Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_{1}$ and $\rho_{2}\left(\rho_{1}=8 \rho_{2}\right)$ and have radii of $1\; \mathrm{mm}$ and $2\; \mathrm{mm}$, respectively. They are made to fall vertically (from rest) in a viscous medum whose coefficient of viscosity equals $\eta$ and whose denstry is $0.1 \mathrm{\rho}_{2} .$ The ratio of their terminal velocitites would be
A small spherical ball of radius $0.1 \,mm$ and density $10^{4} \,kg m ^{-3}$ falls freely under gravity through a a distance $h$ before entering a tank of water. If after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of $h$ wil be $m$. (Given $g =10 \,ms ^{-2}$, viscosity of water $=1.0 \times 10^{-5} \,N - sm ^{-2}$ )
The terminal velocity of a small sphere of radius $a$ in a viscous liquid is proportional to
A spherical solid ball of volume $V$ is made of a material of density $\rho_1$ . It is falling through a liquid of density $\rho_2 (\rho_2 < \rho_1 )$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$, i.e., $F_{viscous}= -kv^2 (k >0 )$,The terminal speed of the ball is
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 ........... $\mathrm{cm} / \mathrm{s}$.