We know, terminal velocity of a sphere of radius R in a liquid of viscosity $\eta$,

$$v = {2 \over 9}{{{R^2}} \over \eta }(\sigma - \rho )$$ .... (1)

where, $\sigma$ = mass of density of sphere

$\rho$ = density of liquid

we can see, $$v \propto {R^2}$$ (for constant $\eta,\sigma$ & $\rho$)

Hence, graph between v and R is parabola.

As v depends on R so the terminal velocities of different diameter balls will be different.

We know, the viscosity of a liquid usually decreases as the temperature increases and $$v \propto {1 \over \eta }$$

So terminal velocity depends on the temperature. $$T \uparrow \Rightarrow \eta \downarrow \Rightarrow v \uparrow $$

As the equation $$v = {2 \over 9}{{{R^2}} \over v}(\sigma - \rho )$$ involves density of liquid $\rho$. So the experiment can be utilized to asses it.

From (1), $$\eta = {2 \over 9}{{{R^2}} \over v}(\sigma - \rho )$$

Here, $\eta$ does not depend on initial speed of the sphere. Hence, option 3 is correct.