JEE MAIN - Physics (2023 - 24th January Morning Shift - No. 19)
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 $$n\times10^{-3}\Omega$$. If the resistivity of the material is $$\mathrm{2.4\times10^{-8}\Omega m}$$. The value of $$n$$ is ___________.
Answer
2
Explanation
Resistance of the hollow cylindrical conductor is given by,
$R=\rho \frac{l}{\pi\left(r_2^2-r_1^2\right)}$
where $r_2=$ outer radius
$$ \begin{gathered} r_1=\text { inner radius } \\\\ \rho=\text { resistivity, } l=\text { length } \\\\ \therefore \rho=\frac{2.4 \times 10^{-8} \times 3.14}{\pi\left(\frac{8^2}{4}-\frac{4^2}{4}\right) \times 10^{-6}} \\\\ =\frac{2.4 \times 10^{-8} \times 4}{48 \times 10^{-6}}=2 \times 10^{-3} \Omega \end{gathered} $$
$R=\rho \frac{l}{\pi\left(r_2^2-r_1^2\right)}$
where $r_2=$ outer radius
$$ \begin{gathered} r_1=\text { inner radius } \\\\ \rho=\text { resistivity, } l=\text { length } \\\\ \therefore \rho=\frac{2.4 \times 10^{-8} \times 3.14}{\pi\left(\frac{8^2}{4}-\frac{4^2}{4}\right) \times 10^{-6}} \\\\ =\frac{2.4 \times 10^{-8} \times 4}{48 \times 10^{-6}}=2 \times 10^{-3} \Omega \end{gathered} $$
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