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The area of cross-section of a wire is $10^{- 5} m^{2}$ , when its length is increased by $0.1 %$ a tension of $1000 N$ is produced. The Young's modulus of the wire will be ----

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$10^{12} N m^{- 2}$

$10^{9} N m^{- 2}$

$10^{10} N m^{- 2}$

$10^{11} N m^{- 2}$

answer is B.

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Detailed Solution

$\frac{Δ ℓ}{ℓ} = \frac{0.1}{100}$

$y = \frac{F ℓ}{A ∆ l} = \frac{F}{A (\frac{Δ ℓ}{ℓ})} = \frac{10^{3} \times 10^{5}}{10^{- 3}} = 10^{11}$

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