A uniform metal wire of length $l$ has a resistance of $20 \Omega$. This wire is uniformly stretched

Question

A uniform metal wire of length $l$ has a resistance of $20 \Omega$. This wire is uniformly stretched to a length $3l$ and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of this circle is

Professor PC Thomas Classes · Answer

(a) : The volume of the wire remains constant during stretching. Since $l' = 3l$, the new area $A' = A/3$. The new resistance is $R' = ho \frac{l'}{A'} = ho \frac{3l}{A/3} = 9 \left(ho \frac{l}{A}ight) = 9 	imes 20 = 180 \Omega$. When bent into a circle, the resistance of each semi-circle is $\frac{180}{2} = 90 \Omega$. The equivalent resistance across the diameter is two such semi-circles in parallel: $R_{eq} = \frac{90 	imes 90}{90 + 90} = 45 \Omega$.