A thin rod having a length of 1 m and area of cross-section 3×10−6 m2 is suspended vertically from one end. The rod is cooled from 210°C to 160°C. After cooling, a mass M is attached at the lower end of the rod such that the length of rod again becomes 1 m. Young's modulus and coefficient of linear expansion of the rod are 2×1011 Nm−2 and 2×10−5 K−1, respectively. The value of M is _____ kg. (Take g=10 m s−2)
Answer: 60
Solution
Cooling by ΔT=50 K would shorten the rod by Δℓ=ℓαΔT=1×2×10−5×50=10−3 m. The hanging mass must stretch it back by the same amount, and Young's modulus relates the stretching force to the extension through Y=AΔℓFℓ, so F=ℓYAΔℓ=12×1011×3×10−6×10−3=600 N. Since F=Mg, the required mass is M=10600=60 kg.
JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Integer answer
