Question

Work done in stretching a wire 1 mm is 2J. What amount of work will be done for elongating another wire of the same material, with half the length and double the radius of the cross section, by 1mm?
A) $2J$
B) $4J$
C) $8J$
D) $16J$

Answer 1

Hint: When a material is subjected to stress and strain, elongation may take place in the material which depends on the Young’s modulus of that material. Young’s modulus is defined as the ratio of stress to strain.

Formula used:
Stress is defined as the force applied per unit area and given as follows:
$Stress = \dfrac{F}{A}{\text{ }}...{\text{(i)}}$
where F is the force applied which stretches the material and A represents the area on which force is being applied.
Strain is defined as change in length per unit original length and is given as follows:
$Strain = \dfrac{{\Delta l}}{l}{\text{ }}...{\text{(ii)}}$
where $\Delta l$ is the change in length due to the applied stress while l signifies the original length of the material under stress.
Young’s modulus is given as the ratio of stress applied and the strain produced in a body. Using equation (i) and (ii), we get
$\begin{gathered}
  Y = \dfrac{{Stress}}{{Strain}} = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta l}}{l}}} = \dfrac{{Fl}}{{A\Delta l}} \\
   \Rightarrow F = \dfrac{{YA\Delta l}}{l}{\text{ }}...{\text{(iii)}} \\
\end{gathered} $

Complete step by step solution:
We can use the equation (iii) to compare the forces stretching the given wire.
In first case, we have following parameters for the wire:
$\Delta {l_1} = 1mm$
We can write
${F_1} = \dfrac{{Y\pi {r_1}^2\Delta {l_1}}}{{{l_1}}}{\text{ }}...{\text{(iv)}}$
In the second case, the new parameters are given as
$\begin{gathered}
  \Delta {l_2} = 1mm \\
  {l_2} = \dfrac{{{l_1}}}{2} \\
  {r_2} = 2{r_1} \\
\end{gathered} $
Therefore we can write
${F_2} = \dfrac{{Y\pi {r_2}^2\Delta {l_2}}}{{{l_2}}} = \dfrac{{Y\pi {{\left( {2{r_1}} \right)}^2}\Delta {l_2}}}{{\left( {\dfrac{{{l_1}}}{2}} \right)}} = 8\dfrac{{Y\pi {r_1}^2\Delta {l_2}}}{{{l_1}}}{\text{ }}...{\text{(v)}}$
Dividing equation (iii) by equation (iv), we get
$\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{1}{8}$ (Here we know that elongation produced is same in both cases)
Work done by stretching the wire by the same length of 1 mm can be compared as follows:
$\begin{gathered}
  \dfrac{{{W_1}}}{{{W_2}}} = \dfrac{{{F_1}\Delta {l_1}}}{{{F_2}\Delta {l_2}}} = \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{1}{8} \\
   \Rightarrow {W_2} = 8{W_1} = 8 \times 2 = 16J \\
\end{gathered} $

Hence, the correct answer is option D.

Note: Elongation is inversely proportional to the Young’s modulus. A stiffer material will have a larger value of Young’s modulus and will resist change in shape when stress is applied to it. Young’s modulus describes deformation along an axis.