Small water droplets of radius $0.01 \mathrm{~mm}$ are formed in the upper atmosphere and falling with a terminal velocity of $10 \mathrm{~cm} / \mathrm{s}$. Due to condensation, if $8 \mathrm{such}$ droplets are coalesced and formed a larger drop, the new terminal velocity will be ........... $\mathrm{cm} / \mathrm{s}$.

Assume that, the drag force on a football depends only on the density of the air, velocity of the ball and the cross-sectional area of the ball. Balls of different sizes but the same density are dropped in an air column. The terminal velocity reached by balls of masses $250 \,g$ and $125 \,g$ are in the ratio

$Assertion :$ Falling raindrops acquire a terminal velocity.
$Reason :$ A constant force in the direction of motion and a velocity dependent force opposite to the direction of motion, always result in the acquisition of terminal velocity.

A water drop of radius $1\,\mu m$ falls in a situation where the effect of buoyant force is negligible. Coefficient of viscosity of air is $1.8 \times 10^{-5}\,Nsm ^{-2}$ and its density is negligible as compared to that of water $10^{6}\,gm ^{-3}$. Terminal velocity of the water drop is________ $\times 10^{-6}\,ms ^{-1}$

(Take acceleration due to gravity $=10\,ms ^{-2}$ )

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 spherical ball of radius $1 \times 10^{-4} \mathrm{~m}$ and density $10^5$ $\mathrm{kg} / \mathrm{m}^3$ falls freely under gravity through a distance $h$ before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of $h$ is approximately:

(The coefficient of viscosity of water is $9.8 \times 10^{-6}$ $\left.\mathrm{N} \mathrm{s} / \mathrm{m}^2\right)$