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Saturday, 31 October

JEE Main 2021 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 31 August 2021, Shift 2, Physics Q11

Question 11 of 90 in this shift, Physics question 11 of 30, Section A.

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50×10350\times10^3 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=2.0×1011Y = 2.0\times10^{11} Pa, g=9.8g = 9.8 m/s2^2]
  1. (1)2.60×10−72.60\times10^{-7}Official answer
  2. (2)3.60×10−83.60\times10^{-8}
  3. (3)1.87×10−31.87\times10^{-3}
  4. (4)7.07×10−47.07\times10^{-4}

Official answer

Option 1

NTA final key.

Idea tested
Hooke's Law