Calculate the percentage increase in length of a wire of diameter 1 mm stretched by a force of half kilo gram weight. Young's modulus of elasticity of wire is \({12}\times{10}^{{{11}}}{\left.{d}{y}\right.}{n}{e}/{c}{m}^{{{2}}}\)

Here, diameter of wire.

\({D}={1}{m}{m}={0.1}{c}{m}\),

Force applied , \({F}=\frac{{1}}{{2}}{k}{g}{w}{t}\)

\(=\frac{{1}}{{2}}\times{1000}\times{980}{\left.{d}{y}\right.}{n}{e}\),

\({Y}={12}\times{10}^{{{11}}}{\left.{d}{y}\right.}{n}{e}/{c}{m}^{{{2}}}\)

\({Y}=\frac{{{F}}}{{{\left(\pi{D}^{{{2}}}/{\left({4}\right)}\right)}}}{\quad\text{or}\quad}\frac{{\Delta{l}}}{{{l}}}=\frac{{{4}{F}}}{{\pi{D}^{{{2}}}{Y}}}\)

\(\therefore\%\) increase in length =\(\frac{{\Delta{l}}}{{{l}}}\times{100}=\frac{{{4}{F}\times{100}}}{{\pi{D}^{{{2}}}{Y}}}\)

\(={4}\times\frac{{1}}{{2}}\times\frac{{{1000}\times{980}\times{100}}}{{{\left(\frac{{22}}{{7}}\right)}\times{\left({0.1}\right)}^{{{2}}}\times{12}\times{10}^{{{11}}}}}\)

\(={5.2}\times{10}^{{-{3}}}={0.0052}\%\).

\({D}={1}{m}{m}={0.1}{c}{m}\),

Force applied , \({F}=\frac{{1}}{{2}}{k}{g}{w}{t}\)

\(=\frac{{1}}{{2}}\times{1000}\times{980}{\left.{d}{y}\right.}{n}{e}\),

\({Y}={12}\times{10}^{{{11}}}{\left.{d}{y}\right.}{n}{e}/{c}{m}^{{{2}}}\)

\({Y}=\frac{{{F}}}{{{\left(\pi{D}^{{{2}}}/{\left({4}\right)}\right)}}}{\quad\text{or}\quad}\frac{{\Delta{l}}}{{{l}}}=\frac{{{4}{F}}}{{\pi{D}^{{{2}}}{Y}}}\)

\(\therefore\%\) increase in length =\(\frac{{\Delta{l}}}{{{l}}}\times{100}=\frac{{{4}{F}\times{100}}}{{\pi{D}^{{{2}}}{Y}}}\)

\(={4}\times\frac{{1}}{{2}}\times\frac{{{1000}\times{980}\times{100}}}{{{\left(\frac{{22}}{{7}}\right)}\times{\left({0.1}\right)}^{{{2}}}\times{12}\times{10}^{{{11}}}}}\)

\(={5.2}\times{10}^{{-{3}}}={0.0052}\%\).

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