Q. Length of a hollow tube is $5m,$ it’s outer diameter is $10\ cm$ and thickness of it’s wall is $5\ mm.$ If resistivity of the material of the tube is $1.7 \times 10^{- 8}\Omega \times m$ then resistance of tube will be
A) $5.6 \times 10^{- 5}\Omega$
B) $2 \times 10^{- 5}\Omega$
C) $4 \times 10^{- 5}\Omega$
D) None of these
Explanation
(a) By using $R = \rho.\frac{l}{A}$; here $A = \pi\left( r_{2}^{2} - r_{1}^{2} \right)$
Outer radius $r_{2} = 5cm$
Inner radius $r_{1} = 5 - 0.5 = 4.5\ cm$
So $R = 1.7 \times 10^{- 8} \times \frac{5}{\pi\{\left( 5 \times 10^{- 2} \right)^{2} - \left( 4.5 \times 10^{- 2} \right)^{2}\}}$
$= 5.6 \times 10^{- 5}\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.