On increasing the length by $0.5\, mm$ in a steel wire of length $2\, m$ and area of cross-section $2\,m{m^2}$, the force required is $[Y$ for steel$ = 2.2 \times {10^{11}}\,N/{m^2}]$
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(d) $F = \frac{{YAl}}{L} = \frac{{2.2 \times {{10}^{11}} \times 2 \times {{10}^{ - 6}} \times 5\, \times {{10}^{ - 4}}}}{2}$$ = 1.1 \times {10^2}N$
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