Question

A hollow copper tube of 5 m length has got external diameter equal to 10 cm and its walls are 5 mm thick The specific resistance of copper is $1.7\times {10}^{-8}\Omega$ m The resistance of the copper tube is approximately equal to

• A. $5.6\times {10}^{-3}\Omega$
• B. $5.6\times {10}^{-9}\Omega$
• C. $5.6\times {10}^{-5}\Omega$
• D. $5.6\times {10}^{-7}\Omega$

Answer 1

Answer: (C)

$5.6\times {10}^{-5}\Omega$

Cross sectional area of the copper tube is:
$A=\frac{\pi }{4}(d_1^2-d_2^2)=\frac{\pi }{4}({10}^2-9^2)=14.9c{m}^2=14.9\ast {10}^{-4}{m}^2$
$l=5m$
$\rho =1.7\ast {10}^{-8}$
Resistance is given by: $R=\frac{\rho l}{A}=5.6\ast {10}^{-5}Ohm$