A wire ring of 5 cm radius is rested on the surface of a liquid and then raised. The pull required is 3.14 gf more before the film breaks than it is after. Find the surface tension of the liquid (in dyne cm-1). Take π=3.14

When the wire ring is resting on the liquid surface, a thin film of liquid is in contact with the ring. Force due to surface tension acts along the circumference. Therefore, the force required to lift the ring is equal to the sum of the weight of the ring and the force due to surface tension. Hence, the extra pull required to break the film is equal to the force on the ring due to surface tension i.e.       pull required to break the film = force on the ring due to surface tensionHere, pull required to break the film     =3.14gf=3.14×1000 dyne and radius of the ring, r = 5 cm.The liquid is in contact with the ring both along its inner and outer circumference. Therefore, force on the ring due to surface tension      =2(2πrT)=2×2π×5×T=20πTHence, 20πT=3.14×1000Or T=3.14×100020π=50.0 dyne cm−1