IIT JEE , AIEEE Forum - Physics , Electricity

johri_anshuman

In the fig there is a cavity of radius 'a' inside a solid cylinder of radius 'b' of a material whose resistivity varies as

=kr where r distance from the axis of the cylinder and k is a constant. Find the resistance between the plane faces. Length of the cylinder is l.

Ans:- kl/2pi(b-a)

Pls post a detailed solution...

spideyunlimited

Ans) k r^2 / 2 pi ( r^2 - a^2 )

spideyunlimited

message me on my nudgebook if u wanna know how to get this answer

spideyunlimited

sorry this is the correct answer( i forgot to consider elements properly)

we can integrate resistance using formula pl/a

integrate krl/2 pi r taking rings as elements
we get k/2 pi [r] from a to b

= kl /2 pi (b-a)

johri_anshuman

i m having problem in integrating.

I m having 3 variables in my equation and my actual problem is that the resistivity is varying in the horizontal plane while we have to integrate elements vertically in order to get the resistance. Pls post the integrations.

spideyunlimited

dR = d p l /A
= d krl/A

R = _{[a ]}

^[b] krl / 2

r (taking horizontal rings of increasing radius as element.. l is constant)

R = kl / 2

_{[a ]}

^{[b ]} r/r
= kl / 2

_a[r]^b
= kl / 2

(b - a)

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