A copper wire of length $4.0m$ and area of cross-section $1.2\,c{m^2}$ is stretched with a force of $4.8 \times {10^3}$ $N.$ If Young’s modulus for copper is $1.2 \times {10^{11}}\,N/{m^2},$ the increase in the length of the wire will be
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(a) $l = \frac{{FL}}{{AY}} = \frac{{4.8 \times {{10}^3} \times 4}}{{1.2 \times {{10}^{ - 4}} \times 1.2 \times {{10}^{11}}}} = 1.33\;mm$
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