MHTCET 17 Apr 2026 Shift 2 Paper

Concept:Excess pressure inside a soap bubble depends inversely on its radius, and volume depends on the cube of the radius.Explanation:For a soap bubble, the excess pressure is$$\Delta P = \frac{4T}{r}$$Here, $T$ is the surface tension and $r$ is the radius of the bubble.Since both bubbles are formed from the same soap solution, $T$ is identical for both.Given that the excess pressure in the first bubble is three times that in the second bubble:$$\Delta P_1 = 3\Delta P_2$$From $\Delta P \propto \frac{1}{r}$, we can write:$$\frac{1}{r_1} = 3 \cdot \frac{1}{r_2}$$Rearranging gives:$$r_2 = 3r_1$$The volume of a spherical bubble is proportional to the cube of its radius:$$V \propto r^3$$Thus, the ratio of volumes is:$$\frac{V_1}{V_2} = \frac{r_1^3}{r_2^3}$$Substitute $r_2 = 3r_1$:$$\frac{V_1}{V_2} = \frac{r_1^3}{(3r_1)^3} = \frac{1}{27}$$Therefore, $V_1 : V_2 = 1 : 27$.Answer:The correct option is D: $1 : 27$.