Calculating Resistance in a Hollow Cylinder with a Pierced Hole

[Mythian]

This problem is about a cylinder, which its length is 'L', its radius 'a' and its resistivity 'p'.

We pierce a hole in the cylinder (this hole has a radius of a/2)

I have now to

a) prove that the resistance equation(between the ends) of this hollow cylinder is

R= 4/3 pL/(pi)a^2

b) prove that the total resistance of the same hollow cylinder filled with a new resistivity material p2 is:

R= 4/3 pL/(pi)a^2 X p2/p2+p

I tried to prove A but it gives me R=pL/pi(b^2 - a^2)...which is not what I want. I have NO idea where does the 4/3 from the equation of the question comes from. If someone could help me with a hint or something, it would be much appreciated.

 

[nrqed]

This problem is about a cylinder, which its length is 'L', its radius 'a' and its resistivity 'p'.

We pierce a hole in the cylinder (this hole has a radius of a/2)

I have now to

a) prove that the resistance equation(between the ends) of this hollow cylinder is

R= 4/3 pL/(pi)a^2

b) prove that the total resistance of the same hollow cylinder filled with a new resistivity material p2 is:

R= 4/3 pL/(pi)a^2 X p2/p2+p

I tried to prove A but it gives me R=pL/pi(b^2 - a^2)...which is not what I want. I have NO idea where does the 4/3 from the equation of the question comes from. If someone could help me with a hint or something, it would be much appreciated.

By $b^2 - a^2$ I think you mean the outer radius squared minus the inner radius squared, right? So you mean $a_{max}^2 - a_{min}^2$. Then just replace a_max by "a" and a_min by "a/2". You will see the 4/3 appear.