question_answer

The moment of inertia of a regular circular disc of mass 0.4 kg and radius 100 cm about the axis    axis perpendicular to the plane of the disc and     passing through its centre is :

A) \[0.002\text{ }kg-{{m}^{2}}\]     

B)  \[0.02\text{ }kg-{{m}^{2}}\]

C) \[2\text{ }kg-{{m}^{2}}\]             

D)  \[0.2\text{ }kg-{{m}^{2}}\]

Correct Answer: D

Solution :

                 Key Idea: We have to find the moment of inertia of disc about its central axis in perpendicular plane. If the mass of a disc is M, its radius is R, then its moment of inertia about an axis passing through its centre of mass and perpendicular to its plane is \[I=\frac{1}{2}M{{R}^{2}}\] Given, \[M=0.4\,\]\[kg,R=100\,cm=1\,m\] \[\therefore \]      \[I=\frac{1}{2}\times 0.4\times {{1}^{2}}=0.2\,\text{kg}-{{\text{m}}^{\text{2}}}\]