Question

A force F doubles the length of wire of cross-section ‘$a$’. What is the Young modulus of wire ?

Answer 1

Step 1: Given data

According to the question, when a force is applied to the wire, then the length of the wire doubles

So, a change in the length of the wire is given as $∆L=2L-L=L$.

Step 2: Formula used

$Y=\frac{F}{a}\times \frac{L}{∆L}$

$F$ is the force applied on the wire, $a$ is the area of the cross-section of the wire, $L$ is the initial length of the wire (before applying the force), $∆L$ is the change in the length of the wire on applying a force and ‘Y’ is the young modulus of elasticity of the wire.

Step 3: Calculating the Young modulus of wire

So, a change in the length of the wire is given as $∆L=2L-L=L$.
Now, substituting $∆L=L$ in the equation $Y=\frac{F}{a}\times \frac{L}{∆L}$ to determine the expression for the young modulus of elasticity of the wire.

$$\begin{gathered} Y=\frac{F}{a}\times \frac{L}{∆L} \\ Y=\frac{F}{a}\times \frac{L}{L} \\ Y=\frac{F}{a} \end{gathered}$$

Hence, the young modulus of the wire such that it elongates when a force $\left(F\right)$ is applied on it is $\frac{F}{a}$