Discussion Forum

A copper wire of cross-sectional area 0.01 cm² is under a tension of 22 N. Find the percentage change in the cross-sectional area, (Young's  modulus of copper =$1.1 \times 10^{11}  N/m^{2}$ and Poisson's ratio =0.32)

Answer:

Explanation:

 Given,  cross section area of copper wire A=0.01 cm² =10^-6 m²  and tension force.F=22N

Posson's  ratio , $\sigma$= 0.32, and Young's modulus  Y= $1.1 \times 10^{11}  N/m^{2}$

Since,Posson's ratio, $\sigma$= lateral strain/ Longitudinal strain

$\sigma=\frac{\frac{\triangle D}{D}}{\frac{\triangle L}{L}}$  ........(i)

$\because$   Area ,A= $\pi D^{2}$

$\Rightarrow$    $  \frac{\triangle A}{A}=2\frac{\triangle D}{D}$  .......(Iii)

 From Eqs,(i) and (ii) , we get

 $\Rightarrow$     $  \frac{\triangle A}{A}=2\sigma\frac{\triangle L}{L}$    ........(iiii)

 Young's modulus ,   $Y= \frac{FL}{A\triangle L}$   .......(iv)

From eqs.(iii) and (iv)  , we get

$\Rightarrow$          $\frac{\triangle A}{A}$=$\frac{2 \sigma F}{YA}$

Now , putting the given values

$\frac{\triangle A}{A}=\frac{2 \times 0.32 \times 22}{1.1 \times 10^{11} \times 10^{-6}}=12.8 \times  10^{-3}$

$\Rightarrow\frac{\triangle A}{A}$% =$12.8 \times 10^{-3}$=$12.6 \times 10^{-3}$

 So, option (a) is correct