An iron sphere of mass $20 \times 10^{-3} \ kg$ falls through a viscous liquid with terminal velocity $0.5 \ ms^{-1}$. The terminal velocity (in $ms^{-1}$) of another iron sphere of mass $54 \times 10^{-2} \ kg$ is (in $.5$)

$A$ 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 $\,m$)

$A$ metal sphere of radius $R$ and density $\varrho_{1}$ moves with terminal velocity $v_{1}$ through a liquid of density $\sigma$. Another sphere of the same radius but of density $\varrho_{2}$ moves through the same liquid. Its terminal velocity will be:

$A$ raindrop with radius $R=0.2 \, mm$ falls from a cloud at a height $h=2000 \, m$ above the ground. Assume that the drop is spherical throughout its fall and the force of buoyancy may be neglected. The terminal speed attained by the raindrop is: (in $m/s$)
[Density of water $\rho_{w}=1000 \, kg/m^3$,density of air $\rho_{a}=1.2 \, kg/m^3$,$g=10 \, m/s^2$,coefficient of viscosity of air $\eta=1.8 \times 10^{-5} \, Ns/m^2$]

If the average terminal velocity of a rain drop is $2 \,m/s$, then the energy transferred by rain to each square metre of the surface at a place which receives $100 \,cm$ of rain in a year is

Two solid spheres of radii $2 \ mm$ and $4 \ mm$ are tied to the two ends of a light string and released in a liquid of specific gravity $1.3$ and coefficient of viscosity $1 \ Pa \cdot s$. The string is just taut when the two spheres are completely submerged in the liquid. If the density of the materials of the two spheres is $2800 \ kg \cdot m^{-3}$,then the terminal velocity of the system of the spheres is (take $g = 10 \ m \cdot s^{-2}$):