A vertical steel post of diameter $25 c m$ and length $2.5 m$ supports a weight of $8000 k g$ . Find the change in length produced.
(Given $Y = 2 \times 10^{11} P a$ )

2.1 cm

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0.21 cm

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0.21 mm

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0.021 mm

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Solution

The correct option is D 0.021 mm
$Δ l = \frac{F l}{A Y} = \frac{4 \times 8000 \times 10 \times 2.5}{π (.25)^{2} \times 2 \times 10^{11}} = \frac{4 \times 16 \times 10^{- 6}}{π} = 0.021 m m$

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