
The resistance if a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be:
(A) \[\dfrac{R}{{{n^2}}}\]
(B) \[nR\]
(C) \[\dfrac{R}{n}\]
(D) \[{n^2}R\]
Answer
232.8k+ views
Hint Given that the resistance of wire R has an initial length \[{l_1}\] and is stretched n times to new length \[{l_2}\]. It is observed that there is no change in radius. This means, assume volume is constant. Find a relation between areas of the wire before and after extension and substitute it in the resistance formula \[R = \dfrac{{\rho l}}{A}\].
Complete Step By Step Solution
It is given that there is a wire of radius r, length \[{l_1}\] having a resistance R. Now this wire is said to be extended to a new length \[{l_2}\], which is n times \[{l_1}\]. Since , there is no change in radius of the wire, let us assume that the volume before and after stretching are constant.
\[{V_1} = {V_2}\], where \[{V_1}\] is volume before stretching and \[{V_2}\]is volume after stretching.
Now we know that volume is a product of area and height. In our case, height is the length of the wire
\[{A_1} \times {l_1} = {A_2} \times {l_2}\]
We know that , \[{l_2} = n \times {l_1}\]. Substituting this , we get
\[ \Rightarrow {A_1} \times {l_1} = {A_2} \times n \times {l_1}\]
\[ \Rightarrow \dfrac{{{A_1}}}{n} = {A_2}\]
We know that resistance R of a material is calculated by using the formula\[R = \dfrac{{\rho l}}{A}\], where \[\rho \]is represented as resistivity of the material , \[A\] is area and \[l\]is length of the wire.
Now , before extension , resistivity is given as , \[{R_1} = \dfrac{{\rho {l_1}}}{{{A_1}}}\]
After extension, resistivity is given as, \[{R_2} = \dfrac{{\rho {l_2}}}{{{A_2}}}\]
Substituting for \[{A_2}\]in the equation for\[{R_2}\], we get
\[ \Rightarrow {R_2} = \dfrac{{\rho {l_2} \times n}}{{{A_1}}}\]
Now substituting\[{l_2} = n \times {l_1}\], we get
\[ \Rightarrow {R_2} = \dfrac{{\rho {l_1} \times {n^2}}}{{{A_1}}}\]
(The term \[\dfrac{{\rho {l_1}}}{{{A_1}}}\]is equal to\[{R_1}\] and \[{R_1}\] value given in the question is \[R\])
\[ \Rightarrow {R_2} = R \times {n^2}\]
Thus, Option (d) is the right answer for the given question.
Note Electrical resistivity is defined as the electrical property of a material which defines its strength to oppose electric current.
Complete Step By Step Solution
It is given that there is a wire of radius r, length \[{l_1}\] having a resistance R. Now this wire is said to be extended to a new length \[{l_2}\], which is n times \[{l_1}\]. Since , there is no change in radius of the wire, let us assume that the volume before and after stretching are constant.
\[{V_1} = {V_2}\], where \[{V_1}\] is volume before stretching and \[{V_2}\]is volume after stretching.
Now we know that volume is a product of area and height. In our case, height is the length of the wire
\[{A_1} \times {l_1} = {A_2} \times {l_2}\]
We know that , \[{l_2} = n \times {l_1}\]. Substituting this , we get
\[ \Rightarrow {A_1} \times {l_1} = {A_2} \times n \times {l_1}\]
\[ \Rightarrow \dfrac{{{A_1}}}{n} = {A_2}\]
We know that resistance R of a material is calculated by using the formula\[R = \dfrac{{\rho l}}{A}\], where \[\rho \]is represented as resistivity of the material , \[A\] is area and \[l\]is length of the wire.
Now , before extension , resistivity is given as , \[{R_1} = \dfrac{{\rho {l_1}}}{{{A_1}}}\]
After extension, resistivity is given as, \[{R_2} = \dfrac{{\rho {l_2}}}{{{A_2}}}\]
Substituting for \[{A_2}\]in the equation for\[{R_2}\], we get
\[ \Rightarrow {R_2} = \dfrac{{\rho {l_2} \times n}}{{{A_1}}}\]
Now substituting\[{l_2} = n \times {l_1}\], we get
\[ \Rightarrow {R_2} = \dfrac{{\rho {l_1} \times {n^2}}}{{{A_1}}}\]
(The term \[\dfrac{{\rho {l_1}}}{{{A_1}}}\]is equal to\[{R_1}\] and \[{R_1}\] value given in the question is \[R\])
\[ \Rightarrow {R_2} = R \times {n^2}\]
Thus, Option (d) is the right answer for the given question.
Note Electrical resistivity is defined as the electrical property of a material which defines its strength to oppose electric current.
Recently Updated Pages
Circuit Switching vs Packet Switching: Key Differences Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

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

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

Electricity and Magnetism Explained: Key Concepts & Applications

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding Uniform Acceleration in Physics

Understanding the Electric Field of a Uniformly Charged Ring

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

Derivation of Equation of Trajectory Explained for Students

