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(D) The resistance of a wire is given by $R = \frac{\rho \ell}{A}$,which implies $R \propto \ell$.For the equilateral triangle,each side has a resista

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$32 / 27$

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(D) The resistance of a wire is given by $R = \frac{ho \ell}{A}$,which implies $R \propto \ell$.For the equilateral triangle,each side has a resistance of $R/3$. When finding the equivalent resistance between the two ends of one edge,we have one resistor of $R/3$ in parallel with the other two resistors in series (which sum to $2R/3$).$(R_{eq})_1 = \frac{(R/3) 	imes (2R/3)}{(R/3) + (2R/3)} = \frac{2R^2/9}{R} = \frac{2R}{9}$.For the square,each side has a resistance of $R/4$. When finding the equivalent resistance between the two ends of one edge,we have one resistor of $R/4$ in parallel with the other three resistors in series (which sum to $3R/4$).$(R_{eq})_2 = \frac{(R/4) 	imes (3R/4)}{(R/4) + (3R/4)} = \frac{3R^2/16}{R} = \frac{3R}{16}$.The ratio of the equivalent resistances is $\frac{(R_{eq})_1}{(R_{eq})_2} = \frac{2R/9}{3R/16} = \frac{2}{9} 	imes \frac{16}{3} = \frac{32}{27}$.

$A$ wire of resistance $R$ is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of the equivalent resistance between the two end points of an edge of the triangle to that of the square is

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