Two metal spheres are falling through a liquid of density $2.5 \times 10^3 \ kg/m^3$ with the same uniform speed. The density of the material of the first sphere and the second sphere is $11.5 \times 10^3 \ kg/m^3$ and $8.5 \times 10^3 \ kg/m^3$ respectively. The ratio of the radius of the first sphere to that of the second sphere is

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6 \,cm \,s^{-1}$. If they coalesce to form one big drop, what will be the terminal speed of the bigger drop? (Neglect the buoyancy of the air)

An air bubble of radius $r$ rises steadily through a liquid of density $\rho$ with velocity $v$. The coefficient of viscosity of the liquid is:

$A$ tiny spherical oil drop carrying a net charge $q$ is balanced in still air, with a vertical uniform electric field of strength $\frac{81}{7} \pi \times 10^5 \,V / m$. When the field is switched $OFF$, the drop is observed to fall with terminal velocity $2 \times 10^{-3} \,m / s$. Here $g=9.8 \,m / s^2$, viscosity of air is $\eta = 1.8 \times 10^{-5} \,N s / m^2$ and density of oil is $\rho = 900 \,kg / m^3$. The magnitude of $q$ is

$A$ 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 $\,m$)

Two rain drops of same radius $r$ falling with terminal velocity $V$ merge and form a bigger drop with radius $R$. The terminal velocity of the bigger drop is: