JEE Main 2023 · 24 January · morning · Single correct · +4 / −1
A 100 m long wire having cross-sectional area 6.25×10⁻⁴ m² and Young's modulus is 10¹⁰ Nm⁻² is subjected to a load of 250 N, then the elongation in the wire will be :
- (a) 6.25×10⁻⁶ m
- (b) 4×10⁻³ m
- (c) 4×10⁻⁴ m
- (d) 6.25×10⁻³ m
Answer and solution
Answer (b) 4×10⁻³ m
Solution
Physics sir · 4 steps
Rearrange Young's modulus for the stretch: elongation is force times length over area times Y.
- Start from Y. Young's modulus is stress over strain: Y = (F/A)/(Δ L/L). Rearranging gives the elongation directly.
Δ L = (F L)/(A Y)
- Put in numbers. The numerator is F L = 250 × 100 = 2.5 × 10⁴ N m. The denominator is A Y = 6.25 × 10⁻⁴ × 10¹⁰ = 6.25 × 10⁶ N.
- Divide. Δ L = (2.5 × 10⁴)/(6.25 × 10⁶) = 0.4 × 10⁻² = 4 × 10⁻³ m.
- So. The wire stretches by 4 × 10⁻³ m, which is 4 mm. The answer is (b).
The idea to keep. For any stretched wire, Δ L = FL/AY. Group the powers of ten separately from the plain numbers, and the arithmetic will not trip you up.
Answer from the NTA final answer key. Solution by PrepWiser.