Question

A tensile force of $2 \times {10^5}dynes$ doubles the length of an elastic cord whose area of cross section is 2 sq. cm. Young's modulus of the material of the card is
(A) $1 \times {10^5}dynes/c{m^2}$
(B) $2 \times {10^5}dynes/c{m^2}$
(C) $0.5 \times {10^5}dynes/c{m^2}$
(D) $4 \times {10^5}dynes/c{m^2}$

Answer 1

Hint: Young’s modulus is the ratio of stress over strain. Strain here will be a linear strain since only the length of the rope is increased without changing its area.

Complete step by step answer:
In this question we must find out the Young’s modulus of the rope which has been put under a force, leading to the lengthening of it to twice its original length. Before that let us understand what Young’s modulus is.
Young's modulus is the in-built property of a substance which determines its elasticity. It is given as
$Y = \dfrac{{stress}}{{strain}}$
Stress is the force acting on the body per unit cross sectional area of the body.
$stress = \dfrac{{Force}}{{Area}}$
Strain is the extension of the body per unit original length of the body when it is subjected to that force. Strain can be measured in different ways depending upon the direction in which the force is being applied. If the force is applied linearly then it is called linear strain, which means only the length is increased.
$strain = \dfrac{{\Delta l}}{{{l_{original}}}}$
If the volume is increased then it is called bulk strain, and so on.
$strain = \dfrac{{\Delta V}}{{{V_{original}}}}$
Now given that the rope is being subjected to the force and its area remains constant only length is doubled. This means that it will have a linear strain and that will be calculated
$strain = \dfrac{{\Delta l}}{{{l_{original}}}}$
${l_{initial = }}{l_{original}},{l_{final}} = 2{l_{original}}$
$\therefore \Delta l = {l_{final}} - {l_{initial}} = {l_{original}}$
$\therefore strain = \dfrac{{{l_{original}}}}{{{l_{original}}}} = 1$
The stress will be
$stress = \dfrac{F}{A}$
$F = 2 \times {10^5}dynes,A = 2c{m^2}$
$\therefore stress = \dfrac{{2 \times {{10}^5}}}{2} = 1 \times {10^5}dynes/c{m^2}$
And therefore, Young's modulus will be
$Y = \dfrac{{stress}}{{strain}} = \dfrac{{1 \times {{10}^5}}}{1} = 1 \times {10^5}dynes/c{m^2}$
Hence, the correct option is (A).

Note:The units of Young's modulus and the stress are the same because strain is a dimensionless quantity and therefore it has no units. Since strain has no units and Young's modulus is the ratio of stress and strain, therefore Young's modulus has the same unit as stress. Stress is defined as force per unit area and therefore it has a similar definition to that of pressure, but what you have to keep in mind is pressure has an SI unit of Pascal but stress has an SI unit of Newton per meter square.