Saved Bookmarks
| 1. |
The work done in blowina siap bubble of surface tension `0.06 N,^(-1)` from 2 cm radius to 5 cm radiu isA. `0.004168 J`B. `0.003168 J`C. `0.003158 J`D. `0.004158 J` |
|
Answer» As given, `s=0.06 Nm^(-1)` `r_(1)=2 cm =0.02 m,r_(2)=5 cm =0.05` Since, bubble has two surface Initial surface areqa of the bubble `=2xx4pir_(2)^(2)` `=2xx4pixx(0.2)^(2)` `=32pixx10^(-4)m^(2)` Final surface area of the bubble `=2xx4pir_(2)^(2)` `=2xx4pixx(0.05)^(2)` `=200pixx10^(-4)m^(2)` So, work done `=s xx` increase in surface `=0.06xx(200pi10^(-4)-32pixx10^(-4))` `0.06xx168pi10^(-4)` `0.0031168 J` |
|