Question:
A bubble rises from the bottom of a lake \(90\) m deep. On reaching the surface, its volume becomes \((\text{Atmospheric pressure is }10\text{ m of water})\)
A bubble rises from the bottom of a lake \(90\) m deep. On reaching the surface, its volume becomes \((\text{Atmospheric pressure is }10\text{ m of water})\)
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Pressure in water increases with depth; total pressure at depth \(h\) equals atmospheric pressure plus liquid pressure.
Updated On: May 7, 2026
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The Correct Option is C
Solution and Explanation
Concept:
For a gas bubble, assuming temperature remains constant, Boyle's law is applied: \[ P_1V_1=P_2V_2 \]
Step 1: Depth of lake is: \[ 90\text{ m} \]
Step 2: Atmospheric pressure is equivalent to: \[ 10\text{ m of water} \]
Step 3: Pressure at the bottom of the lake: \[ P_1=90+10=100\text{ m of water} \]
Step 4: Pressure at the surface: \[ P_2=10\text{ m of water} \]
Step 5: By Boyle's law: \[ P_1V_1=P_2V_2 \] \[ 100V_1=10V_2 \] \[ V_2=10V_1 \]
Step 6: Therefore, the volume becomes \(10\) times. \[ \boxed{10\text{ times}} \]
For a gas bubble, assuming temperature remains constant, Boyle's law is applied: \[ P_1V_1=P_2V_2 \]
Step 1: Depth of lake is: \[ 90\text{ m} \]
Step 2: Atmospheric pressure is equivalent to: \[ 10\text{ m of water} \]
Step 3: Pressure at the bottom of the lake: \[ P_1=90+10=100\text{ m of water} \]
Step 4: Pressure at the surface: \[ P_2=10\text{ m of water} \]
Step 5: By Boyle's law: \[ P_1V_1=P_2V_2 \] \[ 100V_1=10V_2 \] \[ V_2=10V_1 \]
Step 6: Therefore, the volume becomes \(10\) times. \[ \boxed{10\text{ times}} \]
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