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A light rod of cross-sectional area A and Young’s Modulus Y is arranged as shown. The applied force F = 250N. If length of rod is 1m, the extension comes out to be $x \times 10^{- 6} m$ . Find x
Given that: $A = 6.25 \times 10^{- 4} m^{2}$ , $Y = 10^{10} N / m^{2}$

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answer is 40.

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Detailed Solution

$Δ l = \frac{F l}{A Y}$
$Δ l = \frac{250 \times 1}{6.25 \times 10^{- 4} \times 10^{10}} = 40 \times 10^{- 6} m$
X = 40.00

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