Question

. A wire of length l and resistance R is stretched to get

the radius of cross-section halved .What is the new

resistance of the wire after stretching?​

Answer 1

Given:

Length of the wire = l

Initial resistance of the wire = R

Radius of  cross section after stretching (r₁) = r/2

To find:

The new resistance of the wire after stretching

Calculation

Initially the radius of the the wire is r and the length l. So the resistance of the wire is

     $R=\frac{\rho l}{A}$

When the wire is stretched, the length of the wire changes but the volume remains same

     $V=V_{1}$

     $\pi r^{2}\times l=\pi r_{1}^{2}\times l_{1}$

     $r^{2}\times l=(\frac{r}{2} )^{2}\times l_{1}$

     $l_{1}=4l$

The final length of the wire after stretching is 4l

The final resistance of the wire will be

     $R_{1}=\frac{\rho l_{1}}{A_{1}}$

     $R_{1} =\frac{\rho l_{1}} {\pi r_{1}^{2}}$

     $R_{1} =\frac{\rho (4\times l)}{\pi \times(\frac{r}{2})^{2} }$

     $R_{1} =\frac{16\rho l}{A}$

     $R_{1}=16R$

Final answer:

The new resistance of the wire after stretching will be 16R

   

     

     

   

         

Answer 2

final answer is 16R

hope it helps you