
One-fourth length of a spring of force constant K is cut away. The force constant of the remaining spring will be
A. \[\dfrac{3}{4}K \\ \]
B. \[\dfrac{4}{3}K \\ \]
C. 4K
D. K
Answer
217.5k+ views
Hint: Spring constant gives the stiffness of a spring and is equal to the force needed to stretch the spring divided by the distance that the spring is compressed or stretched.
Formula used:
The relation of spring constant K with length can be given by,
\[K \propto \dfrac{1}{l}\]
Complete step by step solution:
A spring of constant K is cut away by \[\dfrac{1}{4}\] of its length, we have to find the spring constant of the remaining part of the spring. We know that spring constant is inversely proportional to length of the spring and let the original length of spring be l then the relation of spring constant K with length can be given by,
\[K \propto \dfrac{1}{l}\,.......(1)\]
As the \[\dfrac{1}{4}\]of length l of spring is cut away then the remaining length l’ of spring will be,
\[l' = \dfrac{3}{4}l\]
Let the spring constant of remaining length be K’ then according to equation (1) it will be,
\[K' \propto \dfrac{4}{{3l}}\,.......(2)\]
On dividing equation (2) by equation (1) we get,
\[\dfrac{{K'}}{K} = \dfrac{4}{{3l}} \times l \\
\Rightarrow K' = \,\dfrac{4}{3}K\].
Hence, the constant of remaining length will be \[\dfrac{4}{3}K\].
Therefore, option B is the correct answer.
Note: Springs having larger spring constant will have smaller displacement and one having smaller spring constant will have larger displacement, it always has positive magnitude because negative spring constant will mean that when a compressive force is applied to spring it will compress itself further, which is against the nature of a spring.
Formula used:
The relation of spring constant K with length can be given by,
\[K \propto \dfrac{1}{l}\]
Complete step by step solution:
A spring of constant K is cut away by \[\dfrac{1}{4}\] of its length, we have to find the spring constant of the remaining part of the spring. We know that spring constant is inversely proportional to length of the spring and let the original length of spring be l then the relation of spring constant K with length can be given by,
\[K \propto \dfrac{1}{l}\,.......(1)\]
As the \[\dfrac{1}{4}\]of length l of spring is cut away then the remaining length l’ of spring will be,
\[l' = \dfrac{3}{4}l\]
Let the spring constant of remaining length be K’ then according to equation (1) it will be,
\[K' \propto \dfrac{4}{{3l}}\,.......(2)\]
On dividing equation (2) by equation (1) we get,
\[\dfrac{{K'}}{K} = \dfrac{4}{{3l}} \times l \\
\Rightarrow K' = \,\dfrac{4}{3}K\].
Hence, the constant of remaining length will be \[\dfrac{4}{3}K\].
Therefore, option B is the correct answer.
Note: Springs having larger spring constant will have smaller displacement and one having smaller spring constant will have larger displacement, it always has positive magnitude because negative spring constant will mean that when a compressive force is applied to spring it will compress itself further, which is against the nature of a spring.
Recently Updated Pages
Impulse Momentum Theorem Explained: Formula, Examples & Applications

Inertial and Non-Inertial Frames of Reference Explained

Ionisation Energy and Ionisation Potential Explained

Addition of Three Vectors: Methods & Examples

Addition of Vectors: Simple Guide for Students

Algebra Made Easy: Step-by-Step Guide for Students

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

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

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

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

Understanding Atomic Structure for Beginners

