JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Elasticity & SHM

Four identical hollow cylindrical columns of mild steel support a big structure of mass

50 \times 1 0^{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 \times 1 0^{11} Pa, g = 9.8 m / s^{2}

]

(A) $3.60 \times 1 0^{- 8}$
(B) $2.60 \times 1 0^{- 7}$ ✓ Correct
(C) $1.87 \times 1 0^{- 3}$
(D) $7.07 \times 1 0^{- 4}$

Answer: (B)

Solution

The weight is shared equally by the four columns, so each carries

F = \frac{50 \times 1 0 ^{3} \times 9.8}{4} = 1.225 \times 1 0^{5} N .

The cross-sectional area of one hollow column is

A = π (r_{2}^{2} - r_{1}^{2}) = π (1^{2} - 0. 5^{2}) = 0.75 π = 2.356 m^{2} .

The compressive strain is

ε = \frac{F}{A Y} = \frac{1.225 \times 1 0 ^{5}}{( 2.356 ) ( 2.0 \times 1 0 ^{11} )} = 2.60 \times 1 0^{- 7} .

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Elasticity & SHM

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