Question

The length of the wire is doubled. Its conductance will be
A. Unchanged
B. Halved
C. Quadrupled
D. \[\dfrac{1}{4}\] of the original value

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. The conductance is the reciprocal of the resistance, and conductivity is the reciprocal of the resistivity.

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

Complete step by step solution:
Let the initial length of the wire is \[{l_1}\] and the area of cross section is \[{A_1}\]. Then the initial conductance of the wire is,
\[{C_1} = \dfrac{{\sigma {A_1}}}{{{l_1}}}\]
When the final length of the wire is \[{l_2}\]and the area of cross section \[{A_2}\]. Then the final conductance of the wire is,
\[{C_2} = \dfrac{{\sigma {A_2}}}{{{l_2}}}\]

As it is given that the length is getting doubled keeping the area of cross section constant.So,
\[{l_2} = 2{l_1}\]and \[{A_2} = {A_1}\]
Dividing the expression for both the conductance, we get
\[\dfrac{{{C_1}}}{{{C_2}}} = \left( {\dfrac{{{A_1}}}{{{A_2}}}} \right) \times \left( {\dfrac{{{l_2}}}{{{l_1}}}} \right) \\ \]
\[\Rightarrow \dfrac{{{C_1}}}{{{C_2}}} = \left( {\dfrac{{{A_1}}}{{{A_1}}}} \right) \times \left( {\dfrac{{2{l_1}}}{{{l_1}}}} \right) \\ \]
\[\Rightarrow \dfrac{{{C_1}}}{{{C_2}}} = 2 \\ \]
\[\therefore {C_2} = \dfrac{{{C_1}}}{2}\]
Hence, the final conductance of the wire is half of the initial value of the conductance of the wire.

Therefore, the correct option is B.

Note: The specific conductivity is the material property. The resistor made of the same material will have the same value of specific conductivity irrespective of their dimensions. We should be careful while setting up the relation for the final dimension of the wire. As in the question it was not mentioned that the wire is stretched, so the area of cross-section will remain as before. If the wire is stretched then the area of cross-section decreases to maintain the volume as before.