Physics · Mechanical Properties Of Solids · NEET
No. Stress is force per unit area on ONE cross-section, not the sum of both external forces. Cut the wire anywhere: the lower part is pulled down by weight W and held up by the tension in the wire above it. For equilibrium that tension = W. So the force acting across that cut is W, and stress = W/A. The two equal-opposite forces (W down at the mass, W up at the ceiling) are what a cross-section must balance, but each cross-section still transmits only W, not W+W.
If we ignore the wire's own weight (the usual NEET assumption), the stress is the SAME at every cross-section = W/A. Every slice below any cut must support only the hanging weight W. Only if the wire is heavy do lower sections carry W and higher sections carry W plus the weight of wire below them - but NEET questions almost always say 'light wire' or ignore this.
W/2A would mean each cross-section shares only half the weight, which is wrong. The wire is a single load path: the entire weight W must pass through every cross-section on its way to the ceiling. There is no second wire to split the load. So stress = full weight / area = W/A.
Yes - the tension in the wire equals W (for a massless wire in equilibrium). Tensile stress is just that tension spread over the area: Stress = Tension/Area = W/A. Tension is a force (newton); stress is force per area (N/m squared or pascal).
Stress alone does not give elongation - you also need Young's modulus Y. Use elongation ΔL = (W L)/(A Y), which comes from Y = stress/strain = (W/A)/(ΔL/L). The stress W/A is the first step; divide by Y to get strain, then multiply by L.
A wire is suspended from the ceiling and stretched by a weight W attached at its free end. The longitudinal stress at any point of cross-sectional area A of the wire is:
When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the internal restoring force per unit area acting along the length of the wire. For a weight W on a wire of cross-section A, tensile stress = W/A, with SI unit pascal (N/m squared).
W/A. Only the weight W is transmitted through each cross-section. The 2W/A answer wrongly adds the ceiling reaction, which is already balanced internally.
Stress behaves like a tensor, not a simple vector, but at the NEET level treat its magnitude as W/A. It is defined per cross-section, so a single force can give different stresses on differently oriented planes.
Young's modulus Y = tensile stress / longitudinal strain = (W/A)/(ΔL/L). Rearranged, elongation ΔL = WL/(AY). So the same W/A stress is the starting point for finding stretch.
Then lower cross-sections carry only W, but a section at height x above the load carries W plus the weight of the wire below it, so stress increases upward. NEET normally ignores wire weight unless it is stated.
