Stoke's Law

A small sphere of mass $m$ is dropped from a height. After it has fallen $100 \mathrm{m},$ it has attained its terminal velocity and continues to fall at that speed. The work done by air friction against the sphere during the first $100 \mathrm{m}$ of fall is :-

Water filled in an empty tank of area $10 A$ through a tap of cross sectional area $A.$The speed of water flowing out of tap is given by $v ( \mathrm{m}/\mathrm{s})=10\left(1-\sin \frac{\mathrm{\pi }}{30}\mathrm{t}\right)$ where $\text{'}t\text{'}$ is in seconds. The height of water level from the bottom of the tank at $t=15$ second will be:

Two drops of equal radius are falling through air with a steady velocity of $5 \mathrm{cm} {\sec }^{-1}.$ If the two drops coalesce, then its terminal velocity will be :-

A copper ball of radius $\text{'}r\text{'}$ travels with a uniform speed ' $v$ ' in a viscous fluid. If the ball is changed with another ball of radius ' $2r$ ', then new uniform speed will be :-

Two liquids of densities $d_1$ and $d_2$ are flowing in identical capillaries under same pressure difference. If $t_1$ and $t_2$ are the time taken for the flow of equal quantities of liquids, then the ratio of coefficients of viscosities of liquids must be :-

A rain drop of radius $0.3 \mathrm{mm}$ has a terminal velocity in air $1 \mathrm{m} {\mathrm{s}}^{-1}$. The viscosity of air is $18\times {10}^{-5} \mathrm{poise}$. The viscous force on it is :-

A small  sphere starts falling form a very large height and after falling a distance of $100\mathrm{m}$ it attains the terminal velocity and continues to fall with this velocity. The work done by the atmosphere during the first fall of $100\mathrm{m}$ is:

The velocity of falling rain drops attains limited value because of :

If the terminal speed of a sphere of Gold (density$=19.5 \mathrm{kg} {\mathrm{m}}^{-3}$) is $0.2 \mathrm{m} {\mathrm{s}}^{-1}$ in a viscous liquid (density$=1.5 \mathrm{kg} {\mathrm{m}}^{-3}$), find the terminal speed of a sphere of Silver (density$=10.5 \mathrm{kg} {\mathrm{m}}^{-3}$) of the same size in the same liquid.

A spherical ball of radius $3.0\times {10}^{–4} \mathrm{m}$ and density ${10}^4 \mathrm{kg} {\mathrm{m}}^{-3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering the water the velocity of the ball does not change, find $h$. Viscosity of water is $9.8\times {10}^{–6} \mathrm{N} \mathrm{s} {\mathrm{m}}^{-2}$. $[g= 9.8 \mathrm{m} {\mathrm{s}}^{-2}, {\rho }_w=1000 \mathrm{kg} {\mathrm{m}}^{-3}]$