
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
162k+ 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
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Class 11 JEE Main Physics Mock Test 2025

Differentiate between audible and inaudible sounds class 11 physics JEE_Main

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
