Answer:
a) 1.609× 10⁻⁵ Ω
b) 9 × 10⁻¹⁰ Ω
Explanation:
length of cylinder = 2.50 m
inner diameter = 2.75 cm
outer diameter = 4.60 cm
as we know,
a) resistance of cylinder across the length
[tex]R = \dfrac{\rho L}{A}[/tex]
area of cylinder = [tex]\pi (d_o^2 - d_1^2)[/tex]
= [tex]\pi (4.6^2 - 2.75^2)[/tex]
= 42.71 × 10⁻⁴ m²
[tex]R = \dfrac{2.75\times 10^{-8} \times 2.5}{42.71 \times 10^{-4}}[/tex]
= 1.609× 10⁻⁵ Ω
b) reading of inner and outer surface in ohmmeter means
assume r is radius of the cylinder and dr be thickness
[tex]dr = \dfrac{\rho}{2\pi r_1}\ dr\\dr = \int\limits^b_a \dfrac{\rho}{2\pi r_1}\ dr\\\\R = \dfrac{\rho}{2\pi l} \ (ln)^b_a\\R = \dfrac{\rho}{2\pi l} \ ln(\dfrac{b}{a})\\R = \dfrac{2.5 \times 10^{-8}}{2\pi \times 2.5} \ ln(\dfrac{4.6}{2.75})\\R=9\times 10^{-10} \Omega[/tex]
