Verified Solution • 2025–26 Session moderate

QJEEM21Ph47ea: JEE Main 2021 Physics - Class 11 JEEM21Ph47ea Solution: Four identical hollow cylindrical columns of mild steel support …

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 ×\times 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 ×\times 1011 Pa, g = 9.8 m/s2]

Multiple Choice Options

  • A 3.60 ×\times 10 −- 8
  • B 2.60 ×\times 10 −- 7
  • C 1.87 ×\times 10 −- 3
  • D 7.07 ×\times 10 −- 4

Step-by-Step Verified Solution

Force on each column = mg4{{mg} \over 4}

Strain = mg4AY{{mg} \over {4AY}}

=50×103×9.84×π(1−0.25)×2×1011= {{50 \times {{10}^3} \times 9.8} \over {4 \times \pi (1 - 0.25) \times 2 \times {{10}^{11}}}}

= 2.6 ×\times 10 −- 7

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