MHTCET 19 Apr 2026 Shift 1 Paper

Concept:A soap bubble has two free surfaces, so its surface area is $2 \times 4\pi r^2$.When the bubble breaks, its volume remains constant.Explanation:Let the radius of the large bubble be $R$ and that of each small bubble be $r$.Given the large bubble diameter is $D$, we get:$$R = \frac{D}{2}$$Initial surface energy of the large soap bubble:$$E_i = 2 \times 4\pi R^2 \times T = 8\pi T R^2$$Since volume is conserved when it splits into 64 bubbles:$$\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3$$$$R^3 = 64r^3 \quad \Rightarrow \quad R = 4r \quad \Rightarrow \quad r = \frac{R}{4}$$Final surface energy of 64 small bubbles:$$E_f = 64 \times 8\pi T r^2 = 64 \times 8\pi T \left(\frac{R}{4}\right)^2 = 32\pi T R^2$$Change in surface energy:$$\Delta E = E_f - E_i = 32\pi T R^2 - 8\pi T R^2 = 24\pi T R^2$$Substitute $R = \frac{D}{2}$:$$\Delta E = 24\pi T \left(\frac{D}{2}\right)^2 = 24\pi T \frac{D^2}{4} = 6\pi T D^2$$Answer:The change in surface energy is $6\pi T D^2$.Therefore, the correct option is C.