
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
164.4k+ views
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.
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.
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

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

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

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
