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

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9/8

27/32

8/9

32/27

answer is D.

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Detailed Solution

$\begin{aligned} R = \frac{ρ ℓ}{A} \\ So, R α ℓ \end{aligned}$

Side length of triangle is 1/3 of total length.

$\begin{array}{ll} {(R_{e q})}_{1} = \frac{2 r / 3 \times r / 3}{2 r / 3 + r / 3} & {(R_{e q})}_{2} = \frac{3 r / 4 \times r / 4}{3 r / 4 + r / 4} \\ {(R_{e q})}_{1} = 2 r / 9 & {(R_{e q})}_{2} = 3 r / 16 \end{array}$

$\frac{{(R_{eq})}_{1}}{{(R_{eq})}_{2}} = \frac{2 r / 9}{3 r / 16} = \frac{32}{27}$

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