QJEEM25Ph5d02: JEE Main 2025 Physics - Class 11 JEEM25Ph5d02 Solution: In the experiment for measurement of viscosity ' ' of given…

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.