TG EAMCET 3 May 2025 Shift 1 Paper

Step 1: Find Surface Area RelationshipThe bubble's surface area at the top is $125\%$ more than at the bottom. This means:$A_{\;\text{top }\;}=A_{\;\text{bottom }\;}+1.25A_{\;\text{bottom }\;}=2.25A_{\;\text{bottom }\;}$Step 2: Connect Surface Area and VolumeThe volume $(V)$ of a bubble is related to its surface area $(A)$ by: $V \propto A^{3/2}$Step 3: Express Volume ChangeLet $V_1$ be the volume at the bottom, and $V_2$ at the top. Then,$\frac{V_2}{V_1} = \left(\frac{A_2}{A_1}\right)^{3/2} = (2.25)^{3/2} = 3.375$Step 4: Use Boyle's Law for Pressure and VolumeSince the water temperature is the same, Boyle's law applies: $p_1 V_1 = p_2 V_2 \; \frac{p_1}{p_2} = \frac{V_2}{V_1} = 3.375$Step 5: Write Out Pressure at Bottom and TopAt the bottom, pressure is $p_1=$ atmospheric pressure + pressure from water column with height $h$ Atmospheric pressure is equal to water pressure from 10 m column. So: $p_1=10+h$ At the top, the pressure is just the atmospheric pressure: $p_2=10$Step 6: Set Up the Equation and Solve for $h$So, $\frac{p_1}{p_2} = \frac{10+h}{10}$ Set equal to $3.375: \frac{10+h}{10}=3.37510+h=33.75 h=23.75\,\text{m}$