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 (integer)
2
Solution
Resistance of the hollow cylindrical conductor is given by,
<br/><br/>$R=\rho \frac{l}{\pi\left(r_2^2-r_1^2\right)}$
<br/><br/>where $r_2=$ outer radius
<br/><br/>$$
\begin{gathered}
r_1=\text { inner radius } \\\\
\rho=\text { resistivity, } l=\text { length } \\\\
\therefore \rho=\frac{2.4 \times 10^{-8} \times 3.14}{\pi\left(\frac{8^2}{4}-\frac{4^2}{4}\right) \times 10^{-6}} \\\\
=\frac{2.4 \times 10^{-8} \times 4}{48 \times 10^{-6}}=2 \times 10^{-3} \Omega
\end{gathered}
$$
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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