Question

A wire of $10$ ohm resistance stretched to thrice its original length. What will be its (a) new resistivity, (b) new resistance.

Answer 1

Hint:We can find the resistivity and resistance of the wire by using the concept of resistivity of the material which states that resistance of the wire is directly proportional to its length and inversely proportional to its cross-sectional area. Also, the volume of the wire always remains.

Complete step by step answer:
(a) We know that resistance of the wire is directly proportional to its length and inversely proportional to its cross-sectional area. So,
$R\alpha \dfrac{l}{A}$
$\therefore R = \dfrac{{\rho l}}{A} - - - - - - - - - - - (1)$
Where, $l$ - length of the wire, $A$ - cross-sectional area of the wire and
$\rho $ - resistivity of the wire (proportional constant)
Resistivity is the property of the material and it will not change with changing the length of the wire, it is always a constant for a particular wire.

(b) Now, the length of the wire is stretched to thrice its original length.
Let us assign some terminologies for better understanding.
$R$ - Original resistance , $R'$ - new resistance
$l$ - original length , $l'$ - new length
$A$ - original area , $A'$ - new area of the wire
$l' = 3l$ and $R = 10\Omega $
Here, the volume of the wire is always remains constant and we know that, the volume is given by
$V = A \times l = A' \times l'$
$\Rightarrow \dfrac{{A'}}{A} = \dfrac{l}{{l'}} = \dfrac{l}{{3l}} = \dfrac{1}{3}$
Using the resistance formula for wire, we have
\[\dfrac{{R'}}{R} = \dfrac{{\dfrac{{\rho l'}}{{A'}}}}{{\dfrac{{\rho l}}{A}}} \\
\Rightarrow \dfrac{{R'}}{R} = \dfrac{{l'}}{l} \times \dfrac{A}{{A'}} \\
\Rightarrow \dfrac{{R'}}{R} = \dfrac{{3l}}{l} \times \dfrac{3}{1}\]
\[\Rightarrow R' = 9R \\
\Rightarrow R' = 9 \times 10\]
$\therefore R' = 90\Omega $

The new resistance of the wire is $90\Omega $.

Note:The resistivity or specific resistance of a material is a characteristic of a material, it depends on temperature also. We should note that the volume of the wire remains constant regardless of its extension. For solving such types of problems, the cross-sectional area of the wire is considered as a circle having area $\pi {r^2}$ , where $r$ is the radius of the wire.