MHTCET 11 Apr 2026 Shift 2 Paper

Concept:Surface energy is directly proportional to surface area, and volume remains constant when a drop breaks into smaller droplets.Explanation:Let the radius of the original drop be $R$.Its initial surface area is $4 \pi R^2$, so its surface energy is $E$.When the drop is sprayed into 512 identical droplets, let the radius of each small droplet be $r$.Since volume is conserved:$$\frac{4}{3}\pi R^3 = 512 \times \frac{4}{3}\pi r^3$$Cancelling common factors gives:$$R^3 = 512 r^3$$Since $512 = 8^3$, we get:$$R = 8r \quad \Rightarrow \quad r = \frac{R}{8}$$Now, the total final surface area is:$$512 \times 4\pi r^2 = 512 \times 4\pi \left(\frac{R}{8}\right)^2$$$$= 512 \times 4\pi \frac{R^2}{64} = 8 \times 4\pi R^2$$Thus, the final surface area is 8 times the original surface area.Since surface energy is proportional to surface area, the final surface energy becomes $8E$.Answer:Final surface energy = $8E$Correct option: B. $8E$