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Longitude stress of \[1{\text{ }}kg/m{m^2}\] is applied on a wire. The percentage increase in length is ($Y= 10^{11} N/m^2$)
A) 0.002
B) 0.001
C) 0.003
D) 0.01

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Last updated date: 03rd Mar 2024
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IVSAT 2024
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.