Physics · Mechanical Properties of Matter
JEE Main 2026 — 4 April, Morning Shift — Question 8
A string A of length 0.314 m, Young's modulus 2×10¹⁰ N/m² is connected to another string B of length and Young's modulus both twice of those of A. This series combination is suspended from a rigid support and its free end is fixed to a load of mass 0.8 kg. The net change in length of the combination is ______ mm. (radius of both strings = 0.2 mm, g = 10 m/s², neglect string masses)
- Option A:
3
- Option B:Correct
2
- Option C:
1.9
- Option D:
1
Answer: B
Step-by-step solution
ΔL = FL/(YA). For A: ΔL_A = (8×0.314)/(2e10×π×(0.2e-3)²) ≈ 1 mm. For B: L_B=0.628, Y_B=4e10, ΔL_B = (8×0.628)/(4e10×π×4e-8) ≈ 1 mm. Total = 2 mm
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2026
- Subject
- Physics
- Chapter
- Mechanical Properties of Matter
- Topic
- Stress,Strain and Modulus of Elasticity