Question

When the resistance wire is passed through a die the cross-section area decreases by 1%, the change in resistance of the wire is:
A) 1% decrease
B) 1% increase
C) 2% decrease
D) 2% increase

Answer 1

Hint: Resistance of a wire depends on its length, area of cross section and resistivity. Resistivity depends on the material. Resistance changes when the length or area of cross section changes.

Complete step by step solution:Since it is clear from the above hint that resistance changes with change in length or area. They are related by the following equation:
\[R=\dfrac{\rho L}{A}\]
Where \[R\] represents the resistance
\[\rho \] represents the resistivity of the material used in the wire
\[L\] represents the Length of the wire
\[A\] represents the cross-section area of the wire
Also we know that by definition volume is area times length, this can be represented as follows:
\[V=A.L\]
Where \[V\] represents the volume.
From above equation we can substitute length in terms of volume, as follows
\[L=\dfrac{V}{A}\]
It is to be noted that, when the wire is elongated or contracted the volume doesn’t changes i.e. the volume will remain constant. Hence we have converted the equation of resistance in terms of volume and cross-section area.
The new equation of resistance will be
\[R=\dfrac{\rho }{A}\left( \dfrac{V}{A} \right)\]
\[R=\dfrac{\rho V}{{{A}^{2}}}\]
Let’s call this resistance as initial resistance and denote it by \[{{R}_{initial}}\]
\[\therefore {{R}_{initial}}=\dfrac{\rho V}{{{A}^{2}}}\]
Now the cross section area decreases by \[1%\] , therefore the new cross-section area will be \[0.99%A\]
Let’s put this values in the equation
\[\therefore {{R}_{final}}=\dfrac{\rho V}{0.99{{A}^{2}}}\]
Therefore the change will be given as
\[\Rightarrow {{R}_{final}}-{{R}_{initial}}=\dfrac{\rho V}{{{(0.99A)}^{2}}}-\dfrac{\rho V}{{{A}^{2}}}\]
\[\Rightarrow {{R}_{final}}-{{R}_{initial}}=\dfrac{\rho V}{{{A}^{2}}}\left[ \dfrac{1}{{{(0.99)}^{2}}}-1 \right]\]
Percentage change = $\dfrac{R_{final}-R_{initial}}{R_{initial}} \times 100 %$
$\Rightarrow (\dfrac{1}{{{(0.99)}^{2}}} - 1)\times 100$
$\Rightarrow$ $\approx$ $2 % $

Therefore there is an increase of 2% in the final resistance of the wire.

Additional Information: The length is dependent on temperature i.e. with increase in temperature the length of metallic wires tend to increase. Resistivity of a material is intrinsic property.

Note: Do not mark the answer as 2% decrease. While solving the problem we calculated the value of final resistance minus the initial resistance and the value was positive hence final resistance is more than initial resistance. It can also be noticed since the cross-sectional area is inversely proportional to resistance.