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:

A common error in percentage problems is misinterpreting "X% more than Y". This means $Y + (X/100)Y$, not just $(X/100)Y$. In this problem, "125% more than $A_1$" means $A_1 + 1.25A_1 = 2.25A_1$. Also, expressing pressure in terms of "meters of water column" simplifies the calculation by allowing you to cancel out $\rho g$.

Answer 1

The Correct Option is D

Step 1: Relate Surface Area and Radius: Let suffix 1 represent the bottom and 2 represent the top. Given \( S_2 = S_1 + 1.25 S_1 = 2.25 S_1 \). Since Surface Area \( S = 4\pi r^2 \implies S \propto r^2 \). \[ \frac{r_2^2}{r_1^2} = \frac{S_2}{S_1} = 2.25 \] \[ \frac{r_2}{r_1} = \sqrt{2.25} = 1.5 = \frac{3}{2} \]
Step 2: Relate Radius and Volume: Volume \( V = \frac{4}{3}\pi r^3 \implies V \propto r^3 \). \[ \frac{V_2}{V_1} = \left( \frac{r_2}{r_1} \right)^3 = (1.5)^3 = 3.375 \]
Step 3: Apply Boyle's Law: Since temperature is uniform (isothermal process), \( P_1 V_1 = P_2 V_2 \). \( P_2 \) (at top) = Atmospheric pressure = 10 m of water. \( P_1 \) (at bottom) = Atmospheric pressure + Pressure due to depth \( h \) = \( (10 + h) \) m of water. \[ (10 + h) V_1 = (10) V_2 \] \[ 10 + h = 10 \left( \frac{V_2}{V_1} \right) \] \[ 10 + h = 10 (3.375) \] \[ 10 + h = 33.75 \] \[ h = 33.75 - 10 = 23.75 \, \text{m} \]