Question

A copper cylindrical tube has inner radius a and outer radius b. The resistivity is $5\Omega$. The resistance of the cylinder between the two ends is

• A. $T_{1}$
• B. $T_{2}$
• C. $T_{1}$
• D. $T_{2}$

Answer 1

Answer: (C)

$T_{1}$

Answer 2

: If one had considered a solid cylinder of radius b, one can suppose that it is made of two concentraic cylinders of radius a and the outer part, joined along the length concentrically one inside the other.

If $I_{q}$and $I_{x}$are the currents flowing through the inner and outer cylinders.

$\because I_{total} = I_{b} = I_{a} + I_{x} \Rightarrow \frac{V}{R_{b}} = \frac{V}{R_{a}} + \frac{V}{R_{x}}$

where $R_{b}$is the total resistance and $R_{x}$is the resistance of the tubular part.

$\therefore\frac{1}{R_{x}} = \frac{1}{R_{b}} - \frac{1}{R_{a}}$

But, $R_{a} = \frac{\rho l}{\pi a^{2}}$ and $R_{b} = \frac{\rho l}{\pi b^{2}}$

$\therefore\frac{1}{R_{x}} = \frac{\pi}{\rho l}(b^{2} - a^{2})\therefore R_{x} = \frac{\rho l}{\pi(b^{2} - a^{2})}$