Stoke’s Law

Eight raindrops each of radius $R$ fall through air with terminal velocity $6 \mathrm{cm}{ \mathrm{s}}^{-1}.$ What is the terminal velocity of the bigger drop formed by coalescing these drops together?

A ball is thrown downward with some velocity, greater than terminal speed into a viscous liquid. Which of the following curves represents correct variation for velocity versus time?

A ball of density $\sigma$ and radius $r$ is dropped on the surface of a liquid of density $\rho$ from certain height. If speed of ball does not change even on entering in liquid and viscosity of liquid is $\eta ,$ then the height from which ball dropped is

A spherical ball is dropped in a long column of viscous liquid. The speed $v$ of the ball varies as function of time as

Two rain drops falling through air have radii in the ratio$1: 2$. They will have terminal velocity in the ratio________

What is the terminal velocity of air bubble of $1.0 \mathrm{mm}$ in diameter rising in liquid of viscosity $0.85 \mathrm{N}-\mathrm{s} {\mathrm{m}}^{-2}$ and density $900 \mathrm{kg} {\mathrm{m}}^{-3}$ ?(air density $\left.1.293 \mathrm{kg} {\mathrm{m}}^{-3}, g=9 \mathrm{m} {\mathrm{s}}^{-2}\right]$

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

Under a constant pressure head, the rate of flow of orderly volume flow of liquid through a capillary tube is $V.$ if the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become

The velocity of rain drop having radius $1 \mathrm{mm}$ is $20 \mathrm{cm} {\mathrm{s}}^{-1}$. The velocity of raindrops of size $3 \mathrm{mm}$ is

Two solid spheres of the same metal but of mass $M$ and $8M$ fall simultaneously in a viscous liquid. The ratio of terminal velocity of the second sphere to that of the first sphere is

Which of the following diagrams correctly shows the relation between the terminal velocity $v_T$ of a spherical body falling in a liquid and viscosity $\eta$ of the liquid?

What will be the approximate terminal velocity of a rain drop of diameter $1.8\times {10}^{-3} \mathrm{m}$, when density of rain water $\approx {10}^3{\mathrm{kgm}}^{-3}$ and the coefficient of viscosity of air $\approx 1.8\times {10}^{-5} \mathrm{N}-\mathrm{sm}$${}^{-2} ?$  (Neglect buoyancy of air)

A small spherical body of radius $r$ and density $\rho$ moves with the terminal velocity $v$ in a fluid of coefficient of viscosity $\eta$ and density $\sigma .$ What will be the net force on the body?

A rain drop of radius $0.3 \mathrm{mm}$ has a $\mathrm{terminal} \mathrm{velocity} \mathrm{in} \mathrm{air}$ $=1 \mathrm{m} {\mathrm{s}}^{-1}$. The viscosity of air is $8\times {10}^{-5} \mathrm{poise}$. The viscous force on it is

In an experiment to verify Stokes law, a small spherical ball of radius $r$ and density $\rho$ falls under gravity through a distance $h$ in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of $h$ is proportional to: (ignore viscosity of air)

A small spherical solid ball is dropped from a great height in a viscous liquid. Its journey in the liquid is best described in the diagram given below by the

A gas bubble of $2 \mathrm{cm}$ diameter rises through a liquid of density $1.75 \mathrm{g} {\mathrm{cm}}^{-3}$ with a fixed speed of $0.35 \mathrm{cm} {\mathrm{s}}^{-1}.$ Neglect the density of the gas. The coefficient of viscosity of the liquid is...$\left(\mathrm{Take} \mathrm{g}=981 \mathrm{cm} {\mathrm{s}}^{-2}\right)$

The terminal velocity of freely falling body is

A drop of radius $2\times {10}^{-5} \mathrm{m}$ and density $1.2\times {10}^3 \mathrm{kg} {\mathrm{m}}^{-3}$ falls through air. The viscosity of air is $1.8\times {10}^{-5} \mathrm{N} \mathrm{s} {\mathrm{m}}^{-2}$ . Neglecting buoyancy due to air, the terminal speed of the drop is

Terminal velocity of a steel ball of diameter $0.2 \mathrm{cm}$ when falls through a tube filled with glycerine is $\left[g=9.8 \mathrm{m} {\mathrm{s}}^{-2},\right.$ $\mathrm{density} \mathrm{of} \mathrm{steel}=8000 \mathrm{kg} {\mathrm{m}}^{-3}$$\mathrm{density} \mathrm{of} \mathrm{glycerine}=1330 \mathrm{kg} {\mathrm{m}}^{-3}$ , $\eta =8.33 \mathrm{poise}$ $]$