Question

If the work done by stretching a wire by $1mm\text{ is 2J}$, the work necessary for stretching another wire of the same material but with half the radius of cross section and half the length by $1mm\text{ is}$
$\begin{align}
& \text{A}\text{. 1J} \\
& \text{B}\text{. 4J} \\
& \text{C}\text{. 8J} \\
& \text{D}\text{.16J} \\
\end{align}$

Answer 1

Hint: When wire gets elongated, potential energy is stored in it. Here the material of both the wires are the same therefore Young’s modulus of both the wires are the same. Use the formula of total amount of work done in stretching wire by a length. Take the ratio and find the value of work done.

Complete step-by-step answer:
In the question it is given that a wire stores 2J of energy to stretch/ elongate $1mm$ of wire. Now the work necessary for stretching another wire of the same material but with half the radius of cross section and half the length by $1mm$.
Basically work done is the product of force and displacement.
When wire gets stretched a potential energy is stored in it.
We know that work done is nothing but energy required to do work.
Work done is given by.
Mathematically,
$\text{work done(W) =}\dfrac{AY}{2L}{{l}^{2}}=\dfrac{\pi {{r}^{2}}Y{{l}^{2}}}{2L}$
Where,
Y= young’s modulus
A= area of cross section
R= radius of cross section
L= length of wire
$l$= elongation in wire
Other quantities are constant
As we can from the above formula,
$\text{work done(W)}\alpha \dfrac{{{r}^{2}}}{L}$
Let, ${{W}_{1}}\And {{W}_{2}}$be the work done by 1st wire and another wire respectively.
Then
${{\text{W}}_{1}}=\dfrac{\pi {{r}_{1}}^{2}{{Y}_{1}}{{l}_{1}}^{2}}{2{{L}_{1}}}---(1)$
And
${{\text{W}}_{2}}=\dfrac{\pi {{r}_{2}}^{2}{{Y}_{2}}{{l}_{2}}^{2}}{2{{L}_{2}}}---(2)$
In the question it is given that material of both the questions are same and stretched of both the wires are also same.
Therefore, ${{Y}_{1}}={{Y}_{2}}\And {{l}_{1}}={{l}_{2}}----(3)$
Now take the ratio of both the work done,
So from (1), (2) and (3)
$\begin{align}
&\dfrac{{{\text{W}}_{1}}}{{{\text{W}}_{2}}}=\dfrac{r_{1}^{2}}{{{L}_{1}}}\times \Rightarrow\dfrac{\dfrac{1}{2}{{L}_{1}}}{{{\left( \dfrac{1}{2}{{r}_{1}} \right)}^{2}}} \\
\Rightarrow & \dfrac{{{\text{W}}_{1}}}{{{\text{W}}_{2}}}=2 \\
\Rightarrow & {{\text{W}}_{2}}=\dfrac{{{\text{W}}_{1}}}{2}=1J \\
\end{align}$
The work necessary for stretching another wire of the same material but with half the radius of cross section an half the length by $1mm\text{ is}$ $1 J$
So the answer is option (A)

Note: You can also define work as $\text{work done=}\dfrac{1}{2}\times \text{load}\times \text{extension}$. This is the expression for work done in stretching a wire. Thus total work done in stretching the wire gets stored in the form of its elastic potential energy. It is sometimes called strain energy. Which is given as, $\text{strain energy=}\dfrac{1}{2}\times \text{load}\times \text{extension=}\dfrac{1}{2}\times \text{F}\times l$.