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Question

A conducting medium is shaped in a form of a quarter annulus of radius b and a (b > a) and thickness t. Resistivity of material is ρ.
[CD = AB = EF = GH = t]

A
Resistance between faces ABCD and EFGH is πρ2t ln(ba)
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B
Resistance between faces BCGF and ADHE is 4ρtπ(b2a2)
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C
Resistance between faces ABFE and CDHG is 2ρπtln(ba)
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D
Resistance between faces ABCD and EFGH is ρπ(a+b)μ(ba)t
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Solution

The correct options are
A Resistance between faces ABCD and EFGH is πρ2t ln(ba)
B Resistance between faces BCGF and ADHE is 4ρtπ(b2a2)
C Resistance between faces ABFE and CDHG is 2ρπtln(ba)
For finding the resistance between faces ABCD and EFGH, let's consider an element of thickness 'dr' as shown in figure.
So, the resistance of the element will be:
dR=ρπr2t(dr)
1R=1dR=baπtdrρπr1R=2tρπln(ba)
R=πρ2t ln(ba)

Similarly, if we consider elements between the faces BCGF and ADHE, we will get R=ρt×4π(b2a2)


Now, for faces ABFE and CDHG, the resistance of the element considered between them will be, dR=ρ drπr2×t
R=dR=2ρπtbadrr=2ρπtln(ba)

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