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Question: When the resistance wire is passed through a die the cross-section area decreases by 1%, the change ...

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

Explanation

Solution

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=ρLAR=\dfrac{\rho L}{A}
Where RR represents the resistance
ρ\rho represents the resistivity of the material used in the wire
LL represents the Length of the wire
AA 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.LV=A.L
Where VV represents the volume.
From above equation we can substitute length in terms of volume, as follows
L=VAL=\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=ρA(VA)R=\dfrac{\rho }{A}\left( \dfrac{V}{A} \right)
R=ρVA2R=\dfrac{\rho V}{{{A}^{2}}}
Let’s call this resistance as initial resistance and denote it by Rinitial{{R}_{initial}}
Rinitial=ρVA2\therefore {{R}_{initial}}=\dfrac{\rho V}{{{A}^{2}}}
Now the cross section area decreases by 11% , therefore the new cross-section area will be 0.990.99%A
Let’s put this values in the equation
Rfinal=ρV0.99A2\therefore {{R}_{final}}=\dfrac{\rho V}{0.99{{A}^{2}}}
Therefore the change will be given as
RfinalRinitial=ρV(0.99A)2ρVA2\Rightarrow {{R}_{final}}-{{R}_{initial}}=\dfrac{\rho V}{{{(0.99A)}^{2}}}-\dfrac{\rho V}{{{A}^{2}}}
RfinalRinitial=ρVA2[1(0.99)21]\Rightarrow {{R}_{final}}-{{R}_{initial}}=\dfrac{\rho V}{{{A}^{2}}}\left[ \dfrac{1}{{{(0.99)}^{2}}}-1 \right]
Percentage change = RfinalRinitialRinitial×100\dfrac{R_{final}-R_{initial}}{R_{initial}} \times 100 %
(1(0.99)21)×100\Rightarrow (\dfrac{1}{{{(0.99)}^{2}}} - 1)\times 100
\Rightarrow \approx 22 %

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.