Answer:
[tex]16\Omega[/tex]
Explanation:
Initial length of wire=[tex]L_0[/tex]
Initial radius of wire=[tex]r_0[/tex]
Resistance of wire=1 ohm
After stretching
Radius of wire=r=[tex]0.25r_0[/tex]
New resistance of wire=R
When the wire drawn under tensile stress then volume remains constant.
[tex]V=A_0L_0=AL[/tex]
[tex]\frac{L_0}{L}=\frac{A}{A_0}[/tex]...(1)
[tex]R=\rho\frac{l}{A}[/tex]
Using the formula
[tex]1=\frac{\rho L_0}{A_0}[/tex]
[tex]R=\frac{\rho L}{A}[/tex]
Using equation (1)
[tex]\frac{1}{R}=\frac{L_0A}{LA_0}=\frac{A}{A_0}\times \frac{A}{A_0}=(\frac{A}{A_0})^2[/tex]
Area , A=[tex]\pi r^2[/tex]
Substitute the values
[tex]\frac{1}{R}=(\frac{\pi (0.25r_0)^2}{\pi(r_0)^2})^2=0.0625[/tex]
[tex]R=\frac{1}{0.0625}=16\Omega[/tex]
