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(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 100Ω. If the area of cross-section of the wire is 3 x 10–7 m2, calculate the resistivity of the metal.
(a) 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’ is:
$R=ρ\frac{l}{A}$
where,
$R-$ Resistance of the conductor.
$ρ(rho)-$ Resistivity (constant).
$l-$ Length of the conductor.
$A-$ Area of a cross-section of the conductor.
Derivation of the S.I. unit of electrical resistivity.
For resistivity $ρ$, the above expression for resistance can be rearranged as-
$ρ=\frac{R\times A}{l}$
Now, putting the units of $R,\ A$ and $l$ in the expression.
$ρ=\frac{\Omega\ m^2}{m}$
$ρ=\Omega -m$
Thus, the S.I unit of resistivity is Ohm-meter $(\Omega -m)$.
(b) Given:
Resistance, $R=100 \Omega$
Length, $l=5m$
Area, $A=3\times {10^{-7}}m^2$
To find: Resistivity of the metal $ρ$.
Solution:
We know that resistivity, $ρ$ is given as-
$ρ=\frac{R\times A}{l}$
Putting the given values, we get-
$ρ=\frac{100\times 3\times {10^{-7}}}{5}$
$ρ=\frac{300\times {10^{-7}}}{5}$
$ρ=60\times {10^{-7}}\Omega -m$
$ρ=6\times {10}\times {10^{-7}}\Omega -m$
$ρ=6\times {10^{-6}}\Omega -m$
Thus, the resistivity of the metal is $6\times {10^{-6}}\Omega -m$.