badge image

Enroll For Free Now & Improve your performance.

User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.

Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 12


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

Answers : (1)

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
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

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

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details