Q95 / 180 · JIPMER 2015 · Physics

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

  • Correct answer
    A
    $2.1 \times 10^{-3} \Omega$
  • B
    $1.3 \times 10^{-4} \Omega$
  • C
    $3.2 \times 10^{-4} \Omega$
  • D
    $4.6 \times 10^{-2} \Omega$

Solution

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

Hence, the correct answer is option: 'A'.

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