Question

A stress of 1 kg mm^-2 is applied to a wire of which youngs modulus is 10^11 Nm^-2. Find the percentage increase in length.

Answer 1

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$stress = \frac{1kg}{ {(mm)}^{2} } \\ make \: it \: in \: {m}^{2} \\ stress = \frac{1kg}{ \frac{1}{10} \times \frac{1}{10} {cm}^{2} } \\ = {10}^{6} \frac{kg}{ {m}^{2} } \\ \gamma = {10}^{11} \\ \gamma = \frac{stress}{strain} \\ strain = \frac{stress}{ \gamma } \\strain = \frac{ {10}^{6} }{ {10}^{11} } \\ strain = \frac{change \: in \: length}{total \: length} = \frac{Δl}{l} \\ {10}^{ - 5} = \frac{Δl}{l} \\ Δl = {10}^{ - 5} l \\ \\ \%increase = \frac{change \: in \: length}{total \: length} \\ = \frac{Δl}{l} \times 100 \\ = \frac{ {10}^{ - 5} l}{l} \times 100 \\ {10}^{ - 3} \% = 0.001\%$
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Answer 2

Question : A stress of 1 N mm^-2 is applied to a wire of whose Young's modulus is 10^11 Nm^-2. Find the percentage increase in length.

Young's Modulus is the ratio of stress and strain.

$\gamma = \frac{ \textbf{stress}}{ \textbf{strain}}$

Given, Young modulus of the wire

$\gamma = {10}^{11} Nm ^{ - 2}$

Stress is the force acting upon Unit Area.

Given,

⇒ Stress = 1 N/mm²

⇒ Stress = 1 N / ( 10^-3 m) ²

⇒ Stress = 10^6 N/m².

Strain can be found by the earlier relation.

$\gamma = \frac{ \textbf{stress}}{ \textbf{strain}} \\ \\ \sf \: strain = \frac{1}{ \gamma } (stress)$

So,

$\sf \: strain = \frac{10 ^{6} }{ {10}^{11} } \\ \\ \sf \: strain = {10}^{ - 5}$

Now, Strain is the change of length upon original length.

$\implies \: strain = \frac{ \Delta \: l}{l}$

Since we know value of strain,

$\implies \: strain = \frac{ \Delta \: l}{l} = {10}^{ - 5}$

Now, Percent change of length is given by,

$\: \% \textbf{change \: in \: length} = \frac{ \Delta \: l}{l} \times 100 \\ \\ \frac{ \Delta \: l}{l} \times 100= {10}^{ - 5} \times 100 \\ \\ \frac{ \Delta \: l}{l} \times 100 = {10}^{ - 3} \%$

Therefore, The percent change in length is 0.001 %.