A uniform metallic wire has a resistance of $18 Ω$ and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is

2 Ω

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4 Ω

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12 Ω

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8 Ω

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Solution

The correct option is B $4 Ω$

As $R \propto l$ , so the resistance of each side of the equilateral triangle will be equal to $6 Ω$ .
Resistance $R_{eq}$ between any two vertices,
$\frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{6} \Rightarrow R_{eq} = 4 Ω$

Hence, $(A)$ is the correct answer.

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A uniform metallic wire has a resistance of 18 Ω and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is

The correct option is B 4 Ω As R∝l, so the resistance of each side of the equilateral triangle will be equal to 6 Ω. Resistance Req between any two vertices, 1Req=112+16⇒Req=4 Ω Hence, (A) is the correct answer.

A uniform metallic wire has a resistance of $18 Ω$ and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is

The correct option is B $4 Ω$

As $R \propto l$ , so the resistance of each side of the equilateral triangle will be equal to $6 Ω$ .
Resistance $R_{eq}$ between any two vertices,
$\frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{6} \Rightarrow R_{eq} = 4 Ω$

Hence, $(A)$ is the correct answer.