JEE Main 2025 Apr 3 Shift 1, Physics Q5: Introduction And Ohm’s Law

JEE Main2025Apr 3, Shift 1Physics

A wire of length 25 m and cross-sectional area 5 mm $^{2}$ having resistivity of $2 \times 1 0^{- 6} Ω$ m is bent into a complete circle. The resistance between diametrically opposite points will be

A
$12.5 Ω$
B
$50 Ω$
C
$100 Ω$
D
$2.5 Ω$

Solution

Total resistance of the wire,

$R = \frac{ρ L}{A} = \frac{2 \times 1 0 ^{- 6} \times 25}{5 \times 1 0 ^{- 6}} = 10 Ω$

On bending the wire into a circle, diametrically opposite points divide it into two equal semicircles.

Resistance of each semicircle $= \frac{10}{2} = 5 Ω$ .

These two resistances are in parallel.

$R_{eq} = \frac{5 \times 5}{5 + 5} = 2.5 Ω$

$2.5 Ω$

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JEE Main 2025 Apr 3 Shift 1, Physics Q5: Introduction And Ohm’s Law

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