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: Understanding the Concept:
Since the temperature of the water is uniform, the process of the air bubble rising is considered isothermal. Therefore, we can apply Boyle's Law, which states that for a fixed mass of gas at constant temperature, the product of pressure and volume is constant ($P_1V_1 = P_2V_2$). We also need to relate the bubble's volume to its surface area and use the formula for pressure at a certain depth in a fluid.
Step 2: Key Formula or Approach:
Let the conditions at the bottom of the tank be denoted by subscript 1 and at the top by subscript 2.
1. Boyle's Law: $P_1V_1 = P_2V_2$.
2. Pressure at depth $h$: $P_1 = P_{atm} + h\rho g$, where $P_{atm}$ is the atmospheric pressure.
3. Pressure at the top (surface): $P_2 = P_{atm}$.
4. For a spherical bubble, surface area $A = 4\pi r^2$ and volume $V = \tfrac{4}{3}\pi r^3$. From these, we can establish the relationship $V \propto A^{3/2}$.
Step 3: Detailed Explanation:
First, let's establish the relationship between the surface areas and volumes.
The surface area at the top ($A_2$) is 125% more than the surface area at the bottom ($A_1$).
\[ A_2 = A_1 + (125% \text{ of } A_1) = A_1 + 1.25 A_1 = 2.25 A_1 \] \[ \frac{A_2}{A_1} = 2.25 \] Since the volume of a sphere is proportional to the $3/2$ power of its surface area ($V \propto A^{3/2}$), the ratio of the volumes is:
\[ \frac{V_2}{V_1} = \left(\frac{A_2}{A_1}\right)^{3/2} = (2.25)^{3/2} = (1.5^2)^{3/2} = (1.5)^3 = 3.375 \] Now, applying Boyle's Law for the isothermal process:
\[ P_1V_1 = P_2V_2 \implies \frac{P_1}{P_2} = \frac{V_2}{V_1} = 3.375 \] Let the depth of the tank be $h$. We are given that the atmospheric pressure $P_{atm}$ is equivalent to a 10 m water column. Let this height be $H_0 = 10$ m.
The pressure at the top (surface) is $P_2 = P_{atm} = H_0 \rho g = 10 \rho g$.
The pressure at the bottom (depth $h$) is $P_1 = P_{atm} + h\rho g = (H_0 + h)\rho g = (10+h)\rho g$.
Now, we substitute these pressure expressions into our ratio:
\[ \frac{P_1}{P_2} = \frac{(10+h)\rho g}{10\rho g} = 3.375 \] The term $\rho g$ cancels out:
\[ \frac{10+h}{10} = 3.375 \] \[ 10 + h = 10 \times 3.375 = 33.75 \] \[ h = 33.75 - 10 = 23.75 \text{ m} \] Step 4: Final Answer:
The depth of the water in the tank is 23.75 m. Therefore, option (D) is the correct answer.