Answer:
Explanation:
Let L and A be the length and area of cross section of rectangular segment
If resistor is melted down to and its [tex]\frac{1}{5}[/tex] material is used to make a new uniform rectangular resistor with same resistance R
volume of original rectangular(V)[tex]=A\times L[/tex]
Volume of new rectangular segment with length [tex]L_0(V_0)=A\times L_0[/tex]
[tex]5V_0=V[/tex]
[tex]5\times A_0\times L_0=A\times L[/tex]----1
Resistance is same
[tex]R=\frac{\rho L}{A}=\frac{\rho L_0}{A_0}[/tex]
[tex]\frac{L}{L_0}=\frac{A}{A_0}[/tex]
substitute [tex]\frac{L}{L_0}[/tex] in 1
[tex]5=\frac{L^2}{L_0^2}[/tex]
[tex]\frac{L}{L_0}=\sqrt{5}[/tex]
[tex]L=\sqrt{5}L_0[/tex]
[tex]L_0=\frac{L}{\sqrt{5}}[/tex]
