Which of the following diagrams correctly shows the relation between the terminal velocity $v_{T}$ of a spherical body falling in a liquid and viscosity $\eta$ of the liquid?

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

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 $.5$)

If the terminal velocity of a metal sphere of mass $8 \ g$ falling through a liquid is $3 \ cm s^{-1}$, then the terminal velocity of another sphere of mass $64 \ g$ made of the same metal falling through the same liquid is (in $cm s^{-1}$)

$A$ small ball of mass $m$ and density $\rho$ is dropped in a viscous liquid of density $\rho_0$. After some time,the ball falls with constant velocity. The viscous force on the ball is:

The velocity of a small ball of mass $0.3 \, g$ and density $8 \, g/cc$ when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is $1.3 \, g/cc$,then the value of the viscous force acting on the ball will be $x \times 10^{-4} \, N$. The value of $x$ is [use $g = 10 \, m/s^2$]. (in $.125$)