Question

A resistor is formed in the shape of a quarter cylinder from a
material of resistivity rho
. The length of cylinder is L, inner
and outer radii are
R1
and
R2
respectively. The resistance of
this resistor between the shown terminals P and Q is​

Answer 1

First, draw the figure and try to visualise the wire as a cylinder.

• Now, we can observe in the fig, that by considering the two terminals given, the effective length will vary with the distance r from the centre.
• And similarly, area will also get vary with breath still remain constant as 'L'

We know that,

$\implies \rm R = \rho \dfrac{Length}{area}$

here, both are variable so we have to integrate,

• element will be a thin flim of cylinder, radius 'r' of thickness dr

since its a quater cylinder,

----effective Length ( ring element)

• dL = $\rm \dfrac{2\pi r}{4}= \dfrac{\pi r}{2}$

---Area (rectangle element)

• $\rm dA = L.dr$

Resistance

$\implies \rm dR =\rho \dfrac{dL}{dA}$

$\implies \rm dR =\rho \dfrac{\dfrac{\pi r}{2}}{Ldr}$

$\implies \rm dR = \dfrac{\rho \pi}{2L}\times \dfrac{r}{dr}$

$\implies \rm \displaystyle \int dR = \dfrac{\rho \pi}{2L}\times \dfrac{1}{\displaystyle \int \dfrac{1}{r} dr}$

$\implies \rm \displaystyle \int \limits_{0}^{R} dR = \dfrac{\rho \pi}{2L}\times \dfrac{1}{\displaystyle \int \limits_{R_1}^{R_2} lnr}$

$\implies \bf R = \dfrac{\rho \pi }{2L\: ln \dfrac{R_2}{R_1}}$

-------------