Question

A bubble rises from the bottom of a lake 90 m deep on reaching the surface, its volume becomes (Atmospheric pressure is 10 m of water)

• A. 4 times
• B. 8 times
• C. 10 times
• D. 3 times

Hint:

When pressure is given in "meters of water," it simplifies calculations. The total pressure at a depth 'h' is simply (Atmospheric pressure in m of water + h). This avoids having to use the formula \(P=\rho g h\) explicitly.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
An air bubble rises from the bottom of a lake to the surface. We need to find the factor by which its volume increases.

Step 2: Key Formula or Approach:
As the bubble rises, the external pressure on it decreases, causing it to expand. Assuming the temperature of the lake water is constant, we can apply Boyle's Law: \(P_1 V_1 = P_2 V_2\). We need to find the pressures at the bottom and at the surface. The pressure is conveniently given in terms of 'meters of water'.

Step 3: Detailed Explanation:
Let the state at the bottom be 1 and at the surface be 2.

Pressure at the surface (\(P_2\)):
This is just the atmospheric pressure.
\[ P_2 = P_{atm} = 10 \text{ m of water} \]

Pressure at the bottom (\(P_1\)):
This is the sum of the atmospheric pressure and the gauge pressure due to the water column.
\[ P_1 = P_{atm} + P_{gauge} = P_{atm} + h \]
\[ P_1 = 10 \text{ m of water} + 90 \text{ m of water} = 100 \text{ m of water} \]
Let the volume at the bottom be \(V_1\) and at the surface be \(V_2\).
According to Boyle's Law:
\[ P_1 V_1 = P_2 V_2 \]
\[ (100) \times V_1 = (10) \times V_2 \]
We want to find the ratio \(\frac{V_2}{V_1}\), which tells us how many times the volume becomes.
\[ \frac{V_2}{V_1} = \frac{100}{10} = 10 \]
So, \(V_2 = 10 V_1\). The volume becomes 10 times its original volume.

Step 4: Final Answer:
The volume of the bubble becomes 10 times larger.