Question

A stress of $2\,kg/m{m^2}$ is applied on a wire. If $Y\, = \,{10^{12\,}}\,dyne/c{m^2}$ , then the percentage increase in its length will be
 A) $0.196\% $
 B) $19.6\% $
 C) $1.96\% $
 D) $0.0196\% $

Answer 1

Hint:Young’s Modulus is the ratio of longitudinal stress to longitudinal strain and within elastic limit, it obeys Hooke’s Law. We will be using Hooke’s Law in this question to get the answer of percentage increase.

Step-by-step explanation:Initially, we shall be converting the given data in S.I. units.
We know that, $1\,mm\, = \,{10^{ - 3}}\,m$
Squaring on both side ${(1\,mm)^{2\,}}\, = \,{10^{ - 6}}\,{m^2}$
Also expressing dyne into Newton, $1\,dyne\, = \,{10^{ - 5}}\,N$
Now, given Stress = $2\,kg/m{m^2}$ , converting it to Pascal (Pa)
We can write ${\text{Stress}}\, = \,\dfrac{{2\, \times \,9.8}}{{{{10}^{ - 6}}}}\,Pa$
$=19.6\, \times \,{10^{ - 5}}\,Pa$
As long as wire is within elastic limit, Hooke’s law is obeyed. So,
${\text{Young's Modulus = }}\dfrac{{{\text{Longitudinal stress}}}}{{{\text{Longitudinal strain}}}}$
Thus, ${\text{Strain = }}\dfrac{{{\text{stress}}}}{{{\text{Young's Modulus}}}}$ -[equation 1]
Now we also know that, $1\,cm\, = \,{10^{ - 2}}\,m$
So ${(1\,cm)^2}\, = \,{10^{ - 4}}\,{m^2}$
Young’s Modulus (Y) is given as $Y\, = \,{10^{12\,}}\,dyne/c{m^2}$
Converting young’s modulus into S.I unit, $Y\, = \,({10^{12\,}}\, \times \,{10^{ - 5}}\, \times \,{10^4}\,)N/{m^2}$
So, $Y\, = \,{10^{11}}N/{m^2}$
Putting values of Young’s modulus and stress in equation [1], we get
${\text{Strain = }}\dfrac{{19.6\, \times \,{{10}^6}}}{{{{10}^{11}}}}$
${\text{Strain = 19}}{\text{.6}}\, \times \,{\text{1}}{{\text{0}}^{ - 5}}$

Also, we know that strain is defined as, ${\text{Strain = }}\dfrac{{{\text{change in length (}}\vartriangle {\text{l)}}}}{{{\text{original length (L)}}}}$

Therefore, percentage change in length = $(19.6\, \times \,{10^{ - 5}}\, \times \,100)\,\% $
Percentage change in length = $(19.6\, \times \,{10^{ - 3}}\,)\,\% $
So finally, it can be written as, percentage change in length = $0.0196\,\% $

Hence option D is the correct answer.

Additional information: Stress is the force acting on the body per unit area. Hence, it has units $N/{m^2}$ .
Strain is the ratio of change in length to original length, i.e. two similar quantities, hence it has no units. There are 3 moduli which give relation between stress and strains. Namely Young’s Modulus, Bulk Modulus and Modulus of Rigidity.

Note:Attention must be paid towards proper conversion of given units into S.I. units. Within elastic limit, Hooke’s law is obeyed which implies stress is directly proportional to strength. Beyond this limit elasticity is reduced and permanent deformation is seen.
Within elastic limit stress is directly proportional to strain, that is ${\text{stress}}\, \propto \,{\text{strain}}$ .