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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Physics
Chapter
Current Electricity

JIPMER 2015 · Physics · Current Electricity

Find the resistance of a hollow cylindrical conductor of length 1.0...

Find the resistance of a hollow cylindrical conductor of length 1.0 mm and 2.0 mm respectively. The resistivity of the material is 2.0×10−8Ωm2.0 \times 10^{-8} \Omega \mathrm{m}.

Options

  1. A2.1×10−3Ω2.1 \times 10^{-3} \Omega
  2. B1.3×10−4Ω1.3 \times 10^{-4} \Omega
  3. C3.2×10−4Ω3.2 \times 10^{-4} \Omega
  4. D4.6×10−2Ω4.6 \times 10^{-2} \Omega

Correct answer

A. 2.1×10−3Ω2.1 \times 10^{-3} \Omega

Step-by-step solution

The area of cross-section of the conductor through which charges will be flow is A=π(20 mm)2−π(1.0 mm)2=3×πmm2\begin{aligned} A & =\pi(20 \mathrm{~mm})^2-\pi(1.0 \mathrm{~mm})^2 \\ & =3 \times \pi \mathrm{mm}^2\end{aligned} The resistance of the wire is therefore R=ρ1A=(2.0×10−8)×(10)3.0×π×10−6=2.1×10−3Ω\begin{aligned} R & =\rho \frac{1}{A}=\frac{\left(2.0 \times 10^{-8}\right) \times(10)}{3.0 \times \pi \times 10^{-6}} \\ & =2.1 \times 10^{-3} \Omega\end{aligned}
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