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JEE Main 202324 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
24 January
Shift
Morning Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2023 · 24 January · Morning Shift · Physics · Current Electricity

A hollow cylindrical conductor has length of 3 . 14m , while its...

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}Ω.
If the resistivity of the material is 2.4×10−8Ωm2 . 4 \times {10}^{- 8}Ωm. The value of nn is _____.

Correct answer

0

Step-by-step solution

Resistance of a conductor with uniform cross-section is given by,

R=ρlAR = ρ\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 = ρ\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}Ω

Hence, n=2n = 2.

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