A circular ring of small thickness and of inner and outer diameters ` 3.8` cm and `4.2` cm respectively , suspended on one side of a common balance is exactly balanced by keeping sufficient mass in the pan on other side .At this instant the lower surface of ring is just resting on the surface of a liquid . Now if the ring is puled up gently , such that it does not loose its contact with the liquid , find how much mass should be changed in other pan , again to keep the balance in balanced position ? (Assume angle of contact is zero and surface tension of liquid is ` (1 //pi) ` N/m)

Explanation

Total force due to surface tension ` = ( 2 pi r_(1) +2pir_(2))T`
` mg = 2piT (r_(1)+r_(2))`
` m = (2piT(r_(1)+r_(2)))/(g) = (2pi xx 10/pi xx 10^(3) (2+2.2))/(1000) = 8 gm `