Question

If the work done by streching a wire by 1mm is 2J, the work necessary for stretching another wire of the same material but with the radius of cross section twice and half length by 1mm is:
a) $\dfrac{1}{4}$J
b)4 J
c)8 J
d)16 J

Answer 1

Hint: When a material is stretched such that there is a change in length the elasticity of the material is given by Young’s modulus. For a given material the Young’s modulus remains constant. Comparing the Young’s modulus for a given material for two different conditions will enable us to determine the work done for one of the conditions such that the other is known.
Formula used:
$Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}$

Complete step-by-step solution:
If a material with Young’s’ modulus ‘Y’ is stretched by a force ‘F’ of length ‘l’ and area of cross section ‘A’ such that there is a change in the length of the material by $\Delta l$ , then the Young’s modulus for the material is given by,
$Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}=\dfrac{Fl}{A\Delta l}$
In first case the wire has length ‘l’ and radius R. When the ${{F}_{1}}$ is applied on the material it gets stretched by 1mm. hence the Young’s modulus for the material is given by,
$Y=\dfrac{{{F}_{1}}l}{\pi {{R}^{2}}(1mm)}.....(1)$
In the second case a wire with the same material is taken but with length ‘l/2’ and radius r=2R and stretched by a force say ${{F}_{2}}$ . Therefore the Young’s modulus for this wire can be written as,
$\begin{align}
  & Y=\dfrac{{{F}_{2}}l}{\pi {{r}^{2}}2\Delta l} \\
 & Y=\dfrac{{{F}_{2}}l}{\pi {{(2R)}^{2}}2mm} \\
 & \therefore Y=\dfrac{1}{8}\left( \dfrac{{{F}_{2}}l}{\pi {{R}^{2}}1mm} \right).....(2) \\
\end{align}$
In both the cases the two wires are made up of same material. Therefore the Young’s modulus in both the cases is the same. Hence equating equation 1 and 2 we obtain,
$\begin{align}
  & \dfrac{{{F}_{1}}l}{\pi {{R}^{2}}1mm}=\dfrac{1}{8}\left( \dfrac{{{F}_{2}}l}{\pi {{R}^{2}}1mm} \right) \\
 & \therefore \dfrac{{{F}_{1}}}{{{F}_{2}}}=\dfrac{1}{8}....(3) \\
\end{align}$
Since work done is the product of force times the extension in the length, From equation 3
$\begin{align}
  & \dfrac{{{F}_{1}}\times \Delta l}{{{F}_{2}}\times \Delta l}=\dfrac{1}{8} \\
 & \Rightarrow \dfrac{{{W}_{1}}}{{{W}_{2}}}=\dfrac{1}{8},\ \because {{W}_{1}}=2J \\
 & \Rightarrow \dfrac{2J}{{{W}_{2}}}=\dfrac{1}{8} \\
 & \therefore {{W}_{2}}=16J \\
\end{align}$
Hence the correct answer of the above question is option d.

Note: It is to be noted that the above formula for Young’s modulus for two different materials only holds valid when both of the wires have the same distribution of matter. Even if the material is the same the distribution of the atoms is also very important. Extension along the length is best represented by Young’s modulus.