Question

If a wire is stretched to make it 0.1 % longer, its resistance will
A. increase by 0.05 %
B. increase by 0.2 %
C. decrease by 0.2 %
D. decrease by 0.05 %

Answer 1

Hint: The length of the wire after stretching is calculated in terms of the length of the wire before stretching. In the same way, the area of the wire after stretching is also calculated in terms of the area of the wire before stretching. Then, the equation for resistance is applied. Finally, by substituting all these factors, the rate of change of resistance is obtained.

Complete answer:
Let the initial length of the wire before stretching be L1
After stretching, the length of the wire is increased by 0.1 %
Let the length of the wire after stretching be L2
Similarly,
Let the initial area of the wire before stretching be A1
After stretching,
let the area of the wire be A2
Let the initial resistance of the wire before stretching be R1
After stretching,
let the resistance of the wire be R2
Let the initial volume of the wire before stretching be V1
After stretching,
 let the volume of the wire be V2
Since the length of the wire is increased by 0.1 % after stretching,
the length of the wire after stretching is equal to the sum of initial length of the wire before stretching and 0.1 % of initial length of the wire before stretching
That is,
\[{{L}_{2}}~=\text{ }{{L}_{1}}+\text{ }0.1\text{ }%\text{ }{{L}_{1}}\]----(1)
Let this be equation (1)
From equation (1), we get
\[{{L}_{2}}~=\text{ }1.1\text{ }%\text{ }{{L}_{1}}\]----(2)
Let this be equation (2)
Even if there is an increase in length, the volume remains constant.
Therefore,
The initial volume of the wire before stretching is equal to the volume of the wire after stretching.
That is,
\[{{V}_{2}}=\text{ }{{V}_{1}}~\]---(3)
Let this be equation (3)
Substituting the values of area and length before and after stretching in equation (3)
We get
\[{{A}_{2}}{{L}_{2}}~=\text{ }{{A}_{1}}{{L}_{1}}\]----(4)
Let this be equation (4)
Substituting the values of length before and after stretching in equation (4)
We get,
\[{{A}_{2}}{{L}_{2}}=\text{ }{{A}_{1}}\text{x }1.1\text{ }%\text{ }{{L}_{1}}\]---(5)
Let this be equation (5)
Cancelling L1 on both sides,
We get,
\[{{A}_{2}}=\text{ }{{A}_{1}}\text{x }1.1\text{ }%\]---(6)
Let this be equation (6)
The new resistance is expressed as
\[\text{ }{{R}_{2}}=\text{ }\dfrac{\rho {{L}_{2}}}{{{A}_{2}}}\]---(7)
Where,
ρ is the density
Let this be equation (7)
Substitute the values of L2 and A2 in equation (7)
We get,
\[\text{ }{{R}_{2}}=\text{ }\dfrac{1.1%\rho {{L}_{1}}}{(\dfrac{{{A}_{1}}}{1.1%})}\]
\[{{R}_{2}}=\text{ }\dfrac{{{(1.1)}^{2}}\rho {{L}_{1}}}{{{A}_{1}}}\]---(8)
Let this be equation (8)
We know that
\[{{R}_{1}}=\dfrac{\rho {{L}_{1}}}{{{A}_{1}}}\] ---(9)
 Let this be equation (9)
Substitute equation (9) in equation (8),
We get,
\[{{R}_{2}}={{(1.1)}^{2}}{{R}_{1}}\]-----(10)
Let this be equation (10)
Equation (10) can be written as
\[{{R}_{2}}=1.002{{R}_{1}}\]---(11)
 Let this be equation (11)
The percentage change in resistance is given by,
\[percentage\text{ }change\text{ }in\text{ }resistance=\dfrac{{{R}_{2}}-{{R}_{1}}}{{{R}_{1}}}\text{x}100%\]----(12)
Let this be equation (11)
Substitute the value of R2 in equation (11)
We get,
\[percentage\text{ }change\text{ }in\text{ }resistance=\dfrac{1.002{{R}_{1}}-{{R}_{1}}}{{{R}_{1}}}\text{x}100%\]
\[percentage\text{ }change\text{ }in\text{ }resistance=0.2%\]
Therefore, the resistance is increased by 0.2%

Therefore, the correct answer is
B. increase by 0.2 %

Note:
Even if the length, area and resistance of the wire changes, certain factors remain constant in the wire no matter how much it is stretched. Such factors include volume and density of the wire.This principle is applicable even if shrinkage takes place, though shrinkage is mostly impossible in a wire.