Question

Resistivity of iron is $1 \times {10^{ - 7}}ohmm$. The resistance of the given wire of a particular thickness and length is $1ohm$. If the diameter and length of the wire both are doubled find the resistivity and resistance of the wire.

Answer 1

Hint Keep in mind that resistivity is an intrinsic property of a conductor and depends upon its material. In the expression relating resistance and resistivity of a conductor, substitute the given values to obtain the new resistance of the wire.
Formula used:
$\rho = \dfrac{{RA}}{l}$ where $\rho $is the resistivity, $A$is the area of cross-section, $l$is the length and $R$is the resistance of the conductor.

Complete step by step answer
The ratio of the potential difference to the current is called the electric resistance $R$ of the conductor.
Resistivity or specific resistance on the other hand is defined as the ratio of the intensity of electric field at any point within the conductor and the current density at that point. Resistivity is an intrinsic property of the conductor and therefore does not depend on its size. Thus if the diameter and length of the wire is changed, the value of the resistivity will remain the same.
The expression relating resistance and resistivity together is given as
$
  \rho = \dfrac{{RA}}{l} \\
   \Rightarrow R = \dfrac{{\rho l}}{A} \\
$
Where $R$is the resistance, $\rho $is the resistivity, $A = \pi {r^2}$$ = \dfrac{{\pi {d^2}}}{4}$is the area of cross-section, $r$is the radius, $d$is the diameter and $l$is the length of the wire.
Now in the question it is given that the both diameter and length of the wire is doubled such that
\[d' = 2d\]and $l' = 2l$
Therefore the new area of cross-section will be
$A' = \dfrac{{\pi d{'^2}}}{4} = \dfrac{{\pi {{\left( {2d} \right)}^2}}}{4} = \pi {d^2}$
$ \Rightarrow A = \dfrac{1}{4}A'$
The new length is $l' = 2l$
Therefore, the new resistance can be written as
$R' = \dfrac{{\rho l'}}{{A'}}$
$R' = \dfrac{{\rho 2l}}{{4A}} = \dfrac{{\rho l}}{{2A}} = 0.5R$
Now as the original resistance is $1ohm$so the new resistance becomes,
$R' = 0.5 \times 1 = 0.5ohm$
Therefore, the new resistance becomes half the original resistance.

Note Since the resistivity of a conductor is its intrinsic property, so it can be used to compare conductors of different materials on their ability to conduct electric current. Higher resistivity designates poor conductors and vice versa.