A ring mass of 1 gm and radius 3 cm is in the surface of a liquid of surface tension 8 × 10–2 N/m. The force required to take out the ring from the surface of liquid will be
(1) 4 × 10–2 N
(2) 3 × 10–2 N
(3) 1.5 × 10–2 N
(4) zero
Correct Answer is: (1) 4 × 10–2 N
The force required to pull out the ring
= weight of ring + force due to surface tension
= mg + 2 × 2πr × T
= 10–3 × 9.8 + 4 × 3.14 × 3 × 10–2 × 8 × 10–2
= 4 × 10–2 N
