Question

A metallic wire of resistance 20 ohm stretched until its length is doubled. Its resistance is
A. $20\Omega$
B. $40\Omega$
C. $80\Omega$
D. $60\Omega$

Answer 1

Metallic wire allows the flow of current when potential difference is applied across the ends of the wire. That wire will be having resistance and heat resistance depends on the type of material and the length and the cross section area of the conductor through which current flows.

Formula used:
$R = \dfrac{{\rho L}}{A}$

Complete step by step solution:
Flow of a quantity with time is known as a current. It can be fluid current or heat current or electric current.
In case of fluid current a pressure difference is maintained and that drives the flow of fluid and fluid always flows from high pressure region to the low pressure region and fluid current is governed by fluid resistance too
Whereas in electric current the electric charge flows with time and the voltage difference and electric resistance combined will govern the electric current. Charge flows from higher voltage to lower voltage naturally.
Similarly in thermal current it is governed by temperature difference and thermal resistance.
Electrical resistance of a conductor of length ‘L’ and cross sectional area ‘A’ and electrical resistivity $\rho$ is given by $R = \dfrac{{\rho L}}{A}$
When the wire is stretched the volume of the wire will be constant as length increases and cross sectional area decreases. By multiplying and dividing with length in the resistor formula we can get the dependence of resistance on length and volume of conductor.
$R = \dfrac{{\rho L}}{A}$
\eqalign{ & \Rightarrow R = \dfrac{{\rho L}}{A} \times \dfrac{L}{L} \cr & \Rightarrow R = \dfrac{{\rho {L^2}}}{V} \cr}
Where ‘V’ is the volume of the conductor.
So resistance will be proportional to square of length as volume is constant.
\eqalign{ & R\propto {L^2} \cr & \Rightarrow {R_1} = k{L^2} \cr & \Rightarrow {R_2} = k{\left( {2L} \right)^2} \cr & \Rightarrow {R_2} = 4{R_1} \cr & \Rightarrow {R_2} = 4\left( {20} \right) \cr & \therefore {R_2} = 80\Omega \cr}

So, the correct answer is “Option C”.

Note: If the wire is not stretched and if it is cut into half, then the volume is not constant. Area of the cross section will be constant and the length will be reduced to half. Then the resistance will be proportional to length only, not to the square of the length. So for these kinds of problems, first we should know which term will be constant and then proceed.