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 N m⁻² and 2 × 10⁻⁵ K⁻¹, respectively. The value of M is __________ kg.
(Take g=10 m s⁻²)

Solution

Physics sir · 4 steps

The weight must stretch the rod back by exactly the amount cooling shrank it, so match the two strains.

1. Shrink on cooling. The temperature drops by Δ T = 210 - 160 = 50 K. The rod shortens by Δ L = LαΔ T = 1 × 2 × 10⁻⁵ × 50 = 1 × 10⁻³ m.
2. Stretch it back. The hanging weight Mg must stretch the rod by the same Δ L. Young's modulus links the stress to that strain, so the mechanical strain (Δ L)/L equals the thermal strain αΔ T.

   Mg/A = Y (Δ L)/L = Yα Δ T

3. Put in numbers. Mg = YαΔ T × A = 2 × 10¹¹ × 2 × 10⁻⁵ × 50 × 3 × 10⁻⁶ = 600 N.
4. So. M = 600/10 = 60 kg.

The common slip: 600. Stopping at the force gives 600, which is Mg in newtons. Divide by g = 10 m s⁻² to get the mass.

The idea to keep. When a load has to undo a thermal change in length, set the stress to YαΔ T. The length cancels, and the force is that stress times the area.

Answer from the NTA final answer key. Solution by PrepWiser.