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A 100 m long wire having cross-sectional area $6.25 \times 10^{- 4} m^{2}$ and Young’s modulus as $10^{10} N m^{- 2}$ is subjected to a load of 250N , then the elongation in the wire will be:

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$6.25 \times 10^{- 3} m$

$4 \times 10^{- 4} m$

$6.25 \times 10^{- 6} m$

$4 \times 10^{- 3} m$

answer is B.

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Detailed Solution

$e = \frac{F l}{Y A} = \frac{250 \times 100}{10^{10} \times 6.25 \times 10^{- 4}}$
$= 4 \times 10^{- 3} m$

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