Question

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

Open in App
Solution

## Step 1: Given dataAccording to the question, when a force is applied to the wire, then the length of the wire doublesSo, a change in the length of the wire is given as $∆L=2L-L=L$. Step 2: Formula used$Y=\frac{F}{a}×\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 wireSo, a change in the length of the wire is given as $∆L=2L-L=L$. Now, substituting $∆L=L$ in the equation $Y=\frac{F}{a}×\frac{L}{∆L}$ to determine the expression for the young modulus of elasticity of the wire.$Y=\frac{F}{a}×\frac{L}{∆L}\phantom{\rule{0ex}{0ex}}Y=\frac{F}{a}×\frac{L}{L}\phantom{\rule{0ex}{0ex}}Y=\frac{F}{a}$Hence, the young modulus of the wire such that it elongates when a force $\left(F\right)$ is applied on it is $\frac{F}{a}$

Suggest Corrections
5