Question

The resistance of a hollow cylindrical copper conductor of length 10m and inner and
outer radius as 2cm and 3cm is: [Pcu = 2 x 10-82 - m]
(A) x 10-32
(B) x 10-22
(C) 10-32
(D) None of these
plss guys help me​

Answer 1

Answer:

R=(ρ)lA=(2.0×10−8(Ω)m)3.0×(π)×10−6m2=2.1×10−3(Ω)

Explanation:

The area of cross section of the conductor through which the charges will flow is

A=(π)(2.0mm)2−(π)(1.0mm)2.

=3.0×(π)mm2.

the resistance of the wire is therefore,

R=(ρ)lA=(2.0×10−8(Ω)m)3.0×(π)×10−6m2=2.1×10−3(Ω).