
Define 1 Ohm resistance. A student has a resistance wire of 1 Ohm. If the length of this wire is 50cm, to what length he should stretch it uniformly so as to obtain a wire of 4 Ohms resistance? Justify your answer.
Answer
154.5k+ views
Hint: From ohm’s law, the definition of 1 ohm resistance can be defined. The resistance of a substance is directly proportional to its length.
Formula used: In this solution we will be using the following formulae;
\[R = \dfrac{V}{I}\] where \[R\] is resistance, \[V\] is voltage and \[I\] is current.
\[R = \rho \dfrac{l}{A}\] where \[\rho \] is the resistivity of the wire, \[l\] is the length and \[A\] is the cross sectional area.
Complete Step-by-Step Solution:
From the ohm law equation, the definition of 1 ohm, could be derived.
The ohm’s law equation is can be written as
\[R = \dfrac{V}{I}\] where \[R\] is resistance, \[V\] is voltage and \[I\] is current.
Hence, for 1 ohms, we can write
\[1\Omega = \dfrac{{1{\text{V}}}}{{1{\text{A}}}}\]
Hence, 1 ohm’s could be defined as the resistance to a 1 A current which causes 1 volt potential difference or voltage drop. Or it can be defined as the resistance value which would allow only 1 A of current to flow when the component is connected to a 1 volt source.
For the second part of the question, a student possesses a resistance wire of 1 ohms which has a length 50 cm. to what length should he stretch it uniformly to obtain a resistance wire of 4 ohms.
Recall that the resistance of a wire can be given as
\[R = \rho \dfrac{l}{A}\] where \[\rho \] is the resistivity of the wire, \[l\] is the length and \[A\] is the cross sectional area.
Hence, if the all other properties are kept constant,
\[\dfrac{R}{l} = k\] where \[k\] is a constant.
Then,
\[\dfrac{{{R_1}}}{{{l_1}}} = \dfrac{{{R_2}}}{{{l_2}}}\]
\[ \Rightarrow \dfrac{1}{{50}} = \dfrac{4}{{{l_2}}}\]
Hence, by cross multiplication,
\[{l_2} = 50 \times 4 = 250cm\]
Hence, the length of the wire has to be stretched to 250 cm.
Note: In actuality, the wire would not need to stretch that far before the resistance actually attains 4 ohms as desired. This is due to the Poisson phenomenon, which is the fact that as the length stretches, the cross sectional area of the wire will reduce, hence creating a double effect of increment of resistivity (the smaller the cross sectional area, the lower the resistance).
Formula used: In this solution we will be using the following formulae;
\[R = \dfrac{V}{I}\] where \[R\] is resistance, \[V\] is voltage and \[I\] is current.
\[R = \rho \dfrac{l}{A}\] where \[\rho \] is the resistivity of the wire, \[l\] is the length and \[A\] is the cross sectional area.
Complete Step-by-Step Solution:
From the ohm law equation, the definition of 1 ohm, could be derived.
The ohm’s law equation is can be written as
\[R = \dfrac{V}{I}\] where \[R\] is resistance, \[V\] is voltage and \[I\] is current.
Hence, for 1 ohms, we can write
\[1\Omega = \dfrac{{1{\text{V}}}}{{1{\text{A}}}}\]
Hence, 1 ohm’s could be defined as the resistance to a 1 A current which causes 1 volt potential difference or voltage drop. Or it can be defined as the resistance value which would allow only 1 A of current to flow when the component is connected to a 1 volt source.
For the second part of the question, a student possesses a resistance wire of 1 ohms which has a length 50 cm. to what length should he stretch it uniformly to obtain a resistance wire of 4 ohms.
Recall that the resistance of a wire can be given as
\[R = \rho \dfrac{l}{A}\] where \[\rho \] is the resistivity of the wire, \[l\] is the length and \[A\] is the cross sectional area.
Hence, if the all other properties are kept constant,
\[\dfrac{R}{l} = k\] where \[k\] is a constant.
Then,
\[\dfrac{{{R_1}}}{{{l_1}}} = \dfrac{{{R_2}}}{{{l_2}}}\]
\[ \Rightarrow \dfrac{1}{{50}} = \dfrac{4}{{{l_2}}}\]
Hence, by cross multiplication,
\[{l_2} = 50 \times 4 = 250cm\]
Hence, the length of the wire has to be stretched to 250 cm.
Note: In actuality, the wire would not need to stretch that far before the resistance actually attains 4 ohms as desired. This is due to the Poisson phenomenon, which is the fact that as the length stretches, the cross sectional area of the wire will reduce, hence creating a double effect of increment of resistivity (the smaller the cross sectional area, the lower the resistance).
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation

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

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

Trending doubts
If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Explain the construction and working of a GeigerMuller class 12 physics JEE_Main

Wheatstone Bridge for JEE Main Physics 2025

The force of interaction of two dipoles if the two class 12 physics JEE_Main

Three charges sqrt 2 mu C2sqrt 2 mu Cand sqrt 2 mu class 12 physics JEE_Main

The potential of A is 10V then the potential of B is class 12 physics JEE_Main

Other Pages
JEE Advanced 2025 Revision Notes for Mechanics

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

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

An uncharged sphere of metal is placed inside a charged class 12 physics JEE_Main

Three mediums of refractive indices mu 1mu 0 and mu class 12 physics JEE_Main

A signal of 5kHz frequency is amplitude modulated on class 12 physics JEE_Main
