The length of a rubber cord is \[{l_1}\]meters when the tension in it is \[4N\] and \[{l_2}\]meters when the tension is \[5N\]. then the length in meters when the tension is \[9N\] is
(A) \[3{l_2} + 4{l_1}\]
(B) \[3{l_2} + 2{l_1}\]
(C) \[5{l_2} - 4{l_1}\]
(D) \[3{l_2} - 2{l_1}\]
Answer
253.5k+ views
Hint Let us assume that the original length of the rubber cord to be l meters. When a force of\[4N\]is applied on the cord it changes to \[{l_1}\]and when \[5N\]is applied it changes to \[{l_2}\]. Now using Young's modulus formula, we can relate the tension and change in length from original to find the change in length when extra force is applied.
Complete Step By Step Answer
It is given that a rubber cord , which originally has a length of \[l\]meters, undergoes two different tensions of variable magnitude and undergoes expansion or change in length. Whenever a material undergoes a tension of particular magnitude, we consider it’s tensile property to analyze whether the material is stiff or undergoes elasticity.
To determine the stiffness of any component, we use Young's modulus of elasticity to determine whether the given material will remain stiff or not under applied forces or tension. Mathematically, it can be represented as ratio between stress and strain and also as :
\[E = \dfrac{{TL}}{{A\Delta L}}\], where T is the tension experienced, L is the length of the cord, A is the area of the cord
Applying this for the first condition where the tension experienced is \[4N\]we get,
\[ \Rightarrow E = \dfrac{{4l}}{{A({l_1} - l)}}\]
Applying this for the second and third conditions respectively , we get
\[ \Rightarrow E = \dfrac{{5l}}{{A({l_2} - l)}}\] and \[ \Rightarrow E = \dfrac{{9l}}{{A({l_3} - l)}}\]
Equating all the E values, we get
\[ \Rightarrow E = \dfrac{{4l}}{{A({l_1} - l)}} = \dfrac{{5l}}{{A({l_2} - l)}} = \dfrac{{9l}}{{A({l_3} - l)}}\]
Equate 1 and 3 initially and 2 and 3 separately, so as to find \[{l_3}\],this implies
\[ \Rightarrow E = \dfrac{{4l}}{{A({l_1} - l)}} = \dfrac{{9l}}{{A({l_3} - l)}}\]
\[ \Rightarrow 4{l_3} + 5l = 9{l_1}\]-----(1)
Equating 2 and 3 we get
\[ \Rightarrow E = \dfrac{{5l}}{{A({l_2} - l)}} = \dfrac{{9l}}{{A({l_3} - l)}}\]
\[ \Rightarrow 5{l_3} + 4l = 9{l_2}\]--(2)
On solving 1 and 2 , we get
\[25{l_3} + 20l = 45{l_2}[(2) \times 5]\]
\[16{l_3} + 20l = 36{l_1}[(1) \times 4]\]
Cancelling out common 20l term, we get
\[ \Rightarrow {l_3} = 5{l_2} - 4{l_1}\]
Thus the length of the cord ,when a tension of 9N is applied is \[5{l_2} - 4{l_1}\]
Hence, Option (c) is the right answer for the given question.
Note In general, elasticity of a material is defined as the property of the material to resist distortion due to constant application of tension , undergo deformation and return back to its original shape and size once the force applied is removed.
Complete Step By Step Answer
It is given that a rubber cord , which originally has a length of \[l\]meters, undergoes two different tensions of variable magnitude and undergoes expansion or change in length. Whenever a material undergoes a tension of particular magnitude, we consider it’s tensile property to analyze whether the material is stiff or undergoes elasticity.
To determine the stiffness of any component, we use Young's modulus of elasticity to determine whether the given material will remain stiff or not under applied forces or tension. Mathematically, it can be represented as ratio between stress and strain and also as :
\[E = \dfrac{{TL}}{{A\Delta L}}\], where T is the tension experienced, L is the length of the cord, A is the area of the cord
Applying this for the first condition where the tension experienced is \[4N\]we get,
\[ \Rightarrow E = \dfrac{{4l}}{{A({l_1} - l)}}\]
Applying this for the second and third conditions respectively , we get
\[ \Rightarrow E = \dfrac{{5l}}{{A({l_2} - l)}}\] and \[ \Rightarrow E = \dfrac{{9l}}{{A({l_3} - l)}}\]
Equating all the E values, we get
\[ \Rightarrow E = \dfrac{{4l}}{{A({l_1} - l)}} = \dfrac{{5l}}{{A({l_2} - l)}} = \dfrac{{9l}}{{A({l_3} - l)}}\]
Equate 1 and 3 initially and 2 and 3 separately, so as to find \[{l_3}\],this implies
\[ \Rightarrow E = \dfrac{{4l}}{{A({l_1} - l)}} = \dfrac{{9l}}{{A({l_3} - l)}}\]
\[ \Rightarrow 4{l_3} + 5l = 9{l_1}\]-----(1)
Equating 2 and 3 we get
\[ \Rightarrow E = \dfrac{{5l}}{{A({l_2} - l)}} = \dfrac{{9l}}{{A({l_3} - l)}}\]
\[ \Rightarrow 5{l_3} + 4l = 9{l_2}\]--(2)
On solving 1 and 2 , we get
\[25{l_3} + 20l = 45{l_2}[(2) \times 5]\]
\[16{l_3} + 20l = 36{l_1}[(1) \times 4]\]
Cancelling out common 20l term, we get
\[ \Rightarrow {l_3} = 5{l_2} - 4{l_1}\]
Thus the length of the cord ,when a tension of 9N is applied is \[5{l_2} - 4{l_1}\]
Hence, Option (c) is the right answer for the given question.
Note In general, elasticity of a material is defined as the property of the material to resist distortion due to constant application of tension , undergo deformation and return back to its original shape and size once the force applied is removed.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Circuit Switching vs Packet Switching: Key Differences Explained

Mass vs Weight: Key Differences Explained for Students

[Awaiting the three content sources: Ask AI Response, Competitor 1 Content, and Competitor 2 Content. Please provide those to continue with the analysis and optimization.]

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

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

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

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Other Pages
JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

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

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

