Question

An air bubble rises from the bottom of a lake of depth \(37\) m. The temperature of the lake is constant from the bottom to the top and atmospheric pressure is equal to the pressure due to \(10\) m of water column. The percentage increase in the volume of the bubble when it rises from a depth of \(20\) m to a depth of \(15\) m is

• A. \(25\%\)
• B. \(10\%\)
• C. \(20\%\)
• D. \(30\%\)

Hint:

For an isothermal process, \[ \boxed{PV=\text{constant}.} \] Pressure at a depth \(h\) in water is \[ \boxed{P=P_{\text{atm}}+\rho gh.} \] When atmospheric pressure is expressed as an equivalent water column, simply add the corresponding height.

Answer 1

The Correct Option is C

Step 1: Apply Boyle's law. Since the temperature remains constant, \[ PV=\text{constant}. \] Atmospheric pressure is equivalent to a water column of \[ 10\text{ m}. \] Hence, At a depth of \(20\) m, \[ P_1=(20+10)=30 \] (in equivalent metres of water). At a depth of \(15\) m, \[ P_2=(15+10)=25. \]

Step 2: Find the ratio of volumes. Using Boyle's law, \[ P_1V_1=P_2V_2. \] Therefore, \[ \frac{V_2}{V_1} = \frac{P_1}{P_2} = \frac{30}{25} = \frac65. \] Thus, \[ V_2=1.2V_1. \]

Step 3: Calculate the percentage increase. Percentage increase \[ = \frac{V_2-V_1}{V_1}\times100 = (1.2-1)\times100 = 20\%. \] Hence, \[ \boxed{20\%}. \] Thus, \[ \boxed{(C)} \] is the correct answer.