JEE Main 31-Aug-2021 Shift 2 Solved Paper

Let inner and outer radii of hollow cylindrical column are r and R, respectively.
Given that, r = 50 cm, R = 100 cm
Mass supported on four columns, M=50×103‌kg
Mass supported on each column, m=

M

4

⇒m=

50×103

4

=12.5×103‌kg
Now, weight, w=mg=12.5×9.8×103N=1225×105N
Area of cross-section of each column
A=π(R2−r2)
=3.14{(100)2−(50)2}×10−4m2=2.35m2
Young's modulus, Y=2.0×1011‌Pa
By using Hooke’s law,
Stress = Y × Strain
∴ Compressive strain =

Stress

Y

=

W

AY

Substituting the values, we get
Compressive strain =

1.225×105

2.35×2.0×1011

=2.60×10−7