Explain the free fall of a spherical body in a viscous liquid and derive the expression for the viscous force acting on it.

A metal block of base area $0.20 {\mathrm{m}}^2$ is placed on a table, as shown in figure. A liquid film of thickness $0.25 \mathrm{mm}$ is inserted between the block and the table. The block is pushed by a horizontal force of $0.1 \mathrm{N}$ and moves with a constant speed. If the viscosity of the liquid is $5.0\times {10}^{–3} \mathrm{Pl}$, the speed of block is ______ $\times {10}^{-3} \mathrm{m} {\mathrm{s}}^{-1}$.

A steel ball is dropped in a viscous liquid. The distance of the steel ball from the top of the liquid is shown below. The terminal velocity of the ball is closest to

A raindrop with radius $\mathrm{R}=0.2 \mathrm{mm}$ falls from a cloud at a height $\mathrm{h}=2000 \mathrm{m}$ above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is : [Density of water $f_{\mathrm{w}}=1000 \mathrm{kg}{ \mathrm{m}}^{-3}$ and Density of air $f_{\mathrm{a}}=1.2 \mathrm{kg}{ \mathrm{m}}^{-3}, \mathrm{g}=10 \mathrm{m}/{\mathrm{s}}^2$ Coefficient of viscosity of air $\left.=1.8\times {10}^{-5}\mathrm{N} \mathrm{s} {\mathrm{m}}^{-2}\right]$

A flat plate of area $10{ \mathrm{cm}}^2$ is separated form a large plate by a layer of Glycerine $1 \mathrm{mm}$ thick. If the coefficient of viscosity of Glycerine is $20$ poise, the force required to keep the plate moving with a velocity of $1 \mathrm{cm} {\mathrm{s}}^{-1}$ is $\underline{ }$