Physics · Mechanical Properties Of Solids · NEET
The force (tension) is the SAME in both wires. Think of the join point: the copper wire pulls the steel wire and the steel wire pulls the copper wire with equal force (Newton's third law). So if a load W hangs at the bottom, every cross-section along the chain carries that same force F = W. What changes from wire to wire is the STRESS (F/A), because the areas may differ, and the strain, because Young's modulus may differ.
You ADD the elongations, not the forces. Force is common (same F in each). Total stretch ΔL = ΔL₁ + ΔL₂ + ... Each individual stretch is found separately using ΔLᵢ = F·Lᵢ / (Aᵢ·Yᵢ), then summed. This is the opposite of wires in PARALLEL (side by side sharing a load), where the elongation is common and the forces add up.
NCERT Example 8.2: copper (L=2.2 m) and steel (L=1.6 m), both diameter 3.0 mm, joined end to end, net elongation 0.70 mm. Since it is series, ΔL_total = F[L_cu/(A·Y_cu) + L_steel/(A·Y_steel)], where A = π(1.5×10⁻³)² m² is the same for both (same diameter). Put Y_cu ≈ 1.1×10¹¹ Pa and Y_steel ≈ 2.0×10¹¹ Pa, set ΔL_total = 0.70×10⁻³ m, and solve for F. This gives a load of about 178 N (≈ 1.8×10² N).
Same material means Y is common. Then ΔL_total = (F/Y)[L₁/A₁ + L₂/A₂]. The force is still the same in both. The thinner wire (smaller A) has higher stress and stretches more per unit length, so it usually contributes most of the total elongation. This is why a chain breaks at its weakest (thinnest) link — highest stress there.
Series (end to end): one common force, elongations add. Parallel (side by side, sharing one load): one common elongation, forces add. A single wire cut into pieces and rejoined end to end behaves like series — same material and area, so total elongation is unchanged for the same force because total length is unchanged. Always first identify which arrangement the question describes before choosing add-force or add-elongation.
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Δl on applying a force F, how much force is needed to stretch the second wire by the same amount?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
ΔL_total = ΔL₁ + ΔL₂ + ... = F·L₁/(A₁Y₁) + F·L₂/(A₂Y₂) + ..., where F is the common force in every wire and each term uses that wire's own length, area and Young's modulus.
Not always. The force is the same, but stress = F/A. If the wires have different cross-sectional areas, the thinner wire has larger stress even though the force is equal in both.
Because the wires are joined tip to tip, the total stretch is simply the stretch of the first plus the stretch of the second (lengths add). The same tension is transmitted through the whole chain, so the force is common, not additive.
The wire with the larger value of L/(A·Y) stretches more. A longer, thinner, or lower-Young's-modulus wire elongates more for the same common force.
Most NEET questions on joined wires just need you to (1) realise the force is common, (2) compute each ΔL with F·L/(A·Y), and (3) add them or set the sum equal to a given total to solve for the unknown load.
