Physics · Mechanical Properties Of Fluids · NEET
The disc's own weight is already balanced when you hold it. The EXTRA force is the pull from the water film. Water molecules cling to the edge of the disc (adhesion) and the surface tension of the film pulls the disc back toward the water. To break free you must supply an extra force equal to this pull. That is why the answer only uses T and the edge length, not the mass or g.
Use F = T x 2(pi)r for a solid flat disc. The water contacts the disc only along its outer circumference, which is a single line of length 2(pi)r. There is only ONE circular contact line. The 4(pi)r case is a common trap that belongs to a thin RING, not a disc.
A thin ring has TWO edges: an inner circle and an outer circle. Water clings along both. For a thin ring where inner and outer radius are nearly equal to r, the total contact length is 2(pi)r + 2(pi)r = 4(pi)r, so F = T x 4(pi)r. If the two radii differ, use F = T x 2(pi)(r_inner + r_outer).
A soap film has two liquid-air faces, so you double it. Here the water is a single body with one free surface pulling on the solid. So you count the number of CONTACT LINES on the solid (one for a disc, two for a ring), not the number of soap-film faces. Do not mix the two rules.
A straight slider on a soap film has film on both sides, giving F = 2TL. A disc lifted from plain water touches water along one closed loop, giving F = T x 2(pi)r. The idea is the same - force per unit length times length - but you must correctly count how many liquid edges pull on the object.
A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If the surface tension of water is 0.07 N/m, then the excess force required to take it away from the surface is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
F = T x 2(pi)r, where T is surface tension of water and r is the disc radius. It is the surface tension times the length of the circular contact line (the circumference).
A thin ring touches water along two edges, so F = T x 4(pi)r for a ring of radius r, or F = T x 2(pi)(r_inner + r_outer) if the radii differ noticeably.
No. The question asks for the EXTRA force to break the water surface. That extra force is only the surface-tension pull. The weight is handled separately when you actually lift the object.
Water molecules stick to the disc's edge (adhesion) and the surface film is under tension. This tension pulls the wetted edge back toward the water, so you must apply extra force to separate them.
Convert radius from cm to metres and keep T in N/m. Then F comes out in newtons. In the NEET 2024 disc problem, 0.0198 N is the same as 19.8 mN.
