Question

An air bubble rises from the bottom to the top of a water tank in which the temperature of the water is uniform. The surface area of the bubble at the top of the tank is 125% more than its surface area at the bottom of the tank. If the atmospheric pressure is equal to the pressure of 10 m water column, then the depth of water in the tank is

• A. 16.25 m
• B. 27 m
• C. 19 m
• D. 23.75 m

Hint:

When dealing with percentage increases, "X% more than Y" means \(Y + (X/100)Y\). In this problem, "125% more than \(A_1\)" means \(A_1 + 1.25A_1 = 2.25A_1\), not \(1.25A_1\). This is a common point of error.

Answer 1

The Correct Option is D

Step 1: Formula for Elastic Potential Energy Density: Energy density (u) = Energy / Volume. \[ u = \frac{1}{2} \times \text{stress} \times \text{strain} \] Since \( Y = \frac{\text{stress}}{\text{strain}} \implies \text{strain} = \frac{\text{stress}}{Y} \). Substitute strain: \[ u = \frac{1}{2} \times \text{stress} \times \frac{\text{stress}}{Y} = \frac{(\text{stress})^2}{2Y} \]
Step 2: Calculate Stress: Force \( F = 9 \times 10^4 \, \text{N} \). Area \( A = 3 \, \text{cm}^2 = 3 \times 10^{-4} \, \text{m}^2 \). Stress \( \sigma = \frac{F}{A} = \frac{9 \times 10^4}{3 \times 10^{-4}} = 3 \times 10^8 \, \text{Nm}^{-2} \).
Step 3: Calculate Energy Density: Young's Modulus \( Y = 2 \times 10^{11} \, \text{Nm}^{-2} \). \[ u = \frac{(3 \times 10^8)^2}{2 \times (2 \times 10^{11})} \] \[ u = \frac{9 \times 10^{16}}{4 \times 10^{11}} \] \[ u = \frac{9}{4} \times 10^5 = 2.25 \times 10^5 \, \text{Jm}^{-3} \]