Question

A hollow cylindrical conductor has length of $314\, m$, 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 $24 \times 10^{-8} \Omega m$ The value of $n$ is ___

Answer 1

Correct Answer: 2

The resistance of a conductor is given by: \[ R = \rho \frac{\ell}{A} \] where: - \(R = 2 \times 10^{-3} \, \Omega\), - \(\rho = 2.4 \times 10^{-8} \, \Omega \, \text{m}\), - \(\ell = 3.14 \, \text{m}\), - \(A\) is the cross-sectional area of the hollow cylinder. The cross-sectional area \(A\) of the hollow cylinder is: \[ A = \pi (b^2 - a^2) \] where: - \(b = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m}\), - \(a = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m}\). Substitute the values: \[ A = \pi \left[(4 \times 10^{-3})^2 - (2 \times 10^{-3})^2\right] = \pi \left[16 \times 10^{-6} - 4 \times 10^{-6}\right] = \pi \cdot 12 \times 10^{-6}. \] Now calculate \(R\): \[ R = \rho \frac{\ell}{A} = \frac{2.4 \times 10^{-8} \cdot 3.14}{\pi \cdot 12 \times 10^{-6}} = \frac{2.4 \times 10^{-8} \cdot 3.14}{3.77 \times 10^{-5}} = 2 \times 10^{-3} \, \Omega. \] This confirms the calculated value, and: \[ n = 2. \]