MHTCET 15 Apr 2026 Shift 1 Paper

Concept:Excess pressure inside a soap bubble is inversely proportional to its radius.Explanation:For a soap bubble, the excess pressure is given by$$\Delta P=\frac{4T}{R}$$where $T$ is the surface tension and $R$ is the radius of the bubble.For the first bubble,$$\Delta P_{1}=\frac{4T}{R_{1}}$$For the second bubble,$$\Delta P_{2}=\frac{4T}{R_{2}}$$Given that the excess pressure in the first bubble is three times that in the second bubble:$$\Delta P_{1}=3\Delta P_{2}$$Substituting the expressions:$$\frac{4T}{R_{1}}=3\left(\frac{4T}{R_{2}}\right)$$Cancelling $4T$ from both sides:$$\frac{1}{R_{1}}=\frac{3}{R_{2}}$$This gives:$$R_{2}=3R_{1}$$The volume of a spherical bubble is proportional to the cube of its radius:$$V\propto R^{3}$$Thus,$$\frac{V_{1}}{V_{2}}=\frac{R_{1}^{3}}{R_{2}^{3}}$$Substituting $R_{2}=3R_{1}$:$$\frac{V_{1}}{V_{2}}=\frac{R_{1}^{3}}{(3R_{1})^{3}}$$$$\frac{V_{1}}{V_{2}}=\frac{1}{27}$$Therefore, the ratio of volumes is:$$V_{1}:V_{2}=1:27$$Answer:The correct option is $1:27$, i.e., Option C.