A conductor of resistance 3 Omega is stretche | vedclass

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(B) When a wire is stretched,its volume remains constant. Resistance $R = \rho \frac{l}{A} = \rho \frac{l^2}{V}$. Since $\rho$ and $V$ are constant,$R

Vedclass · Accepted Answer

$\frac{8}{3}$

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(B) When a wire is stretched,its volume remains constant. Resistance $R = ho \frac{l}{A} = ho \frac{l^2}{V}$. Since $ho$ and $V$ are constant,$R \propto l^2$.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. Each side will have a resistance of $R_{side} = \frac{12 \Omega}{3} = 4 \Omega$.Let the vertices be $A, B, C$. The sides are $AB, BC, CA$ each of $4 \Omega$.To find the effective resistance between the ends of any side (e.g.,across $B$ and $C$),the side $BC$ $(4 \Omega)$ is in parallel with the series combination of $AB$ and $AC$ $(4 \Omega + 4 \Omega = 8 \Omega)$.$R_{eq} = \frac{R_{BC} 	imes (R_{AB} + R_{AC})}{R_{BC} + (R_{AB} + R_{AC})} = \frac{4 	imes 8}{4 + 8} = \frac{32}{12} = \frac{8}{3} \Omega$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

$A$ 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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