JEE Physics 2023 Question with Solution

Given: Eight equal water drops fall through air with steady speed $10 \, \text{cm/s}$.

Find: The new steady speed when they collapse into one bigger drop.

For eight small spherical drops merging into one large spherical drop, volume is conserved:

$$8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$$

So,

$$8r^3 = R^3$$

Hence,

$$R = 2r$$

The terminal velocity of a small sphere in a fluid is proportional to the square of its radius:

$$V \propto r^2$$

Therefore,

$$\frac{V_1}{V_2} = \left(\frac{r}{R}\right)^2$$

Substituting $R = 2r$,

$$\frac{V_1}{V_2} = \left(\frac{r}{2r}\right)^2 = \frac{1}{4}$$

So,

$$V_2 = 4V_1$$

Now using $V_1 = 10 \, \text{cm/s}$,

$$V_2 = 4 \times 10 = 40 \, \text{cm/s}$$

Therefore, the new velocity is $40 \, \text{cm/s}$ and the correct option is B.

The solution shows km/s in one step, but that is inconsistent with the question and options; the correct unit here is cm/s.