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One end of a wire 2m long and 0.2 cm^2 in cross section is fixed in a ceilign and a load of 4.8 kg is attached to the free end. Find the extension of the wire. Young modulus of steel $$={2.0}\times{10}^{{11}}{N}{m}^{{-{{2}}}}$$. Take $${g}={10}{m}{s}^{{-{{2}}}}$$.
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## QuestionPhysicsClass 11

One end of a wire 2m long and 0.2 cm^2 in cross section is fixed in a ceilign and a load of 4.8 kg is attached to the free end. Find the extension of the wire. Young modulus of steel $$={2.0}\times{10}^{{11}}{N}{m}^{{-{{2}}}}$$. Take $${g}={10}{m}{s}^{{-{{2}}}}$$.

We have
$${y}={s}{t}{r}{e}{s}\frac{{s}}{{s}}{t}{r}{a}\in=\frac{{\frac{{T}}{{A}}}}{{\frac{{l}}{{L}}}}$$
with symbols having their usual meanings. The extension is
$${l}=\frac{{{T}{L}}}{{{A}{Y}}}$$
As the load is in equilibrium after the extension, the tension in the wire is equal to the weight of the load
$$={4.8}{k}{g}\times{10}{m}{s}^{{-{{2}}}}={48}{N}$$
Thus, $${l}=\frac{{{\left({48}{N}\right)}{\left({2}{m}\right)}}}{{{\left({0.2}\times{10}^{{-{{4}}}}{m}^{{2}}\right)}\times{\left({2.0}\times{10}^{{11}}{N}{m}^{{-{{2}}}}\right)}}}$$
$$={2.4}\times{10}^{{-{{5}}}}{m}$$
4.6
4.6
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