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

An air bubble of radius $1 \ cm$ rises from the bottom portion through a liquid of density $1.5 \ g/cc$ at a constant speed of $0.25 \ cm \ s^{-1}$. If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (in $Pa \ s$):

$A$ 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

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 density $\rho$ and radius $0.003 \ m$ is dropped into a tube containing a viscous fluid filled up to the $0 \ cm$ mark as shown in the figure. The viscosity of the fluid $\eta = 1.260 \ N \cdot s \cdot m^{-2}$ and its density $\rho_L = \rho/2 = 1260 \ kg \cdot m^{-3}$. Assume the ball reaches a terminal speed by the $10 \ cm$ mark. The time taken by the ball to traverse the distance between the $10 \ cm$ and $20 \ cm$ mark is $(g = 10 \ m \cdot s^{-2})$

Two drops of equal radius are falling through air with a steady velocity of $5\,cm/s$. If the two drops coalesce,then its terminal velocity will be