Question

a wire of length 3m extends by 3mm when a force of 2N is applied on it.Calculate stress produced in it,i f Y=2×10 11N/m2​

Answer 1

I have attached pic of solution please see it. (Ignore the spelling errors If any)

wait I forgot to do 2nd part I attaching it also

Hope it helps.

Answer 2

Stress produced in wire is$2\times 10^{8} \ \left ( \frac{N}{m^{2}} \right )$. Cross-sectional area of wire is$0.01 \ mm^{2}$.

Explanation:

1. Given data

  Length of wire (L) =3 m

  Change in length of wire $\mathbf{(\Delta L)=3 \ mm=0.003 \ m}$

  Young's modulus of elasticity $\mathbf{(Y)=2\times 10^{11} \ \frac{N}{m^{2}}}$

2. $\mathbf{Strain(\varepsilon )=\frac{\Delta L}{L}=\frac{0.003}{3}=0.001}$

3. From formula of stress and strain relationship    

  $\mathbf{Stress(\sigma )=Youngs \ modulus \ of\ elasticity(Y)\times Strain(\varepsilon )}$

  So

  $\mathbf{Stress(\sigma )=2\times 10^{11}\times 0.001}$

  $\mathbf{Stress(\sigma )=2\times 10^{8} \ \frac{N}{m^{2}}}$

4. From formula

   $\mathbf{Area(A)=\frac{Force(F)}{Stress(\sigma )}}$

   $\mathbf{Area(A)=\frac{2}{2\times 10^{8}}=10^{-8}\ m^{2}=0.01 \ mm^{2}}$