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: Understanding the Concept:
Since the temperature of the water is uniform, the process is isothermal. We can apply Boyle's Law (\(P_1V_1 = P_2V_2\)) to the air bubble. We also need to relate the surface area of the bubble to its volume and understand how pressure varies with depth in a fluid.
Step 2: Key Formula or Approach:
Let the conditions at the bottom be denoted by subscript 1 and at the top by subscript 2.
- Boyle's Law: \(P_1V_1 = P_2V_2\)
- Pressure at depth \(h\): \(P_1 = P_{atm} + h\rho g\), where \(P_{atm}\) is the atmospheric pressure.
- Pressure at the top: \(P_2 = P_{atm}\).
- For a spherical bubble, surface area \(A = 4\pi r^2\) and volume \(V = \frac{4}{3}\pi r^3\). From this, we can deduce that \(V \propto A^{3/2}\).
Step 3: Detailed Explanation:
First, let's find the relationship between the volumes.
We are given that the surface area at the top (\(A_2\)) is 125% more than at the bottom (\(A_1\)).
\[ A_2 = A_1 + 1.25 A_1 = 2.25 A_1 \implies \frac{A_2}{A_1} = 2.25 \] Since \(V \propto A^{3/2}\), 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, let's apply Boyle's Law:
\[ \frac{V_2}{V_1} = \frac{P_1}{P_2} \implies 3.375 = \frac{P_1}{P_2} \] Let the depth of the tank be \(h\). We are given that atmospheric pressure \(P_{atm}\) is equivalent to a 10 m water column. Let \(H = 10\) m.
\[ P_2 = P_{atm} = H\rho g = 10\rho g \] The pressure at the bottom is:
\[ P_1 = P_{atm} + h\rho g = 10\rho g + h\rho g = (10+h)\rho g \] Substituting these into the pressure ratio:
\[ 3.375 = \frac{(10+h)\rho g}{10\rho g} = \frac{10+h}{10} \] Now, solve for \(h\):
\[ 3.375 \times 10 = 10 + h \] \[ 33.75 = 10 + h \] \[ h = 33.75 - 10 = 23.75 \text{ m} \] Step 4: Final Answer:
The depth of the water in the tank is 23.75 m. Therefore, option (D) is correct.