Two circular discs of the same mass and thickness are made from metals having densities \[{\rho _1}\]and \[{\rho _2}\]respectively. The ratio of their moment of inertia about an axis passing through their centre is.
A.\[{\rho _1}:{\rho _2}\]
B. \[{\rho _1}{\rho _2}:1\]
C. \[{\rho _2}:{\rho _1}\]
D. \[1:{\rho _1}{\rho _2}\]
Answer
297.3k+ views
Hint:The moment of inertia of disc is proportional to the product of mass and the square of the radius. More will be the radius, more will be the moment of inertia for the bodies of the same mass.
Formula used:
\[\rho = \dfrac{m}{V}\]
Here\[\rho \] is the density, m is the mass and V is the volume of the body.
\[V = \pi {r^2}t\]
Here V is the volume of the disc, r is the radius and t is the thickness.
\[{I_{disc}} = \dfrac{{m{r^2}}}{2}\]
Here \[{I_{disc}}\] is the moment of inertia of the disc about the axis passing through its centre, m is the mass and r is the radius.
Complete step by step solution:
Let the mass of the circular discs be m, thickness t, and the radius of the discs are \[{r_1}\] and \[{r_2}\] respectively. The volumes of the discs will be,
\[{V_1} = \pi r_1^2t\]
\[\Rightarrow {V_2} = \pi r_2^2t\]
Using the formula of density,
\[{V_1} = \dfrac{m}{{{\rho _1}}} \Rightarrow \pi r_1^2t = \dfrac{m}{{{\rho _1}}}\]
\[\Rightarrow {V_2} = \dfrac{m}{{{\rho _2}}} \Rightarrow \pi r_2^2t = \dfrac{m}{{{\rho _2}}}\]
Then the ratio of the moment of inertia of the two discs about the axis of rotation passing through the centre will be,
\[\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{\left( {\dfrac{{{m_1}r_1^2}}{2}} \right)}}{{\left( {\dfrac{{{m_2}r_2^2}}{2}} \right)}} \\ \]
\[\Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{\left( {\dfrac{m}{2}\left( {\dfrac{m}{{{\rho _1}\pi t}}} \right)} \right)}}{{\left( {\dfrac{m}{2}\left( {\dfrac{m}{{{\rho _1}\pi t}}} \right)} \right)}} = \dfrac{{{\rho _2}}}{{{\rho _1}}}\]
Hence, the ratio of the moment of inertia of two discs about the axis of rotation passing through the centre is \[{\rho _2}:{\rho _1}\].
Therefore,the correct option is C.
Note: We must be careful while calculating the moment of inertia. If the axis of rotation changes then the mass distribution also changes with respect to the axis of rotation. As the moment of inertia is proportional to the product of the mass and the square of the distance from the axis of rotation which leads to change in moment of inertia.
Formula used:
\[\rho = \dfrac{m}{V}\]
Here\[\rho \] is the density, m is the mass and V is the volume of the body.
\[V = \pi {r^2}t\]
Here V is the volume of the disc, r is the radius and t is the thickness.
\[{I_{disc}} = \dfrac{{m{r^2}}}{2}\]
Here \[{I_{disc}}\] is the moment of inertia of the disc about the axis passing through its centre, m is the mass and r is the radius.
Complete step by step solution:
Let the mass of the circular discs be m, thickness t, and the radius of the discs are \[{r_1}\] and \[{r_2}\] respectively. The volumes of the discs will be,
\[{V_1} = \pi r_1^2t\]
\[\Rightarrow {V_2} = \pi r_2^2t\]
Using the formula of density,
\[{V_1} = \dfrac{m}{{{\rho _1}}} \Rightarrow \pi r_1^2t = \dfrac{m}{{{\rho _1}}}\]
\[\Rightarrow {V_2} = \dfrac{m}{{{\rho _2}}} \Rightarrow \pi r_2^2t = \dfrac{m}{{{\rho _2}}}\]
Then the ratio of the moment of inertia of the two discs about the axis of rotation passing through the centre will be,
\[\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{\left( {\dfrac{{{m_1}r_1^2}}{2}} \right)}}{{\left( {\dfrac{{{m_2}r_2^2}}{2}} \right)}} \\ \]
\[\Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{\left( {\dfrac{m}{2}\left( {\dfrac{m}{{{\rho _1}\pi t}}} \right)} \right)}}{{\left( {\dfrac{m}{2}\left( {\dfrac{m}{{{\rho _1}\pi t}}} \right)} \right)}} = \dfrac{{{\rho _2}}}{{{\rho _1}}}\]
Hence, the ratio of the moment of inertia of two discs about the axis of rotation passing through the centre is \[{\rho _2}:{\rho _1}\].
Therefore,the correct option is C.
Note: We must be careful while calculating the moment of inertia. If the axis of rotation changes then the mass distribution also changes with respect to the axis of rotation. As the moment of inertia is proportional to the product of the mass and the square of the distance from the axis of rotation which leads to change in moment of inertia.
Recently Updated Pages
How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Maths Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 2) Chemistry Question Paper with Answer Key

Hydrogen and Its Type Important Concepts and Tips for JEE Exam Preparation

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

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

