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

Current Electricity PYQ — JEE Main 2025 · 28 Jan · Shift 1 · Q59 with Solution

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Correct answer: C

Given

A wire of total resistance $R$ is bent into an equilateral triangle.

An identical wire of total resistance $R$ is bent into a square.

We need the ratio:

\frac{\text{resistance between the ends of one edge of triangle}}{\text{resistance between the ends of one edge of square}}

Since the wires are identical, each full loop has total resistance $R$ .

Triangle case

The wire is bent into an equilateral triangle, so each side has resistance

\frac{R}{3}

Let the two endpoints of one side be $A$ and $B$ .

Between $A$ and $B$ , there are two possible paths:

directly along one side: resistance $\dfrac{R}{3}$
along the other two sides: resistance $\dfrac{2R}{3}$

These two paths are in parallel.

So equivalent resistance is

R_T = \frac{\left(\frac{R}{3}\right)\left(\frac{2R}{3}\right)}{\frac{R}{3}+\frac{2R}{3}}

R_T = \frac{\frac{2R^2}{9}}{R} = \frac{2R}{9}

Square case

The wire is bent into a square, so each side has resistance

\frac{R}{4}

Let the endpoints of one side be $A$ and $B$ .

Between $A$ and $B$ , there are again two paths:

directly along one side: resistance $\dfrac{R}{4}$
along the remaining three sides: resistance $\dfrac{3R}{4}$

These are in parallel.

So equivalent resistance is

R_S = \frac{\left(\frac{R}{4}\right)\left(\frac{3R}{4}\right)}{\frac{R}{4}+\frac{3R}{4}}

R_S = \frac{\frac{3R^2}{16}}{R} = \frac{3R}{16}

Required ratio

\frac{R_T}{R_S} = \frac{\frac{2R}{9}}{\frac{3R}{16}} = \frac{2R}{9}\cdot \frac{16}{3R} = \frac{32}{27}

Option check

A: $\dfrac{8}{9}$
B: $\dfrac{9}{8}$
C: $\dfrac{32}{27}$
D: $\dfrac{27}{32}$

So the correct option is:

\boxed{\text{C } \frac{32}{27}}

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.

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View written solutionFree

Correct answer: C

Given

A wire of total resistance $R$ is bent into an equilateral triangle.

An identical wire of total resistance $R$ is bent into a square.

We need the ratio:

\frac{\text{resistance between the ends of one edge of triangle}}{\text{resistance between the ends of one edge of square}}

Since the wires are identical, each full loop has total resistance $R$ .

Triangle case

The wire is bent into an equilateral triangle, so each side has resistance

\frac{R}{3}

Let the two endpoints of one side be $A$ and $B$ .

Between $A$ and $B$ , there are two possible paths:

directly along one side: resistance $\dfrac{R}{3}$
along the other two sides: resistance $\dfrac{2R}{3}$

These two paths are in parallel.

So equivalent resistance is

R_T = \frac{\left(\frac{R}{3}\right)\left(\frac{2R}{3}\right)}{\frac{R}{3}+\frac{2R}{3}}

R_T = \frac{\frac{2R^2}{9}}{R} = \frac{2R}{9}

Square case

The wire is bent into a square, so each side has resistance

\frac{R}{4}

Let the endpoints of one side be $A$ and $B$ .

Between $A$ and $B$ , there are again two paths:

directly along one side: resistance $\dfrac{R}{4}$
along the remaining three sides: resistance $\dfrac{3R}{4}$

These are in parallel.

So equivalent resistance is

R_S = \frac{\left(\frac{R}{4}\right)\left(\frac{3R}{4}\right)}{\frac{R}{4}+\frac{3R}{4}}

R_S = \frac{\frac{3R^2}{16}}{R} = \frac{3R}{16}

Required ratio

\frac{R_T}{R_S} = \frac{\frac{2R}{9}}{\frac{3R}{16}} = \frac{2R}{9}\cdot \frac{16}{3R} = \frac{32}{27}

Option check

A: $\dfrac{8}{9}$
B: $\dfrac{9}{8}$
C: $\dfrac{32}{27}$
D: $\dfrac{27}{32}$

So the correct option is:

\boxed{\text{C } \frac{32}{27}}

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.

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