
The mass per unit length of a non uniform rod of length L varies as \[{\text{m} = {\lambda x}}\]. Where \[\lambda\] is a constant. The center of mass of the rod will lie at
(A) \[\dfrac{{\text{2}}}{{\text{3}}}{\text{L}}\]
(B) \[\dfrac{3}{2}{\text{L}}\]
(C) \[\dfrac{1}{2}{\text{L}}\]
(D) \[\dfrac{4}{3}{\text{L}}\]
Answer
218.7k+ views
Hint The center of mass of a body is the summation of mass of the body multiplied by their respective distances divided by the total mass of the body
Complete step-by-step solution
We know that position of center of mass:
\[{{\text{X}}_{{\text{cm}}}}{\text{ = }}\dfrac{{\smallint {\text{xdm}}}}{{\smallint {\text{dm}}}}\]
Where x is the distance of dm mass from one end of the body.
In the question, we are given that the body is non uniform and its mass distribution is given by \[{\text{m} = {\lambda x}}\], differentiating the equation we get,
\[{\text{dm} = {\lambda dx}}\],
\[{{\text{X}}_{{\text{cm}}}}{\text{ = }}\dfrac{{\smallint {\text{x}\lambda dx}}}{{\smallint {\lambda dx}}}\]
\[{{\text{X}}_{{\text{cm}}}}{\text{ = }}\dfrac{{\dfrac{{{\text{|}}{{\text{x}}^{\text{2}}}{\text{|}}_{\text{0}}^{\text{L}}}}{{\text{2}}}}}{{{\text{|x|}}_{\text{0}}^{\text{L}}}}{\text{ = }}\dfrac{{{{\text{L}}^{\text{2}}}}}{{{\text{2L}}}}{\text{ = }}\dfrac{{\text{L}}}{{\text{2}}}\]
Therefore, the correct answer is option D.
Note Whenever we have to find the center of mass of a non uniform section, we have to follow this procedure. In case you are given masses of different bodies and their respective location, then you have to use the summation symbol but if the body is continuous, we have to use the integration.
Complete step-by-step solution
We know that position of center of mass:
\[{{\text{X}}_{{\text{cm}}}}{\text{ = }}\dfrac{{\smallint {\text{xdm}}}}{{\smallint {\text{dm}}}}\]
Where x is the distance of dm mass from one end of the body.
In the question, we are given that the body is non uniform and its mass distribution is given by \[{\text{m} = {\lambda x}}\], differentiating the equation we get,
\[{\text{dm} = {\lambda dx}}\],
\[{{\text{X}}_{{\text{cm}}}}{\text{ = }}\dfrac{{\smallint {\text{x}\lambda dx}}}{{\smallint {\lambda dx}}}\]
\[{{\text{X}}_{{\text{cm}}}}{\text{ = }}\dfrac{{\dfrac{{{\text{|}}{{\text{x}}^{\text{2}}}{\text{|}}_{\text{0}}^{\text{L}}}}{{\text{2}}}}}{{{\text{|x|}}_{\text{0}}^{\text{L}}}}{\text{ = }}\dfrac{{{{\text{L}}^{\text{2}}}}}{{{\text{2L}}}}{\text{ = }}\dfrac{{\text{L}}}{{\text{2}}}\]
Therefore, the correct answer is option D.
Note Whenever we have to find the center of mass of a non uniform section, we have to follow this procedure. In case you are given masses of different bodies and their respective location, then you have to use the summation symbol but if the body is continuous, we have to use the integration.
Recently Updated Pages
Two discs which are rotating about their respective class 11 physics JEE_Main

A ladder rests against a frictionless vertical wall class 11 physics JEE_Main

Two simple pendulums of lengths 1 m and 16 m respectively class 11 physics JEE_Main

The slopes of isothermal and adiabatic curves are related class 11 physics JEE_Main

A trolly falling freely on an inclined plane as shown class 11 physics JEE_Main

The masses M1 and M2M2 M1 are released from rest Using class 11 physics JEE_Main

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

Understanding Uniform Acceleration in Physics

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

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

