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Q.

A steel wire of length 1m has cross sectional area $1 c m^{2} .$ If Young’s modulus of steel is $10^{11} \frac{N}{m^{2}},$ then force required to increase the length of wire by 1mm will be

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a

$10^{11} N$

b

$10^{2} N$

c

$10^{4} N$

d

$10^{7} N$

answer is C.

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Detailed Solution

L=1m, A= $10^{- 4} m^{2}, Δ L = 1 \times 10^{- 3} m$

$Δ L = \frac{F L}{A Y}$

$F = A Y \frac{Δ L}{L} = 10^{- 4} \times 10^{11} \times \frac{10^{- 3}}{1} = 10^{4} N .$

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