A swimming pool (Figure) is conical in shape of diameter 20 m. To measure the depth of the pool, a p — States of Matter and Gaseous State Chemistry Question
Question
A swimming pool (Figure) is conical in shape of diameter 20 m. To measure the depth of the pool, a pipe line having cross sectional area $\pi \text{ mm}^2$ has been introduced through the base centre to the depth of the pool. Other end of the pipe line is connected with the gas cylinder. When gases are passed through the pipe line, it has been measured that the volume of each bubble across the pipe line at the surface of the pool is $2\pi \text{ mm}^3$. Assume the temperature of water is same as that of atmosphere and 1 atm pressure is equal to 10 m of water column. If the water holding capacity of the swimming pool is $V \text{ m}^3$, then the value of $\frac{3V}{100\pi}$ is
💡 Solution & Explanation
Bubble initially assumes a spherical shape matching the pipe's radius. Area $\pi r^2 = \pi \text{ mm}^2 \implies r = 1\text{ mm}$. Initial volume $V_1 = \frac{4}{3}\pi r^3 = \frac{4\pi}{3} \text{ mm}^3$. Surface volume $V_2 = 2\pi \text{ mm}^3$. Let depth be $h$. Pressure at depth $P_1 = (10 + h)$ m of water. Surface pressure $P_2 = 10$ m of water. $P_1 V_1 = P_2 V_2 \implies (10 + h)(\frac{4\pi}{3}) = 10(2\pi) \implies 10 + h = 15 \implies h = 5\text{ m}$. Volume of conical pool $V = \frac{1}{3}\pi R^2 h = \frac{1}{3}\pi (10)^2(5) = \frac{500\pi}{3} \text{ m}^3$. Thus, $\frac{3V}{100\pi} = \frac{3(500\pi/3)}{100\pi} = 5$. Therefore, correct answer is 5.