JEE Mains · Physics · STD 12 - 3. current electricity
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 \(2.4 \times 10^{-8}\,\Omega\,m\). The value of \(n\) is \(..........\)
- A \(2\)
- B \(4\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(R =\rho \frac{\ell}{ A }\), the cross-sectional area is \(\pi\left( b ^2- a ^2\right)\) \(R =\rho \frac{\ell}{\pi\left( b ^2- a ^2\right)}=\frac{2.4 \times 10^{-8} \times 3.14}{3.14 \times\left(4^2-2^2\right) \times 10^{-6}}\) \(=2 \times 10^{-3}\,\Omega\) \(\rightarrow n =2\)
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