JEE Physics 2023 Question with Solution

Given: radius of mercury drop $R = 10^{-3} \, \text{m}$, number of droplets $n = 125$, surface tension $T = 0.45 \, \text{N m}^{-1}$.

Find: gain in surface energy after breaking the drop into equal droplets.

Initial surface energy is

$$E_i = T \cdot 4\pi R^2$$

So,

$$E_i = 0.45 \times 4\pi \times (10^{-3})^2$$

By conservation of volume,

$$\frac{4}{3}\pi (10^{-3})^3 = 125 \times \frac{4\pi}{3} R_{\text{new}}^3$$

Therefore,

$$10^{-3} = 5R_{\text{new}}$$

Hence,

$$R_{\text{new}} = \frac{10^{-3}}{5} \, \text{m}$$

Final surface energy is

$$E_f = 0.45 \times 125 \times 4\pi \left(\frac{10^{-3}}{5}\right)^2$$

Increase in energy is

$$\Delta E = 0.45 \times 4\pi \times (10^{-3})^2 \left[\frac{125}{25} - 1\right]$$ $$= 4 \times 0.45 \times 4\pi \times 10^{-6}$$ $$= 2.26 \times 10^{-5} \, \text{J}$$

Therefore, the gain in surface energy is $2.26 \times 10^{-5} \, \text{J}$.

The solution working gives this value, but the solution labels the correct option as C. The numerical value matches option A.

Given: one initial drop of radius $10^{-3} \, \text{m}$ is converted into $125$ equal droplets.

Find: change in surface energy.

Surface energy is proportional to surface area:

$$E = T A$$

For the original drop,

$$A_i = 4\pi R^2$$

For each small drop, radius becomes $R_{\text{new}}$ and total area is

$$A_f = 125 \times 4\pi R_{\text{new}}^2$$

Using volume conservation,

$$R^3 = 125R_{\text{new}}^3$$ $$R_{\text{new}} = \frac{R}{5}$$

Hence,

$$A_f = 125 \times 4\pi \left(\frac{R}{5}\right)^2 = 5 \times 4\pi R^2$$

So the increase in area is

$$\Delta A = A_f - A_i = (5-1)4\pi R^2 = 4 \cdot 4\pi R^2$$

Therefore,

$$\Delta E = T\Delta A = 0.45 \times 4 \times 4\pi \times (10^{-3})^2$$ $$= 2.26 \times 10^{-5} \, \text{J}$$

Thus the correct numerical result is $2.26 \times 10^{-5} \, \text{J}$.