$A$ spherical ball of radius $1 \ mm$ and density $10.5 \ g/cc$ is dropped in glycerine of coefficient of viscosity $9.8 \ \text{poise}$ and density $1.5 \ g/cc$. The viscous force on the ball when it attains constant velocity is $3696 \times 10^{-x} \ N$. The value of $x$ is $\text{(Given, } g = 9.8 \ m/s^2 \text{ and } \pi = \frac{22}{7}\text{)}$.

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

$A$ 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

$A$ 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 $\eta=1.8 \times 10^{-5} \, Ns/m^2$]

$A$ 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 $(g = 10 \ m \cdot s^{-2})$

$A$ small spherical ball of radius $0.1 \,mm$ and density $10^{4} \,kg \,m^{-3}$ 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 and it continues to fall with the same constant velocity inside the water,then the value of $h$ will be $m$. (Given $g = 10 \,m \,s^{-2}$,viscosity of water $\eta = 1.0 \times 10^{-5} \,N \,s \,m^{-2}$,density of water $\rho_w = 10^3 \,kg \,m^{-3}$)