Question

A brass wire of diameter $1mm$ and of length $2m$ is stretched by applying a force of $20N$. If the increase in length is $0.51mm$. Then the Young's modulus of the wire is

• A. $8.848\times {10}^{10}N/m^2$
• B. $7.984\times {10}^{10}N/m^2$
• C. $6.984\times {10}^{10}N/m^2$
• D. $9.984\times {10}^{10}N/m^2$

Answer 1

The correct option is D $9.984\times {10}^{10}N/m^2$

Given,

$F=20N,l=2m,d=1mm,\Delta l=0.51mm$

We know,

$Y=\frac{Fl}{A\Delta l}$

$Y=\frac{20\times 2}{\pi 5^2\times {10}^{-8}\times 0.51\times {10}^{-3}}$

$Y=\frac{40}{25\pi \times 0.51}\times {10}^{1}1$

$Y=0.9984\times {10}^{1}1=9.984\times {10}^{10}$

Option $\text{D}$ is the correct answer