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QJEEM23Pha226: JEE Main 2023 Physics - Class 12 JEEM23Pha226 Solution: A hollow cylindrical conductor has length of 3.14 m, while its…

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×10−3Ωn\times10^{-3}\Omega . If the resistivity of the material is 2.4×10−8Ωm\mathrm{2.4\times10^{-8}\Omega m} . The value of nn is ___________.

Step-by-Step Verified Solution

Resistance of the hollow cylindrical conductor is given by,

R=ρlπ(r22−r12)R=\rho \frac{l}{\pi\left(r_2^2-r_1^2\right)}

where r2=r_2= outer radius

r1= inner radius ρ= resistivity, l= length ∴ρ=2.4×10−8×3.14π(824−424)×10−6=2.4×10−8×448×10−6=2×10−3Ω\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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