Question

A wire of resistance of $10$ ohm is stretched so as to increase its length by $20 \%$. Its resistance then would be:
A.$10$ ohm
B.$12$ ohm
C.$14.4$ ohm
D.$10.2$ ohm

Answer 1

Hint: Resistance is a measure of the opposition of flow of electrons. It is measured in ohms. Resistance of a material is dependent on the length of the material, area of the material and resistivity of the material. When the nature of material will not change then resistivity will remain constant. Then, resistance will depend on length and area of the material.
Formula Used: $R = \rho \dfrac{L}{A}$
Where, R is the resistance of a wire.
$\rho$ is the resistivity of a material.
L is the length of the wire.
A is the cross-sectional area.

Complete answer:
Resistance of wire is $R = \rho \dfrac{L}{A}$.
Let L is the length of the wire.
R = $10$ ohm
Volume of the wire remains constant and V=AL .
Put $A=\dfrac{V}{L}$ in the formula of resistance.
We will get, $R = \rho \dfrac{L^{2}}{V}$.
As the material is the same, resistivity is not changed.
Therefore, $R=L^{2}$.
In the question, given that length is increased by $20 \%$. so, new length $L’ = L +0.20L$
$L’ = 1.20L$
So, by using $R\propto L^{2}$, we get $\dfrac{R’}{R}=\dfrac{ L’^{2}}{ L^{2}} $
By putting the value of R and L’. We get
$R’ = \dfrac{10 \times 1.20L^{2}}{L^{2}}$ = $14.4$ ohm
When length will change by $20 \%$ then resistance will be $14.4$ ohm.

So, option C is correct.

Additional information:
Resistivity is dependent on the nature of the material only. When the nature of material will change then resistivity will no longer be constant and comes into the formula of resistance. When the stretching of wire will take place then length and area will both change to make the volume constant.

Note:
We can solve the above question by-
 As volume is constant, so $A \times L = A’ \times L’$.
$\dfrac{A’}{A}=\dfrac{ L}{ L’} $
Putting the value of $L’ = 1.20L$, we get the ratio of A’ and A.
$\dfrac{A’}{A}=\dfrac{ 1}{ 1.20} $
As resistivity is constant so, by this formula $R = \rho \dfrac{L}{A}$, we get
$R \propto \dfrac{L}{A}$
$ \dfrac{R’}{R} = \dfrac{L’}{L} \times \dfrac{A}{A’}$
$R’ = 1.20 \times 1.20 \times 10 = 14.4$ ohm