
Resistance of a conductor of length $75\,cm$ is $3.25\Omega $. What will be the length of a similar conductor whose resistance is $13.25\Omega $ $?$
(A) $305.76\,cm$
(B) $503.76\,cm$
(C) $200\,cm$
(D) $610\,cm$
Answer
147k+ views
Hint We know the value of conductor length and resistance of any conductive material here, so by using this formula we calculate the end-to-end resistance of a conductor given length and area, we also know that specific resistance is given in units of ohms for materials.
Useful formula
Resistance of conductor is given below,
$R = \rho * \left( {\dfrac{L}{A}} \right)$
$R = $ resistance in ohms
$\rho = $ material resistivity in ohms per meter
$L = $ conductor length in meters
$A = $ cross-sectional area in square meters
Complete step by step procedure
Given by,
Resistance of a conductor of length $75\,cm$is \[3.25{\text{ }}ohm\]
\[R = 3.25{\text{ }}ohm\] and \[I = 75\,cm\]
Let length of another conductor whose resistance is \[16.25{\text{ }}ohm\]be $x$
\[R{\text{ }} = {\text{ }}16.25{\text{ }}ohm\] \[I = \,x\,cm\]
A conductor's resistance and the conductor's length are directly proportional to one another. The resistance of the conductor also increases as the length of the conductor increases.
The relationship between a conductor's length and resistance is given as $R = \rho * \left( {\dfrac{L}{A}} \right)$
The material of the conductor remains the same, so there is constant resistivity $(\rho )$.
According to that,
The given value in substituting the equation.
So,
$3.25 = \rho \dfrac{{75}}{A}$……………………………. $(1)$
For the resistance to be $13.25$ ohms.
The equation is written as,
$13.25 = \rho \dfrac{x}{A}$……………………………$(2)$
Solving the both equations,
We get,
\[x = 305.76\,cm\]
Hence,
The length of the conductor for it have $13.25\Omega $ resistance is $305.76\,cm$
Thus, option A is the correct answer.
Note As electrons pass through a conductor, such as a metal wire, an electrical current flow. With the ions in the metal, the moving electrons will collide. It makes it harder for the current to flow. All materials have some resilience, but some materials are more or less resistant to electric current flow than other materials.
Useful formula
Resistance of conductor is given below,
$R = \rho * \left( {\dfrac{L}{A}} \right)$
$R = $ resistance in ohms
$\rho = $ material resistivity in ohms per meter
$L = $ conductor length in meters
$A = $ cross-sectional area in square meters
Complete step by step procedure
Given by,
Resistance of a conductor of length $75\,cm$is \[3.25{\text{ }}ohm\]
\[R = 3.25{\text{ }}ohm\] and \[I = 75\,cm\]
Let length of another conductor whose resistance is \[16.25{\text{ }}ohm\]be $x$
\[R{\text{ }} = {\text{ }}16.25{\text{ }}ohm\] \[I = \,x\,cm\]
A conductor's resistance and the conductor's length are directly proportional to one another. The resistance of the conductor also increases as the length of the conductor increases.
The relationship between a conductor's length and resistance is given as $R = \rho * \left( {\dfrac{L}{A}} \right)$
The material of the conductor remains the same, so there is constant resistivity $(\rho )$.
According to that,
The given value in substituting the equation.
So,
$3.25 = \rho \dfrac{{75}}{A}$……………………………. $(1)$
For the resistance to be $13.25$ ohms.
The equation is written as,
$13.25 = \rho \dfrac{x}{A}$……………………………$(2)$
Solving the both equations,
We get,
\[x = 305.76\,cm\]
Hence,
The length of the conductor for it have $13.25\Omega $ resistance is $305.76\,cm$
Thus, option A is the correct answer.
Note As electrons pass through a conductor, such as a metal wire, an electrical current flow. With the ions in the metal, the moving electrons will collide. It makes it harder for the current to flow. All materials have some resilience, but some materials are more or less resistant to electric current flow than other materials.
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

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

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

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

Electric field due to uniformly charged sphere class 12 physics JEE_Main

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

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

Degree of Dissociation and Its Formula With Solved Example for JEE

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

Ideal and Non-Ideal Solutions Raoult's Law - JEE
