Question

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

Answer 1

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 $$Strain = \dfrac{{1{\text{ }}kg/m{m^2}}}{{{{10}^{11}}}} = \dfrac{1}{{{{10}^{ - 6}}}} \times \dfrac{1}{{{{10}^{11}}}}$$($$On conversion ofmm^2$to$m^2$$)$$ \Rightarrow $Strain$ = $$\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.