JEE Main 2025 Jan 28 Shift 1, Physics Q15: Wheatstone Bridge, Ammeter, Voltmeter And Potentiometer

JEE Main2025Jan 28, Shift 1Physics

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

A
$9/8$
B
$8/9$
C
$27/32$
D
$32/27$

Solution

Both wires have the same total length $L$ and total resistance $R$ . The resistance of any piece is proportional to its length.

Triangle. Each side has length $L /3$ and resistance $R /3$ . Between two endpoints of one side, the two paths are: the side itself ( $R /3$ ) in parallel with the two remaining sides in series ( $R /3 + R /3 = 2 R /3$ ).

$R_{△} = \frac{( R /3 ) ( 2 R /3 )}{R /3 + 2 R /3} = \frac{2 R ^{2} /9}{R} = \frac{2 R}{9}$ .

Square. Each side has length $L /4$ and resistance $R /4$ . Between two adjacent vertices, paths are: one side ( $R /4$ ) in parallel with three sides in series ( $3 R /4$ ).

$R_{□} = \frac{( R /4 ) ( 3 R /4 )}{R /4 + 3 R /4} = \frac{3 R ^{2} /16}{R} = \frac{3 R}{16}$ .

Ratio: $\frac{R _{△}}{R _{□}} = \frac{2 R /9}{3 R /16} = \frac{2}{9} \cdot \frac{16}{3} = \frac{32}{27}$ .

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JEE Main 2025 Jan 28 Shift 1, Physics Q15: Wheatstone Bridge, Ammeter, Voltmeter And Potentiometer

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