
Consider a uniform square plate of side $a$ and mass $m$. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corner is:
A) $\dfrac{5}{6}m{a^2}$
B) $\dfrac{1}{{12}}m{a^2}$
C) $\dfrac{7}{{12}}m{a^2}$
D) $\dfrac{2}{3}m{a^2}$
Answer
214.2k+ views
Hint: To solve this question we have to apply a parallel axis theorem. According to this theorem, the moment of inertia along an axis parallel to the original axis will be the sum of the moment of inertia along the perpendicular axis and the product of mass and the distance between the perpendicular axis and parallel axis.
Formulae used:
${I_{parallel}} = {I_{perpendicular}} + M{R^2}$
Here ${I_{parallel}}$ is the moment of inertia along the parallel axis, ${I_{perpendicular}}$ is the moment of inertia along the axis through the centre of mass, $M$ is the mass of the object and $R$ is the distance between the centre of mass and the parallel axis.
Complete step by step answer:
In the question, a uniform square plate of side $a$ and mass $m$ is given. Let’s draw a figure.

From the above figure, we can easily find $R$ using the Pythagoras theorem,
$ \Rightarrow R = \sqrt {{a^2} - {{\left( {\dfrac{a}{2}} \right)}^2}} = \dfrac{a}{{\sqrt 2 }}$
We know that for a square plate, the moment of inertia along a perpendicular axis passing through the centre of mass is,
$ \Rightarrow {I_{perpendicular}} = \dfrac{{m{a^2}}}{6}$
So, using the parallel axis theorem, we get
$ \Rightarrow {I_{parallel}} = {I_{perpendicular}} + M{R^2}$
Here ${I_{parallel}}$ is the moment of inertia along the parallel axis, ${I_{perpendicular}}$ is the moment of inertia along the axis through the centre of mass, $M$ is the mass of the object and $R$ is the distance between the centre of mass and the parallel axis.
Substituting the value of $R$ and ${I_{perpendicular}}$ we get
$ \Rightarrow {I_{parallel}} = {I_{perpendicular}} + M{R^2}$
$ \therefore {I_{parallel}} = \dfrac{{m{a^2}}}{6} + \dfrac{{m{a^2}}}{{{{\left( {\sqrt 2 } \right)}^2}}} = \dfrac{2}{3}m{a^2}$
So the required answer is $\dfrac{2}{3}m{a^2}$. Hence option (D) is correct.
Note: While solving questions related to moment of inertia, make sure to apply the correct formulae. There are two different theorems i.e. parallel axis theorem and perpendicular axis theorem. Always use the correct theorem. The parallel axis theorem is used for axes parallel to the centroidal axis of the body. However, the perpendicular axis theorem is used for axes that are perpendicular to the centroidal axis of the body.
Formulae used:
${I_{parallel}} = {I_{perpendicular}} + M{R^2}$
Here ${I_{parallel}}$ is the moment of inertia along the parallel axis, ${I_{perpendicular}}$ is the moment of inertia along the axis through the centre of mass, $M$ is the mass of the object and $R$ is the distance between the centre of mass and the parallel axis.
Complete step by step answer:
In the question, a uniform square plate of side $a$ and mass $m$ is given. Let’s draw a figure.

From the above figure, we can easily find $R$ using the Pythagoras theorem,
$ \Rightarrow R = \sqrt {{a^2} - {{\left( {\dfrac{a}{2}} \right)}^2}} = \dfrac{a}{{\sqrt 2 }}$
We know that for a square plate, the moment of inertia along a perpendicular axis passing through the centre of mass is,
$ \Rightarrow {I_{perpendicular}} = \dfrac{{m{a^2}}}{6}$
So, using the parallel axis theorem, we get
$ \Rightarrow {I_{parallel}} = {I_{perpendicular}} + M{R^2}$
Here ${I_{parallel}}$ is the moment of inertia along the parallel axis, ${I_{perpendicular}}$ is the moment of inertia along the axis through the centre of mass, $M$ is the mass of the object and $R$ is the distance between the centre of mass and the parallel axis.
Substituting the value of $R$ and ${I_{perpendicular}}$ we get
$ \Rightarrow {I_{parallel}} = {I_{perpendicular}} + M{R^2}$
$ \therefore {I_{parallel}} = \dfrac{{m{a^2}}}{6} + \dfrac{{m{a^2}}}{{{{\left( {\sqrt 2 } \right)}^2}}} = \dfrac{2}{3}m{a^2}$
So the required answer is $\dfrac{2}{3}m{a^2}$. Hence option (D) is correct.
Note: While solving questions related to moment of inertia, make sure to apply the correct formulae. There are two different theorems i.e. parallel axis theorem and perpendicular axis theorem. Always use the correct theorem. The parallel axis theorem is used for axes parallel to the centroidal axis of the body. However, the perpendicular axis theorem is used for axes that are perpendicular to the centroidal axis of the body.
Recently Updated Pages
Chemical Equation - Important Concepts and Tips for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

Conduction, Transfer of Energy Important Concepts and Tips for JEE

JEE Analytical Method of Vector Addition Important Concepts and Tips

Atomic Size - Important Concepts and Tips for JEE

JEE Main 2022 (June 29th Shift 1) Maths Question Paper with Answer Key

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

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

Equation of Trajectory in Projectile Motion: Derivation & Proof

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

Hybridisation in Chemistry – Concept, Types & Applications

Angle of Deviation in a Prism – Formula, Diagram & Applications

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

Collision: Meaning, Types & Examples in Physics

