MHTCET 5 May 2025 Shift 2 Paper

Concept:Surface energy of a liquid drop depends on its surface area and surface tension.For a given liquid, surface tension $T$ remains constant.Explanation:Let the radius of the original drop be $R$ and that of each small drop be $r$.Since volume is conserved during splitting,$$\frac{4}{3}\pi R^{3}=729 \times \frac{4}{3}\pi r^{3}$$$$R^{3}=729r^{3}$$Taking the cube root gives:$$r=\frac{R}{9}$$The surface energy of the original drop is:$$E=T \times 4\pi R^{2}$$The total surface energy of the 729 small drops is:$$U=729 \times T \times 4\pi r^{2}$$$$U=729 \times T \times 4\pi \left(\frac{R}{9}\right)^{2}$$$$U=729 \times T \times \frac{4\pi R^{2}}{81}$$$$U=9T \times 4\pi R^{2}$$Thus,$$\frac{E}{U}=\frac{T \times 4\pi R^{2}}{9T \times 4\pi R^{2}}=\frac{1}{9}$$Given that $\frac{E}{U}=\frac{1}{x}$, comparing the denominators gives $x=9$.Answer:$$x=9$$Correct option: A (9)