Question

In a water tank, an air bubble rises from the bottom to the top surface of the water. If the depth of the water in the tank is 7.28 m and atmospheric pressure is 10 m of water, then the ratio of the radii of the bubble at the bottom of the tank and at the top surface of the water is (Temperature of the water in the tank is constant)

• A. 2:3
• B. 5:6
• C. 3:4
• D. 4:5

Hint:

Boyle's law: $P_1V_1 = P_2V_2$ (for constant temperature). Pressure at depth h: $P = P_{atm} + \rho gh$.

Answer 1

The Correct Option is B

Let $P_1$ and $r_1$ be the pressure and radius of the bubble at the bottom of the tank, and $P_2$ and $r_2$ be the pressure and radius at the top surface. At the bottom, $P_1 = P_{atm} + \rho gh = 10 + 7.28 = 17.28$ m of water (where $\rho$ is the density of water and $h$ is the depth). At the top, $P_2 = P_{atm} = 10$ m of water. Since the temperature is constant, we can use Boyle's law: $P_1V_1 = P_2V_2$. $P_1(\frac{4}{3}\pi r_1^3) = P_2(\frac{4}{3}\pi r_2^3)$ $17.28 r_1^3 = 10 r_2^3$ $\frac{r_1^3}{r_2^3} = \frac{10}{17.28} = \frac{10}{10+7.28}$ $\frac{r_1}{r_2} = \sqrt[3]{\frac{10}{17.28}} = \frac{\sqrt[3]{10}}{2.579} \approx \frac{5}{6}$ If $P_{atm}=10$ m of water, and $h=7.28$ m, then $P_1 = 10\rho g + 7.28\rho g$ and $P_2=10 \rho g$. $P_1 V_1 = P_2 V_2$ so $(17.28 \rho g) (\frac{4}{3}\pi r_1^3) = (10\rho g)(\frac{4}{3}\pi r_2^3)$ $17.28 r_1^3 = 10r_2^3$ $\frac{r_1}{r_2} = \sqrt[3]{\frac{10}{17.28}} \approx 0.833 \approx \frac{5}{6}$.