Current Electricity

JEE Main level, numerical answer, medium

62 of 380

Question 62 of 380

A hollow cylindrical conductor has length of 3.14m3.14m, while its inner and outer diameters are 4mm4mm and 8mm8mm respectively. The resistance of the conductor is n×10−3Ωn\times {10}^{-3}\Omega.
If the resistivity of the material is 2.4×10−8Ωm2.4\times {10}^{-8}\Omega m. The value of nn is _____.

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Answer: 2

Resistance of a conductor with uniform cross-section is given by,
R=ρlAR=\rho \dfrac{l}{A}
In this case the cross-sectional area is π(b2−a2)\pi \left({b}^{2}-{a}^{2}\right).
Therefore,
R=ρlπ(b2−a2)=(2.4×10−8)×3.143.14×[(42−22)×10−6]=2×10−3ΩR=\rho \dfrac{l}{\pi \left({b}^{2}-{a}^{2}\right)}=\dfrac{\left(2.4\times {10}^{-8}\right)\times 3.14}{3.14\times \left[\left({4}^{2}-{2}^{2}\right)\times {10}^{-6}\right]}=2\times {10}^{-3}\Omega
Hence, n=2n=2.

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