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 × 10⁻³ Ω. If the resistivity of the material is 2.4 × 10⁻⁸ Ω m, the value of n is ___________. Numerical answer type. Work it out, then open the answer.

Answer: 2. Area A = π(r₂² - r₁²) = π(16-4)×10⁻⁶ = 12π×10⁻⁶ m². R = (ρ l)/(A) = 2.4×10⁻⁸×3.143.14×12×10⁻⁶ = 2×10⁻³ Ω So n = 2.

A hollow cylindrical conductor has length of 3.14 m, while its inner… | JEE Main 2023 (24 Jan, Shift 1)

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Ohm's Law, Resistivity & Temperature

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 $n \times 10^{-3}\ \Omega$ . If the resistivity of the material is $2.4 \times 10^{-8}\ \Omega\text{ m}$ , the value of $n$ is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

Area $A = \pi(r_2^2 - r_1^2) = \pi(16-4)\times10^{-6} = 12\pi\times10^{-6}\ \text{m}^2$ .
$R = \dfrac{\rho l}{A} = \dfrac{2.4\times10^{-8}\times3.14}{3.14\times12\times10^{-6}} = 2\times10^{-3}\ \Omega$
So $n = 2$ .

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