Answer:
So after stretching new resistance will be 0.1823 ohm
Explanation:
We have given initially length of the wire [tex]l_1=150m[/tex]
Radius of the wire [tex]r_1=0.32mm=0.32\times 10^{-3}m[/tex]
Resistance of the wire initially [tex]R_1=90ohm[/tex]
We know that resistance is equal to [tex]R=\frac{\rho l}{A}[/tex] ,here [tex]\rho[/tex] is resistivity, l is length and A is area
From the relation we can say that [tex]\frac{R_1}{R_2}=\frac{l_1}{l_2}\times \frac{A_2}{A_1}[/tex]
Now length of wire become 6.75 m
Volume will be constant
So [tex]A_1l_1=A_2l_2[/tex]
So [tex]\pi \times (0.32)^2\times150=\pi \times r_2^2\times 6.75[/tex]
[tex]r_2=1.508mm[/tex]
So [tex]\frac{90}{R_2}=\frac{150}{6.75}\times \frac{1.508^2}{0.32^2}[/tex]
[tex]R_2=0.1823ohm[/tex]