mechanics again!!!!!!!!

for any segment of the rope, at an angle 'a' to the starting radius perpendicular to T1,

we have dT=dF
where dT is the slight increment in the tension, for an increment of a by da,

we have dF=kdN=dT

we also have dN=Tda

so kTda=dT

or Kda=dT/T

\int_{0}^{A}{da}= 1/K \int_{T1}^{T2}{dT/T}

so AK=log(T2/T1)
or T2=T1e^kA

cheers!!

http://targetiit.com/iit_jee_forum/posts/30_5_09_easy_pulleys_9308.html

look at #7

probably this is what you are talking about.....just redrawing the same thing for convenience