
In a rectangle ABCD, \[BC = 2AB\]. The moment of inertia is minimum along an axis through?

A. \[CD\]
B. \[BC\]
C. \[HF\]
D. \[EG\]
Answer
163.8k+ views
Hint:To calculate the moment of inertia about an axis through an object, two theorems are used. One is the parallel axis theorem and another is the perpendicular axis theorem. Both are used depending upon the situations given in the question. The moment of inertia due to rotation about an axis passing through one of the ends of a rod of length L and mass M and perpendicular to the rod is \[\dfrac{{M{L^2}}}{3}\].
Formula Used:
The moment of inertia, \[I\] due to rotation about an axis passing through one of the ends of a rod of length L and mass M and perpendicular to the length of the rod is,
\[I = \dfrac{{M{L^2}}}{3}\] ---- (1)
The moment of inertia due to rotation about an axis passing through the centre of the rod of length L and mass M and in a direction perpendicular to the length of the rod is,
\[{I_C} = \dfrac{{M{L^2}}}{{12}}\]--- (2)
Complete step by step solution:
Given: A rectangle ABCD where \[BC = 2AB\]
Let M be the mass of the rectangle ABCD and length of AB = L ---- (3) .
Then, \[AB = CD = L\]
Then, \[BC = 2AB = 2L\]--- (4).
Then, \[BC = AC = 2L\]
The moment of inertia about the axis CD is:
Using the condition from equation (1),
\[{I_{ < CD > }} = \dfrac{1}{3}M{(BC)^2}\]
\[{I_{ < CD > }} = \dfrac{1}{3}M{(2L)^2}\]
\[\Rightarrow {I_{ < CD > }} = \dfrac{8}{3}M{L^2}\]--- (5)
The moment of inertia about the axis BC is:
Using the condition from equation (1),
\[{I_{ < BC > }} = \dfrac{1}{3}M{(AB)^2}\]
\[\Rightarrow {I_{ < BC > }} = \dfrac{1}{3}M{(L)^2}\]---- (6)
The moment of inertia about the axis HF is:
As HF passes through the centre of edge AB, using the condition from equation (2),
\[{I_{ < HF > }} = \dfrac{1}{{12}}M{(AB)^2}\]
\[\Rightarrow {I_{ < HF > }} = \dfrac{1}{{12}}M{L^2}\]---- (7)
The moment of inertia about the axis EG is:
As EG passes through the centre of edge BC, using the condition from equation (2),
\[{I_{ < EG > }} = \dfrac{1}{{12}}M{(BC)^2}\]
\[\Rightarrow {I_{ < EG > }} = \dfrac{1}{{12}}M{(2L)^2}\]
\[\Rightarrow {I_{ < EG > }} = \dfrac{1}{3}M{L^2}\]---- (8)
From equations (5), (6), (7) and (8),
The moment of inertia about HF, \[{I_{ < HF > }} = \dfrac{1}{{12}}M{L^2}\] has the minimum value.
Hence, option C is the correct answer.
Note: If we have to calculate the moment of inertia about an axis parallel to the axis passing through the centre of the object, we use the theorem of the parallel axis. When the body has symmetry in about two out of three axes in a 3D plane, the perpendicular axis theorem is used. The complete rectangle is considered a single entity of mass M. If the axis of rotation is any of the diagonals, then the moment of inertia is \[\dfrac{1}{6}M{L^2}\].
Formula Used:
The moment of inertia, \[I\] due to rotation about an axis passing through one of the ends of a rod of length L and mass M and perpendicular to the length of the rod is,
\[I = \dfrac{{M{L^2}}}{3}\] ---- (1)
The moment of inertia due to rotation about an axis passing through the centre of the rod of length L and mass M and in a direction perpendicular to the length of the rod is,
\[{I_C} = \dfrac{{M{L^2}}}{{12}}\]--- (2)
Complete step by step solution:
Given: A rectangle ABCD where \[BC = 2AB\]
Let M be the mass of the rectangle ABCD and length of AB = L ---- (3) .
Then, \[AB = CD = L\]
Then, \[BC = 2AB = 2L\]--- (4).
Then, \[BC = AC = 2L\]
The moment of inertia about the axis CD is:
Using the condition from equation (1),
\[{I_{ < CD > }} = \dfrac{1}{3}M{(BC)^2}\]
\[{I_{ < CD > }} = \dfrac{1}{3}M{(2L)^2}\]
\[\Rightarrow {I_{ < CD > }} = \dfrac{8}{3}M{L^2}\]--- (5)
The moment of inertia about the axis BC is:
Using the condition from equation (1),
\[{I_{ < BC > }} = \dfrac{1}{3}M{(AB)^2}\]
\[\Rightarrow {I_{ < BC > }} = \dfrac{1}{3}M{(L)^2}\]---- (6)
The moment of inertia about the axis HF is:
As HF passes through the centre of edge AB, using the condition from equation (2),
\[{I_{ < HF > }} = \dfrac{1}{{12}}M{(AB)^2}\]
\[\Rightarrow {I_{ < HF > }} = \dfrac{1}{{12}}M{L^2}\]---- (7)
The moment of inertia about the axis EG is:
As EG passes through the centre of edge BC, using the condition from equation (2),
\[{I_{ < EG > }} = \dfrac{1}{{12}}M{(BC)^2}\]
\[\Rightarrow {I_{ < EG > }} = \dfrac{1}{{12}}M{(2L)^2}\]
\[\Rightarrow {I_{ < EG > }} = \dfrac{1}{3}M{L^2}\]---- (8)
From equations (5), (6), (7) and (8),
The moment of inertia about HF, \[{I_{ < HF > }} = \dfrac{1}{{12}}M{L^2}\] has the minimum value.
Hence, option C is the correct answer.
Note: If we have to calculate the moment of inertia about an axis parallel to the axis passing through the centre of the object, we use the theorem of the parallel axis. When the body has symmetry in about two out of three axes in a 3D plane, the perpendicular axis theorem is used. The complete rectangle is considered a single entity of mass M. If the axis of rotation is any of the diagonals, then the moment of inertia is \[\dfrac{1}{6}M{L^2}\].
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

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

Degree of Dissociation and Its Formula With Solved Example for JEE

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

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 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
