Question:
A bubble rises from the bottom of a lake \(90\text{ m}\) 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 \(90\text{ m}\) deep. On reaching the surface, its volume becomes \((\text{Atmospheric pressure is }10\text{ m of water})\)
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Pressure at depth \(h\) includes atmospheric pressure also. Use \(P_1V_1=P_2V_2\) for bubbles rising in water.
Updated On: May 7, 2026
- \(4\) times
- \(8\) times
- \(10\) times
- \(3\) times
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The Correct Option is C
Solution and Explanation
The bubble is at the bottom of a lake of depth:
\[
90\text{ m}.
\]
Atmospheric pressure is equivalent to:
\[
10\text{ m of water}.
\]
So, pressure at the bottom of the lake is:
\[
P_1=90+10.
\]
\[
P_1=100\text{ m of water}.
\]
At the surface, pressure is only atmospheric pressure:
\[
P_2=10\text{ m of water}.
\]
Assuming temperature remains constant, Boyle's law applies:
\[
P_1V_1=P_2V_2.
\]
So:
\[
\frac{V_2}{V_1}=\frac{P_1}{P_2}.
\]
Substitute values:
\[
\frac{V_2}{V_1}=\frac{100}{10}.
\]
\[
\frac{V_2}{V_1}=10.
\]
Therefore, the volume becomes:
\[
10\text{ times}.
\]
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