MHTCET 13 Apr 2026 Shift 1 Paper

Concept:Surface energy of a liquid drop is the product of surface tension and surface area, and volume is conserved when drops coalesce.Explanation:Let the radius of each small drop be $r$.Let the radius of the big drop formed be $R$.Volume before coalescence is $2 \times \frac{4}{3} \pi r^3$.Volume after coalescence is $\frac{4}{3} \pi R^3$.Equating the two volumes:$$\frac{4}{3} \pi R^3 = 2 \times \frac{4}{3} \pi r^3$$This gives $R^3 = 2r^3$, so $R = 2^{1/3} r$.Surface energy before the change is $E_{\text{before}} = T \times 2(4\pi r^2) = 8\pi r^2 T$.Surface energy after the change is $E_{\text{after}} = T \times 4\pi R^2$.Substitute $R = 2^{1/3} r$:$$E_{\text{after}} = 4\pi T (2^{1/3}r)^2 = 4\pi T \cdot 2^{2/3} r^2$$Therefore, the required ratio is:$$\frac{E_{\text{after}}}{E_{\text{before}}} = \frac{4\pi T \cdot 2^{2/3} r^2}{8\pi r^2 T} = \frac{2^{2/3}}{2} = 2^{-1/3}$$Hence the ratio of total surface energies after and before is $2^{-1/3} : 1$.Answer:$2^{-1/3} : 1$Correct option: C.