Home
Class 11
PHYSICS
The work done in blowing a soap bubble o...

The work done in blowing a soap bubble of volume `V` is `W`. The work done in blowing a soap bubble of volume `2V` is

A

W

B

`2^((2)/(3))W`

C

`3^((2)/(3))W`

D

2W

Text Solution

Verified by Experts

The correct Answer is:
B

`W=8pir^(2)T, V=(4)/(3)pir^(3), V alpha r^(3),(W_(2))/(W_(1))=[(V_(2))/(V_(1))]^((2)/(3))`
Promotional Banner

Topper's Solved these Questions

  • MECHANICAL PROPERTIES OF FLUIDS

    NARAYNA|Exercise EXERCISE - II (C.W) (CAPILLARITY & CAPILLARY RISE)|6 Videos
  • MECHANICAL PROPERTIES OF FLUIDS

    NARAYNA|Exercise EXERCISE - II (C.W) (EXCESS PRESSURE INSIDE A LIQUID DROP AND IN A SOAP BUBBLE)|5 Videos
  • MECHANICAL PROPERTIES OF FLUIDS

    NARAYNA|Exercise EXERCISE - II (C.W) (FORCE DUE TO SURFACE TENSION)|2 Videos
  • MATHEMATICAL REVIEW & PHYSICAL WORLD

    NARAYNA|Exercise C.U.Q|13 Videos
  • MECHANICAL PROPERTIES OF SOLIDS

    NARAYNA|Exercise LEVEL-II (H.W)|24 Videos

Similar Questions

Explore conceptually related problems

The work done is blowing a soap bubble of volume V is W . The work done in blowing a soap bubble of volume 2V is

The work done is blowing a soap bubble of volume V is W the work done in blowing a soap bubble of volume of 2V is

The work done in blowing a bubble of volume V is W , then what is the work done in blowing a soap bubble of volume 2V ?

The work done in blowing a soap bubble of radius 5 cm is

The work done in blowing a soap bubble of radius 0.02 cm is ,

Work done is blowing a soap bubble of diameter 2 cm , is

If W is amount of work done in forming a soap bubble of volume V , then the amount of work done In forming a bubble of volume 2V from the same solution will be

Work done in forming a soap bubble of radius R is W. Then workdone is forming a soap bubble of radius '2R' will be :

The excess pressure inside a soap bubble of volume V is P . Then excess pressure inside a soap bubble of volume 2V is