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A wire of length 10 m and of cross -section $1 \times 10^{- 4}$ sq.m is fixed at one end and a load of 1200 kg is attached at the other end. Then the extension produced is $(Y = 9.8 \times 10^{10} N / m^{2})$

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$12 \times 10^{- 3} m$

$12 \times 10^{- 5} m$

$12 \times 10^{- 4} m$

$12 \times 10^{- 2} m$

answer is A.

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Detailed Solution

$e = \frac{F ℓ}{A y}$ $= \frac{1200 \times 9.8 \times 10}{10^{- 4} \times 9.8 \times 10^{10}} = 12 \times 10^{- 3} m$

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