InterviewSolution
Saved Bookmarks
| 1. |
The work done in increasing the radius of a soap bubble from 4 cm to 5 cm is Joule (given surface tension of soap water to be `25xx10^(-3)N//m)`A. `0.5657xx10^(-3)`B. `5.657xx10^(-3)`C. `56.5xx10^(-3)`D. `565xx10^(-3)` |
|
Answer» Correct Answer - A `W=8pi(r_(2)^(2)-r_(1)^(2))T` |
|