Question

A tensile force of $2 \times {10^3}N$ doubles the length of a rubber band of cross-sectional area \[2 \times {10^{ - 4}}{m^2}\]. What is the Young's modulus of elasticity of the rubber band?
\[
  A)\;4 \times {10^7}N{m^2} \\
  B)2 \times {10^7}N{m^2} \\
  C){10^7}N{m^2} \\
  D)\;0.5 \times {10^7}N{m^2} \\
 \]

Answer 1

Hint: Young’s modulus of elasticity for a body is the ratio of stress to strain in the body where stress is the force acting per unit area and strain is the change in length divided by original length.

Complete step by step answer:
Young’s modulus of elasticity for a body is the ratio of stress to strain in the body. So we have to find the stress and strain in the body. According to the question,
Tensile force(F)= $2 \times {10^3}N$
Cross-sectional area (A)= \[2 \times {10^{ - 4}}{m^2}\]
Initial length= L
Final length=2L

Now, to find strain we find the force acting per unit area. Mathematically,
$Stress = \dfrac{F}{A} = \dfrac{{2 \times {{10}^3}}}{{2 \times {{10}^{ - 4}}}} = {10^7}N/{m^2}$
Also, strain is the ratio of change in length to original length. Change in length is mathematically equal to final length minus initial length. So,
$Strain = \dfrac{{\Delta L}}{L} = \dfrac{{2L - L}}{L} = \dfrac{L}{L} = 1$

Now we can find the Young’s modulus of elasticity(Y) for the rubber band as
$Y = \dfrac{{Stress}}{{Strain}} = \dfrac{{{{10}^7}}}{1} = {10^7}N/{m^2}$
Therefore, the correct option is C) $10^7$ N/$m^2$

Note:We have to learn that the change in length is not equal to the final length of the object. One of the common mistakes committed here is taking the strain as 2L/L = 2 which is wrong. This must be taken care of while solving stress-strain questions.