Two spherical rain drops reach the surface of the earth with terminal velocities having ratio $16: 9$. The ratio of their surface area is

$A$ ball rises to the surface at a constant velocity in a liquid whose density is $4$ times greater than that of the material of the ball. The ratio of the force of friction acting on the rising ball to its weight is

Eight identical small drops of water are falling down vertically through a medium,each with terminal velocity $v$. If they combine to form a single drop,then its terminal velocity will be (in $v$)

If the terminal speed of a sphere of gold (density $= 19.5 \times 10^3 \ kg/m^3$) is $0.2 \ m/s$ in a viscous liquid (density $= 1.5 \times 10^3 \ kg/m^3$), find the terminal speed (in $m/s$) of a sphere of silver (density $= 10.5 \times 10^3 \ kg/m^3$) of the same size in the same liquid.

In an experiment to calculate the coefficient of viscosity of a liquid by measuring terminal velocity, it was observed that a ball of mass $(\frac{\pi}{2}) \ gm$ and radius $0.5 \ cm$ takes $5 \ s$ to fall steadily through a height of $50 \ cm$ inside a long column of liquid of density $1.2 \ gm/cc$. The coefficient of viscosity will be given as: (in $\text{poise}$)

$A$ spherical ball of radius $1 \times 10^{-4} \,m$ and density $10^5 \,kg/m^3$ 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, then the value of $h$ is approximately: (The coefficient of viscosity of water is $9.8 \times 10^{-6} \,N s/m^2$) (in $\,m$)