Answer:
The expression for resistance is   [tex]R = \frac{\rho}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]
Explanation:
Generally flow of charge at that point is mathematically given as
       [tex]J = \frac{I}{2 \pi r L}[/tex]
Where L is length of the cylinder as given the question
The potential difference that is between the cylinders is Â
         [tex]\delta V = -E dr[/tex]
 Where is the radius
Where E is the electric field that would be experienced at that point which is mathematically represented as
        [tex]E = \rho J[/tex]
Where is the [tex]\rho[/tex] is the resistivity as given the question
considering the formula for potential difference we have
        [tex]\delta V = -[\frac{\rho I}{2 \pi r L} ]dr[/tex]
To get V we integrate both sides
      [tex]\int\limits^V_0 {\delta V} \, = \int\limits^{R_2}_{R_1} {\frac{\rho I}{2 \pi L r} } \, dr[/tex]
        [tex]V = \frac{\rho I}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]
According to Ohm law
      [tex]V= IR[/tex]
Now making R the subject we have
     [tex]R = \frac{V}{I}[/tex]
        Substituting for V
     [tex]R = \frac{\rho}{2 \pi L} ln[\frac{R_2}{R_1} ][/tex]
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