Question

If the work done by stretching a wire by $1\text{mm}$ is $2\text{J}$ , the work necessary for stretching another wire of the same material and area of cross-section and stretching length is halved is :
(A) 1/4J
(B) 4J
(C) 8J
(D) 16J

Answer 1

We know that the formula of work done in stretching the wire is given by : $w=\dfrac{1}{2}\text{ }\dfrac{\text{YAl}_{{}}^{2}}{L}$. We are provided with the values of W and l for the first wire.
For the second wire , the value of A and l is halved.
On comparing the equations for the two cases , we can get the required value of work .

Formula Used:
$w=1/2\times {\text{stretching force }}\times {\text{change in length }}\to { (1)} \\\ {\text{we know that Y=}} \dfrac{F}{A}\centerdot \dfrac{L}{l}\text{ or F=}\dfrac{YAl}{L} \to (2)\\\$
l= Change in length
L=Original Length
Putting values of (2) in (1) we get
$w=\dfrac{1}{2}\times \text{ }\dfrac{\text{YAl}_{{}}^{2}}{L}\text{ }\times \text{ l} \\\$

Complete Step by step solution:
Given:- Work done in stretching a wire is
$w_1 = 2J \\\ l=1mm \\\$
Now, when are of cross section and stretching length is halved, we get: -
${{w}_{2=}}\text{ }\dfrac{1}{2}\times Y\left( \dfrac{A}{2} \right)\times \dfrac{1}{L}\times \dfrac{1mm}{2}\times \dfrac{1mm}{2}$
We get,
$w_2 =\dfrac{1}{8}\times w_1\\\ w_2 =\dfrac{2}{8}=\dfrac{1}{4}J\\\ w_2 =\dfrac{1}{4}J\\\$
Correct answer is (A).

Additional information:
In stretching the wire, work is done against internal restoring forces. This work is stored as elastic potential energy or strain energy.
In force $\text{F}$ acts along the length $\text{L}$ of the wire of cross section $\text{A}$ and stretches it by $x$, then:
Young’s modulus $\gamma =\dfrac{\text{Stress}}{\text{Strain}}=\dfrac{\dfrac{\text{F}}{\text{A}}}{\dfrac{x}{\text{L}}}$
So work done for additional small increase $dx$ in length,
$dw=\text{F}dx=\dfrac{\gamma \text{A}}{\text{L}}x.dx$
Hence, total work done in increasing the length $l$,
$w=\int{dw=\int{\text{F}\text{.}dx=\dfrac{1}{2}\dfrac{\gamma \text{A}}{\text{L}}{{l}^{2}}}}$
This work done is stored in the wire.
Hence, energy stored in the wire
$\text{U = }\dfrac{1}{2}\dfrac{\gamma \text{A}{{l}^{2}}}{\text{L}}=\dfrac{1}{2}\text{F}l$ $\left( \text{As F =}\dfrac{\gamma \text{A}l}{\text{L}} \right)$

Note: Young modulus is a mechanical property that measures the tensile stiffness of a solid material.
$Y=\dfrac{F}{A}\centerdot \dfrac{L}{l}$
Where F= stretching force
L=original length of body
A= area of body
l=change in length of body after stretching.