Question

(a) Write the relation between resistance and electrical resistivity of the material of a conductor in the shape of a cylinder of length 'l' and area of cross-section 'A'. Hence derive the S.I. unit of electrical resistivity.
(b) Resistance of a metal wire of length 5 m is 100 ohm.If the area of cross-section of the wire is 3 ×10^-7 m^2 ,calculate the resistivity of the metal.

Answer 1

You can see the answer in the image

Answer 2

Answer:

(a). The unit of resistivity is $\Omega-m$

(b). The resistivity of the metal is $\rho= 6\times 10^{-6}\ \Omega-m$

Explanation:

(a). Relation between resistance and electrical resistivity of the material of conductor  in the shape of a cylinder of length and area of cross section  is

$R =\dfrac{\rho l}{A}$

Here, R = resistance

$\rho=electrical resistivity of material</p><p>l = length </p><p>A = area of cross-section</p><p>The unit of electrical resistivity is </p><p>[tex]R =\dfrac{\rho l}{A}$

$\rho = \dfrac{RA}{l}$.....(I)

Put the unit of R,A and l in equation (I)

$\rho= \dfrac{\Omega-m^2}{m}$

$\rho= \Omega-m$

Hence, The unit of resistivity is $\Omega-m$.

(b). Given that,

Resistance R = 100 ohm

Length l =5 m

Area $A =3\times10^{-7}\ m^2$

The resistivity of the metal is

$\rho = \dfrac{RA}{l}$

$\rho = \dfrac{100\times3\times10^{-7}}{5}$

$\rho=0.000006\ \Omega-m$

$\rho=6\times10^{-6}\ \Omega-m$

Hence, The resistivity of the metal is $\rho= 6\times 10^{-6}\ \Omega-m$