Copper wire of length $3 m$ and area of cross-section $1 m m^{2}$ , passes through an arrangement of two frictionless pulleys, $P_{1}$ and $P_{2}$ . One end of the wire is rigidly clamped and a mass of $1 k g$ is hanged from the other end. If Young's modulus for copper is $10 \times 10^{10} N / m^{2}$ , the elongation in the wire is :

0.05 m m

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0.1 m m

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0.2 m m

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0.3 m m

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Solution

The correct option is D $0.3 m m$
$Δ l = \frac{F L}{A Y}$
$Δ l = \frac{10 \times 3}{10 \times 10^{10} \times 10^{- 6}} = 3 \times 10^{- 4} = 0.3 m m$

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