a viscous force $F$. If a solid ball of volume $2V$ of the same material is dropped in the same fluid,then the viscous force acting on it will be

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$A$ metal ball of radius $9 \times 10^{-4} \ m$ and density $10^4 \ kg/m^3$ falls freely under gravity through a distance $h$ and enters a tank of water. Considering that the metal ball has constant velocity in water,the value of $h$ is [coefficient of viscosity of water $= 8.1 \times 10^{-4} \ Pa \cdot s, g = 10 \ m/s^2$,density of water $= 10^3 \ kg/m^3$]. (in $m$)

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The terminal velocity of a small-sized spherical body of radius $r$ falling vertically in a viscous liquid is given by the proportionality:

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