JEE Main 2023 · 30 January · evening · Single correct · +4 / −1
A force is applied to a steel wire 'A', rigidly clamped at one end. As a result elongation in the wire is 0.2 mm. If same force is applied to another steel wire ' B ' of double the length and a diameter 2.4 times that of the wire ' A ', the elongation in the wire ' B ' will be (wires having uniform circular cross sections)
- (a) 6 .9 × 10⁻² mm
- (b) 6.06 × 10⁻² mm
- (c) 2.77 × 10⁻² mm
- (d) 3.0 × 10⁻² mm
Answer and solution
Answer (a) 6 .9 × 10⁻² mm
Solution
Physics sir · 4 steps
Same force, same steel, so elongation goes as length divided by diameter squared. Take the ratio.
- Elongation formula. From Young's modulus, Y = (F/A)/(Δ L/L), so Δ L = FL/AY. Both wires are steel, so Y is the same, and the force is the same too.
- Area uses diameter squared. A circular cross section has A = (π d²)/4, so A ∝ d². That makes Δ L ∝ L/d².
(Δ L_B)/(Δ L_A) = L_B/L_A×(d_A/d_B)² = 2×1/(2.4)² = 2/5.76
- Put in numbers. Δ L_B = 0.2 mm×2/5.76 = 0.4/5.76 mm = 0.0694 mm.
- So. The elongation of wire B is about 6.9×10⁻² mm, which is option (a).
The idea to keep. For wires of the same material under the same force, Δ L ∝ L/d². Do not forget to square the diameter, because the area depends on d².
Answer from the NTA final answer key. Solution by PrepWiser.