Question

If a resistance R is melted and recast to half of its length, then the new resistance of the wire will be
A. \[\dfrac{R}{2}\]
B. 2R
C. R
D. \[\dfrac{R}{4}\]

Answer 1

Hint: When the wire is melted and recast then the volume of the material of which the wire is made 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 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,
\[V = Al\]

When the wire is recasted then the volume of the material of which the wire is made remains constant. 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,
\[{V_i} = {V_f}\]
\[\Rightarrow {A_i}{l_i} = {A_f}{l_f}\]
\[\Rightarrow \dfrac{{{A_i}}}{{{A_f}}} = \dfrac{{{l_f}}}{{{l_i}}}\]

Now the wire is melted and recast to half of its length i.e. \[{l_f} = \dfrac{{{l_i}}}{2} \Rightarrow \dfrac{{{l_f}}}{{{i_i}}} = \dfrac{1}{2}\]
So, the ratio of the final area to the initial area of cross-section of the wire is,
\[\dfrac{{{A_f}}}{{{A_i}}} = \dfrac{{{l_i}}}{{{l_f}}} = 2\]
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 R. 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
\[\dfrac{{{R_i}}}{{{R_f}}} = \dfrac{{\dfrac{{\rho {l_i}}}{{{A_i}}}}}{{\dfrac{{\rho {l_f}}}{{{A_f}}}}} \\ \]
\[\Rightarrow \dfrac{R}{{{R_f}}} = \left( {\dfrac{{{A_f}}}{{{A_i}}}} \right) \times \left( {\dfrac{{{l_i}}}{{{l_f}}}} \right) \\ \]
\[\Rightarrow \dfrac{R}{{{R_f}}} = \left( 2 \right) \times \left( 2 \right) \\ \]
\[\Rightarrow \dfrac{R}{{{R_f}}} = 4 \\ \]
\[\therefore {R_f} = \dfrac{R}{4}\]
Hence, the final resistance of the wire is \[\dfrac{R}{4}\].

Therefore, the correct option is D.

Note: The resistivity is the material property. The resistor made of the same material will have the same resistivity irrespective of their dimensions. The resistance depends on the length and area of the cross section of the conductor while resistivity is not.