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```
An air bubble of radius 0.1 cm in a liquid having surface tension 0.06N/m and density 1000kg/m3. The pressure inside the bubble is 1100Pa greater than the atmospheric pressure. At what depth below the surface of the liquid is the bubble ? (g=9.8m/s2)
An air bubble of radius 0.1 cm in a liquid having surface tension 0.06N/m and density 1000kg/m3. The pressure inside the bubble is 1100Pa greater than the atmospheric pressure. At what depth below the surface of the liquid is the bubble ? (g=9.8m/s2)

```
6 years ago

Sanchayan Ghosh
27 Points
```							The answer requires the usage of two concepts:
1. Gauge pressure.
2. Excess prssure in an air bubble.

Imagine an air bubble at a depth h, having a radius r, in a liquid of density p.

Using gauge pressure, we know,
P - Pa = pgh (Pa = atmospheric pressure)
So, P = pgh + Pa.
Now, using excess pressure,
Pi - Po = 4S/R [Two surfaces in contact).
Pi = Pa + 1100.
Po = Pa + pgh
Pi - Po = 1100 - pgh
so, 1100 - pgh = 4S/R.
h = 1100/pg - 4S/Rpg.
Solving, you get 8.7cm. Just check the answer, you get it like this.
```
6 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions