JEE Physics 2026 Question with Solution

Given: radius of first ball $r_1 = 6 \, \text{mm}$, terminal velocity $v_1 = 20 \, \text{cm/s}$, radius of second ball $r_2 = 3 \, \text{mm}$.

Find: terminal velocity $v_2$ of the second ball.

For a spherical ball moving through a viscous fluid, terminal velocity is given by Stokes' law:

$$v_t = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta}$$

So, for the same material and the same fluid,

$$v_t \propto r^2$$

Therefore,

$$\frac{v_2}{v_1} = \left(\frac{r_2}{r_1}\right)^2$$

Substituting the values,

$$\frac{v_2}{20} = \left(\frac{3}{6}\right)^2$$ $$\frac{v_2}{20} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ $$v_2 = \frac{20}{4} = 5 \, \text{cm/s}$$

Therefore, the terminal velocity of the second ball is $5 \, \text{cm/s}$.

Given: the radius changes from $6 \, \text{mm}$ to $3 \, \text{mm}$.

Find: the new terminal velocity.

Since terminal velocity varies as the square of radius,

$$v_t \propto r^2$$

The radius becomes half:

$$\frac{r_2}{r_1} = \frac{3}{6} = \frac{1}{2}$$

So the terminal velocity becomes one-fourth:

$$\frac{v_2}{v_1} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$

Hence,

$$v_2 = \frac{20}{4} = 5 \, \text{cm/s}$$

This works because all other factors in Stokes' law remain constant for the same material and the same fluid. Therefore, the required value is $5$.