A copper cylindrical tube has inner radius `a` and outer radius `b`. The resistivity is `rho`. The resistance of the cylinder between the two ends is

Options

A. `(rhol)/(b^(2)-a^(2))`
B. `(rhol)/(2pi(b-a)`
C. `(rhol)/(pi(b^(2)-a^(2))`
D. `(pi(b^(2)-a^(2)))/(rhol)`

Answer

C

Explanation

If one ahd considered a solid cylinder of radius b, one can suppose that it is made of two concentric cylinders of radius a nd the outer part, jointd along the length concentrically on inside the other.
If `I_(a) and I_(x)` are the currents flowing throught the innerand outer cylinders
`therefore I_("total")=I_(b)=I_(a)+I_(x)Rightarrow (V)/(R_(b))=(V)/(R_(a))+(V)/(R_(x))`
Where, `R_(b)` is the total resitance and `R_(x)` is the resistance of the tublular part.
`therefore (1)/(R_(x))=(1)/(R_(b))-(1)/(R_(a))`
`But R_(a)=(rhol)/(pia^(2))and R_(b)=(rhol)/(pib^(2))`
`therefore (1)/(R_(x))=(pi)/(rhol)(b^(2)-a^(2)) therefore R_(x)=(rhol)/(pi(b^(2)-a^(2)))`