Four identical hollow cylindr..|Electrostatics|JEE-Main 2021

JEE MAIN 2021

31-08-2021 S1

Question

Four identical hollow cylindrical columns of mild steel support a big structure of mass $50 \times 10^3 \mathrm{~kg}$, 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 $Y=2.0 \times 10^{11} \mathrm{~Pa}, g=9.8 \mathrm{~m} / \mathrm{s}^2$ ]

Select the correct option:

$3.60 \times 10^{-8}$

$2.60 \times 10^{-7}$

$1.87 \times 10^{-3}$

$7.07 \times 10^{-4}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & \text { Force on each column }=\frac{\mathrm{mg}}{4}, \text { Strain }=\frac{\mathrm{mg}}{4 \mathrm{AY}} \\ & =\frac{50 \times 10^3 \times 9.8}{4 \times \pi(1-0.25) \times 2 \times 10^{11}} \\ & =2.6 \times 10^{-7}\end{aligned}$

Question Tags

JEE Main

Physics

Easy

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