A conductor of resistance 3 Omega is stretche | vedclass

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(B) When a wire is stretched,its volume remains constant. Since $R = ho \frac{l}{A}$ and $V = A 	imes l$,we have $R = ho \frac{l^2}{V}$. Thus,$R

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$\frac{8}{3}$

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(B) When a wire is stretched,its volume remains constant. Since $R = ho \frac{l}{A}$ and $V = A 	imes l$,we have $R = ho \frac{l^2}{V}$. Thus,$R \propto l^2$.Given initial resistance $R_1 = 3 \Omega$ and length $l_1 = l$. After stretching,$l_2 = 2l$.$\frac{R_2}{R_1} = \left(\frac{l_2}{l_1}ight)^2 = \left(\frac{2l}{l}ight)^2 = 4$.So,$R_2 = 4 	imes R_1 = 4 	imes 3 = 12 \Omega$.The wire of resistance $12 \Omega$ is bent into an equilateral triangle,so each side has a resistance of $R_{side} = \frac{12}{3} = 4 \Omega$.Let the vertices be $A, B, C$. The resistance of each side is $R_{AB} = 4 \Omega$,$R_{BC} = 4 \Omega$,and $R_{CA} = 4 \Omega$.To find the effective resistance between the ends of any side (e.g.,between $B$ and $C$),we see that $R_{AB}$ and $R_{AC}$ are in series,and their combination is in parallel with $R_{BC}$.Resistance of the series branch $R_{series} = R_{AB} + R_{AC} = 4 + 4 = 8 \Omega$.Now,$R_{series}$ is in parallel with $R_{BC} = 4 \Omega$.$R_{eff} = \frac{R_{series} 	imes R_{BC}}{R_{series} + R_{BC}} = \frac{8 	imes 4}{8 + 4} = \frac{32}{12} = \frac{8}{3} \Omega$.

$A$ conductor of resistance $3 \Omega$ is stretched uniformly until its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in ohms is

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