A thin square plate of side $2\ m$ is moving at the interface of two very viscous liquids of viscosities ${\eta _1} = 1$ poise and ${\eta _2} = 4$ poise respectively as shown in the figure. Assume a linear velocity distribution in each fluid. The liquids are contained between two fixed plates. $h_1 + h_2 = 3\ m$ . A force $F$ is required to move the square plate with uniform velocity $10\ m/s$ horizontally then the value of minimum applied force will be ........ $N$

Which falls faster, big rain drops or small rain drops ?

Spherical balls of radius $ 'r'$  are falling in a viscous fluid of viscosity '$\eta$' with a velocity $ 'v'. $ The retarding viscous force acting on the spherical ball is

Two drops of the same radius are falling through air with a steady velocity of $5 cm per sec.$  If the two drops coalesce, the terminal velocity would be

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

State stokes’ law. By using it deduce the expression for :

$(i)$ initial acceleration of smooth sphere and

$(ii)$ equation of terminal velocity of sphere falling freely through the viscous medium.

$(iii)$ Explain : Upward motion of bubbles produced in fluid.