A weight of $200 \,kg$ is suspended by vertical wire of length $600.5\, cm$. The area of cross-section of wire is $1\,m{m^2}$. When the load is removed, the wire contracts by $0.5 \,cm$. The Young's modulus of the material of wire will be
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(a) $F = 2000N,\;L = 6m,\;l = 0.5\;cm,A = {10^{ - 6}}{m^2}$
$Y = \frac{{FL}}{{Al}} = \frac{{2000 \times 6}}{{{{10}^{ - 6}} \times 0.5 \times {{10}^{ - 2}}}} = 2.35 \times {10^{12}}\;N/{m^2}$
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