JEE Main 2021 · 31 August · evening · Single correct · +4 / −1
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 × 10³ kg. The inner and outer radii of each column are 50 cm and 100 cm respectively. Assuming uniform local distribution, calculate the compression strain of each column. [Use Y = 2.0 × 10¹¹ Pa, g = 9.8 m/s²]
- (a) 3.60 × 10⁻⁸
- (b) 2.60 × 10⁻⁷
- (c) 1.87 × 10⁻³
- (d) 7.07 × 10⁻⁴
Answer and solution
Answer (b) 2.60 × 10⁻⁷
Solution
Physics sir · 4 steps
Split the weight over four columns, use the ring area of the hollow column, then divide stress by Y.
- Load per column. Total weight is 50 × 10³ × 9.8 = 4.9 × 10⁵ N. The load is shared equally, so each column carries F = 1.225 × 10⁵ N.
- Area of the ring. A hollow column bears load only on its ring of metal. With R = 1 m and r = 0.5 m, A = π(R² - r²) = π × 0.75 = 2.356 m².
strain = stress/Y = F/AY = (1.225 × 10⁵)/(2.356 × 2.0 × 10¹¹)
- Work it out. Stress is (1.225 × 10⁵)/2.356 = 5.2 × 10⁴ Pa. Dividing by Y gives a strain of (5.2 × 10⁴)/(2.0 × 10¹¹) = 2.6 × 10⁻⁷.
- So. The compression strain of each column is 2.60 × 10⁻⁷, which is option (b).
The idea to keep. For hollow columns, the load-bearing area is the ring π(R² - r²), not the full circle. Convert cm to m and share the load before finding stress.
Answer from the NTA final answer key. Solution by PrepWiser.