Question

Length of a hollow tube is `5 m`, its over diamter is `10 cm` and thickness of its wall is 5 mm. If resistivity of the material of the tube is `1.7 xx 10^(-8) Omega xx m` then resistance of tube will be
A. `5.6 xx 10^(-5) Omega`
B. `2 xx 10^(-5) Omega`
C. `4 xx 10^(-5) Omega`
D. None of these

Answer 1

Correct Answer - A
(a) By using `R = rho. (l)/(A)`, here `A = pi (r_(2)^(2) - r_(1)^(2))`
Outer radius `r_(2) = 5 cm`
Inner radius `r_(1) = 5 - 0.5 = 4.5 cm`
So `R = 1.7 xx 10^(-8) xx (5)/(pi {(5 xx 10^(-2))^(2) - (4.5 xx 10^(-2))^(2)})`
`= 5.6 xx 10^(-5) Omega`