Question

A hollow cylinder $(\rho =2.2\times {10}^{-3}\Omega m)$ of length 3m has inner and outer as 2mm and 4mm respectively. Potential difference is set up between the inner and outer surfaces of the cylinder. The resistance of the cylinder is

• A. $0.35\times {10}^{-3}\Omega$
• B. $3\times {10}^{-3}\Omega$
• C. $8\times {10}^{-5}\Omega$
• D. $3.1\times {10}^{-3}\Omega$

Answer 1

The correct option is C $8\times {10}^{-5}\Omega$

Give,

Resistivity, $\rho =2.2\times {10}^{-3}\Omega m$

Inner and outer radius, $a=2mmandb=4mm$

Length, $L=3m$

We recall that$J=\frac{I}{A}$, $E=\frac{dV}{dr}$. The area, A will be the surface area of the cylinder. Since,

$J=\frac{1}{\rho }E$

$\frac{I}{A}=\frac{1}{\rho }E$

$\frac{I}{2\pi rL}=\frac{1}{\rho }\frac{dV}{dr}$

$dV=I\frac{\rho }{2\pi rL}dr$

Since, $dR=\frac{dV}{I}$,

$dR=\frac{\rho }{2\pi rL}dr$

Integrating from a to b,

$R=\underset{a}{\overset{b}{\int }}dR=\underset{a}{\overset{b}{\int }}\rho \frac{1}{2\pi rL}dr=\frac{\rho }{2\pi L}\underset{a}{\overset{b}{\int }}\frac{1}{r}dr$

$R=\frac{\rho }{2\pi L}\text{ln}\frac{b}{a}=\frac{2.2\times {10}^{-3}}{2\pi \times 3}\ln \left(\frac{4}{2}\right)$

$R=8.08\times {10}^{-5}\Omega$

Hence, resistance is $8\times {10}^{-5}\Omega$