
Two wires of the same material are given. The first wire is twice as long as the second and has twice the diameter of the second. The resistance of the first will be.
A. Twice of the second
B. Half of the second
C. Equal of the second
D. Four times of the second
Answer
164.1k+ views
Hint:For this question we will use the relationship between the resistance, the specific resistance (or the resistivity), the length and the area of cross section of the conductor (here, the wire). We will write the expression of area in terms of the diameter of the wire because we are given the information about diameter of the wire in this question. Then we will substitute these values in the formula for both the cases to get the required answer.
Formula used:
Resistance, $R = \dfrac{{\rho l}}{A}$
where $\rho $ is the specific resistance (or resistivity), $l$ is the length of the conductor (here, wire), and $A$ is the area of the cross section of the wire.
Area of cross section of the wire,
\[A = \pi {r^2}\]
where \[r\] is the radius of the cross section of the wire.
And $r = \dfrac{d}{2}$ where $d$ is the diameter of the cross section of the wire.
Complete step by step solution:
Let the subscripts $1$ and $2$ be used for ${1^{st}}$ and ${2^{nd}}$ wires respectively.
Given: ${l_1} = 2{l_2}$ and ${d_1} = 2{d_2}$ ...(1)
Since, $A \propto {r^2}$ and $r = \dfrac{d}{2} \\ $
This implies that $A \propto {d^2}$ ...(2)
Also, $R = \dfrac{{\rho l}}{A} \\ $ ...(3)
Which means $R \propto l$ and $R \propto \dfrac{1}{{{d^2}}}$ (from equation (2) and (3))
Thus, we get, ${R_1} \propto \dfrac{{{l_1}}}{{{d_1}^2}} \\ $...(4)
And ${R_2} \propto \dfrac{{{l_2}}}{{{d_2}^2}}$ ...(5)
The value of $\rho $ is taken the same because it has the same value for a given material at a given temperature. Substituting equation (1) in equation (4), we get,
${R_1} \propto \dfrac{{2{l_2}}}{{{{\left( {2{d_2}} \right)}^2}}} \\ $
Solving this, we get,
${R_1} \propto \dfrac{{{l_1}}}{{2{d_1}^2}} \\ $ ...(6)
Dividing the equations (5) and (6), we get,
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\left( {\dfrac{{{l_2}}}{{2{d_2}^2}}} \right)}}{{\left( {\dfrac{{{l_2}}}{{{d_2}^2}}} \right)}} \\ $
Thus, $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{1}{2} \\ $
This implies that, ${R_1} = \dfrac{1}{2}{R_2}$
Hence, option B is the correct answer.
Note: While solving this question an important point to keep in mind is that the value of resistivity should be taken the same because both the wires are made of the same material. We can also solve this question by equating the value of resistivity obtained by putting the values of the known physical quantities for both the wires in the formula used.
Formula used:
Resistance, $R = \dfrac{{\rho l}}{A}$
where $\rho $ is the specific resistance (or resistivity), $l$ is the length of the conductor (here, wire), and $A$ is the area of the cross section of the wire.
Area of cross section of the wire,
\[A = \pi {r^2}\]
where \[r\] is the radius of the cross section of the wire.
And $r = \dfrac{d}{2}$ where $d$ is the diameter of the cross section of the wire.
Complete step by step solution:
Let the subscripts $1$ and $2$ be used for ${1^{st}}$ and ${2^{nd}}$ wires respectively.
Given: ${l_1} = 2{l_2}$ and ${d_1} = 2{d_2}$ ...(1)
Since, $A \propto {r^2}$ and $r = \dfrac{d}{2} \\ $
This implies that $A \propto {d^2}$ ...(2)
Also, $R = \dfrac{{\rho l}}{A} \\ $ ...(3)
Which means $R \propto l$ and $R \propto \dfrac{1}{{{d^2}}}$ (from equation (2) and (3))
Thus, we get, ${R_1} \propto \dfrac{{{l_1}}}{{{d_1}^2}} \\ $...(4)
And ${R_2} \propto \dfrac{{{l_2}}}{{{d_2}^2}}$ ...(5)
The value of $\rho $ is taken the same because it has the same value for a given material at a given temperature. Substituting equation (1) in equation (4), we get,
${R_1} \propto \dfrac{{2{l_2}}}{{{{\left( {2{d_2}} \right)}^2}}} \\ $
Solving this, we get,
${R_1} \propto \dfrac{{{l_1}}}{{2{d_1}^2}} \\ $ ...(6)
Dividing the equations (5) and (6), we get,
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\left( {\dfrac{{{l_2}}}{{2{d_2}^2}}} \right)}}{{\left( {\dfrac{{{l_2}}}{{{d_2}^2}}} \right)}} \\ $
Thus, $\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{1}{2} \\ $
This implies that, ${R_1} = \dfrac{1}{2}{R_2}$
Hence, option B is the correct answer.
Note: While solving this question an important point to keep in mind is that the value of resistivity should be taken the same because both the wires are made of the same material. We can also solve this question by equating the value of resistivity obtained by putting the values of the known physical quantities for both the wires in the formula used.
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

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

Instantaneous Velocity - Formula based Examples for JEE
