The external diameter of $5 \mathrm{~m}$ long hollow tube is $0.1 \mathrm{~m}$ and thickness of its wall is…

The external diameter of $5 \mathrm{~m}$ long hollow tube is $0.1 \mathrm{~m}$ and thickness of its wall is $0.005 \mathrm{~m}$. If $\rho=1.7 \times 10^{-8} \Omega-\mathrm{m}$, then its resistance will be
  1. $5.7 \times 10^{-5} \Omega$
  2. $2.7 \times 10^{-5} \Omega$
  3. $2 \times 10^{-5} \Omega$
  4. $5 \times 10^{-5} \Omega$

Solution

Given, Iength, $l=5 \mathrm{~m}$ External diameter, $d_1=0.1 \mathrm{~m}$ $\therefore$ External radius, $r_1=\frac{d_1}{2}=\frac{0.1}{2}=0.05 \mathrm{~m}$ Thickness, $t=0.005 \mathrm{~m}$ $\therefore$ Internal radius, $r_2=r_1-t$ $ =0.05-0.005=0.045 \mathrm{~m} $ $\therefore$ Area of cross-section of hollow tube, $ \begin{aligned} A & =\pi\left(r_1^2-r_2^2\right) \\ & =3.14\left[(0.05)^2-(0.045)^2\right] \\ & =3.14 \times 4.75 \times 10^{-4} \\ & =14.915 \times 10^{-4}=1.49 \times 10^{-3} \mathrm{~m}^2 \end{aligned} $ $\therefore$ Resistance, $R=\rho \cdot \frac{l}{A}=1.7 \times 10^{-8} \times \frac{5}{1.49 \times 10^{-3}}$ $ =5.7 \times 10^{-5} \Omega $

Asked in: AP EAMCET 2020 (22 Sep Shift 1)

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