The area of cross section of a steel wire $(Y = 2.0 \times {10^{11}}N/{m^2})$ is $0.1\;c{m^2}$. The force required to double its length will be
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(d) When the length of wire is doubled then $l = L$ and
strain $= 1$$\Rightarrow\, Y$ $=$ strain $=\frac{F}{A}$
Force $= Y \times A $ $= 2 \times {10^{11}} \times 0.1 \times {10^{ - 4}}$$ = 2 \times {10^6}N$
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