JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Current Electricity

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A hollow cylindrical conductor has length of 3.14 m, while its inner and outer diameters are 4 mm and 8 mm respectively. The resistance of the conductor is

n \times 1 0^{- 3} Ω

.
If the resistivity of the material is

2.4 \times 1 0^{- 8} Ω m

. The value of

n

is ______.

Answer: 2

Solution

The resistance of the tube is

R = \frac{ρ ℓ}{π ( b ^{2} - a ^{2} )}

, where the current crosses the annular area between outer radius

b = 4 mm

and inner radius

a = 2 mm

. Here

b^{2} - a^{2} = (16 - 4) \times 1 0^{- 6} = 12 \times 1 0^{- 6} m^{2}

, and with

ℓ = 3.14 m \approx π

the factor of

π

cancels, leaving

R = \frac{2.4 \times 1 0 ^{- 8}}{12 \times 1 0 ^{- 6}} = 2 \times 1 0^{- 3} Ω

. The value of

n

2

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Integer answer

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