
A solid rectangular door of uniform thickness is 2 m wide and has a mass of 60 kg. it is hinged about a vertical axis along one of its longer edges. Assume the thickness of the door is small. If the moment of inertia about its hinges of the door is 10n kg $m^2$, find n.
Answer
219.3k+ views
Hint: We will calculate the moment of inertia of the given solid rectangular door of mass M about the centre of mass of door using the formula ${I_{cm}} = \dfrac{{M{b^2}}}{{12}}$ and then we will use the parallel axis theorem to calculate the moment of inertia about the hinges of the door by using the formula: $I = {I_{cm}} + M{d^2}$ where I is the total moment of inertia and Icm is the moment of inertia about the centre of mass and d is the distance between the vertical axis and the centre of mass. After that, we will compare the given value to the obtained value to find the value of n.
Complete step by step solution:
We are given that a solid rectangular door is of uniform thickness.
Its width is given as 2 m.
The mass of the door is also provided which is 60 kg.
The distance between the centre of mass and the vertical axis of the door will be 1 m as we know that the width of the door is 2 m and hence the distance will be halved.
Now, the moment of inertia about the centre of the mass of the rectangular door will be given as: ${I_{cm}} = \dfrac{{M{b^2}}}{{12}}$
We have M = 60 kg and b = 2 m. on putting these values in the above equation, we get
$ \Rightarrow {I_{cm}} = \dfrac{{60 \times {2^2}}}{{12}} = \dfrac{{240}}{{12}} = 20kg{m^2}$
We have calculated the value of Icm and now we will calculate the value of the moment of inertia about the hinges of the door.
We have $I = {I_{cm}} + M{d^2}$, on putting values in this equation, we get
$ \Rightarrow I = 20 + 60 \times {1^2} = 20 + 60 = 80$
Therefore, the moment of inertia about the hinges of the door I = 80 kg$m^2$.
Comparing it with the given value of moment of inertia which is 10n kg$m^2$, we get n = 8.
Note: You may get confused while reading the given problem because both the moment of inertia is given i.e., the moment of inertia about the centre of mass and the moment of inertia about the vertical axis at a certain distance from the centre of mass of the body.
Complete step by step solution:
We are given that a solid rectangular door is of uniform thickness.
Its width is given as 2 m.
The mass of the door is also provided which is 60 kg.
The distance between the centre of mass and the vertical axis of the door will be 1 m as we know that the width of the door is 2 m and hence the distance will be halved.
Now, the moment of inertia about the centre of the mass of the rectangular door will be given as: ${I_{cm}} = \dfrac{{M{b^2}}}{{12}}$
We have M = 60 kg and b = 2 m. on putting these values in the above equation, we get
$ \Rightarrow {I_{cm}} = \dfrac{{60 \times {2^2}}}{{12}} = \dfrac{{240}}{{12}} = 20kg{m^2}$
We have calculated the value of Icm and now we will calculate the value of the moment of inertia about the hinges of the door.
We have $I = {I_{cm}} + M{d^2}$, on putting values in this equation, we get
$ \Rightarrow I = 20 + 60 \times {1^2} = 20 + 60 = 80$
Therefore, the moment of inertia about the hinges of the door I = 80 kg$m^2$.
Comparing it with the given value of moment of inertia which is 10n kg$m^2$, we get n = 8.
Note: You may get confused while reading the given problem because both the moment of inertia is given i.e., the moment of inertia about the centre of mass and the moment of inertia about the vertical axis at a certain distance from the centre of mass 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 Explained: Definition, Examples & Science for Students

Analytical Method of Vector Addition Explained Simply

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

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

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

