Answer
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Hint: Stress is defined as the force experienced by the object which causes a change in the object. The strain is defined as the change in the shape of an object when stress is applied. Stress is measurable and has a unit. The strain is a dimensionless quantity and has no unit.
Complete step by step solution:
Given data:
Stress $ = $ \[1{\text{ }}kg/m{m^2}\]
Young’s Modulus, $Y = 10^{11} N/m^2$
Percentage increase in length $ = $\[?\]
We know that
$Y$ $ = $ $\dfrac{{Stress}}{{Strain}}$
Substituting the values of Stress and Young’s Modulus, we get
$ \Rightarrow $ $10^{11} = \dfrac{1}{{strain}}$
$\therefore $ $Strain = \dfrac{{1{\text{ }}kg/m{m^2}}}{{{{10}^{11}}}} = \dfrac{1}{{{{10}^{ - 6}}}} \times \dfrac{1}{{{{10}^{11}}}}$ \[(\]On conversion of $mm^2$ to $m^2$ \[)\]
$ \Rightarrow $ Strain $ = $ $\dfrac{1}{{{{10}^5}}}$
$ \Rightarrow $$\dfrac{{\Delta l}}{l}$$ = $ $\dfrac{1}{{{{10}^5}}}$
$\therefore $ Percentage increase in length
$ \Rightarrow $$\dfrac{{\Delta l}}{l}$\[ \times \]100 $ = $ $\dfrac{1}{{{{10}^5}}}$\[ \times \]100
$ \Rightarrow $ Percentage increase in the length of wire$ = $0.001
Hence the correct option for the problem is B.
Note: 1) Stress can also be defined as the restoring force per unit area of the material.
2) Strain can also be considered as a fractional change in either length (when tensile stress is considered) or volume (when bulk stress is considered).
3) Young’s modulus is also termed as the modulus of elasticity.
Complete step by step solution:
Given data:
Stress $ = $ \[1{\text{ }}kg/m{m^2}\]
Young’s Modulus, $Y = 10^{11} N/m^2$
Percentage increase in length $ = $\[?\]
We know that
$Y$ $ = $ $\dfrac{{Stress}}{{Strain}}$
Substituting the values of Stress and Young’s Modulus, we get
$ \Rightarrow $ $10^{11} = \dfrac{1}{{strain}}$
$\therefore $ $Strain = \dfrac{{1{\text{ }}kg/m{m^2}}}{{{{10}^{11}}}} = \dfrac{1}{{{{10}^{ - 6}}}} \times \dfrac{1}{{{{10}^{11}}}}$ \[(\]On conversion of $mm^2$ to $m^2$ \[)\]
$ \Rightarrow $ Strain $ = $ $\dfrac{1}{{{{10}^5}}}$
$ \Rightarrow $$\dfrac{{\Delta l}}{l}$$ = $ $\dfrac{1}{{{{10}^5}}}$
$\therefore $ Percentage increase in length
$ \Rightarrow $$\dfrac{{\Delta l}}{l}$\[ \times \]100 $ = $ $\dfrac{1}{{{{10}^5}}}$\[ \times \]100
$ \Rightarrow $ Percentage increase in the length of wire$ = $0.001
Hence the correct option for the problem is B.
Note: 1) Stress can also be defined as the restoring force per unit area of the material.
2) Strain can also be considered as a fractional change in either length (when tensile stress is considered) or volume (when bulk stress is considered).
3) Young’s modulus is also termed as the modulus of elasticity.
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