
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
163.5k+ views
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.
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.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Wheatstone Bridge for JEE Main Physics 2025

Charging and Discharging of Capacitor

Instantaneous Velocity - Formula based Examples for JEE
