Question

The resistance of a 5 cm long wire is \[10\,\Omega \]. It is uniformly stretched so that its length becomes 20 cm. The resistance of the wire is
A. \[160\,\Omega \]
B. \[80\,\Omega \]
C. \[40\,\Omega \]
D. \[20\,\Omega \]

Answer 1

Hint:When the wire is stretched then the volume of the material of which the wire is made of remains constant as well as the resistivity of the material of the wire.

Formula used:
\[R = \dfrac{{\rho l}}{A}\]
where R is the resistance of the wire of length l and cross-sectional area A, \[\rho \] is the resistivity of the material of the wire.

Complete step by step solution:
Resistance of the wire is given as \[R = \dfrac{{\rho l}}{A}\].
The volume of the wire is the product of the length and the cross-section.​
Also volume of wire having length l and the area of cross-section A will be,
\[V = Al\]
If the final length is \[{l_f}\] and cross-section is \[{A_f}\] then the final volume will be equal to the initial volume of the wire,
\[{V_i} = {V_f}\]
\[\Rightarrow {A_i}{l_i} = {A_f}{l_f}\]
\[\Rightarrow \dfrac{{{A_i}}}{{{A_f}}} = \dfrac{{{l_f}}}{{{l_i}}}\]

Initially the length of the wire is 5 cm.
Now the wire is stretched then the length of the wire is 20 cm.
\[{l_f} = 4{l_i} \Rightarrow \dfrac{{{l_f}}}{{{i_i}}} = 4\]
So, the ratio of the final area to the initial area of cross-section of the wire is,
\[\dfrac{{{A_i}}}{{{A_f}}} = \dfrac{{{l_f}}}{{{l_i}}} = 4\]
The initial resistance of the wire is,
\[{R_i} = \dfrac{{\rho {l_i}}}{{{A_i}}}\]
The final resistance of the wire is,
\[{R_f} = \dfrac{{\rho {l_f}}}{{{A_f}}}\]
The initial resistance of the wire is given as \[10\Omega \]. we need to find the final resistance of the wire.

Taking the ratio of the initial resistance of the wire to the final resistance, we get
\[\dfrac{{{R_i}}}{{{R_f}}} = \dfrac{{\dfrac{{\rho {l_i}}}{{{A_i}}}}}{{\dfrac{{\rho {l_f}}}{{{A_f}}}}} \\ \]
\[\Rightarrow \dfrac{{10\Omega }}{{{R_f}}} = \left( {\dfrac{{{A_f}}}{{{A_i}}}} \right) \times \left( {\dfrac{{{l_i}}}{{{l_f}}}} \right) \\ \]
\[\Rightarrow \dfrac{{10\Omega }}{{{R_f}}} = \left( {\dfrac{1}{4}} \right) \times \left( {\dfrac{1}{4}} \right) \\ \]
\[\Rightarrow \dfrac{{10\Omega }}{{{R_f}}} = \dfrac{1}{{16}} \\ \]
\[\therefore {R_f} = 160\Omega \]
Hence, the final resistance of the wire is \[160\Omega \].

Therefore, the correct option is A.

Note: The material property is resistivity. Regardless of size, a resistor manufactured of the same material will have the same resistance. Resistance is proportional to the length and area of the conductor's cross section, but resistivity is not.