question_answer

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 comers is

A) \[\frac{5}{6}m{{a}^{2}}\]

B) \[\frac{1}{12}m{{a}^{2}}\]

C) \[\frac{7}{12}m{{a}^{2}}\]

D) \[\frac{2}{3}m{{a}^{2}}\]

Correct Answer: D

Solution :

[d]\[{{I}_{nn'}}=\frac{1}{12}M({{a}^{2}}+{{a}^{2}})=\frac{M{{a}^{2}}}{6}\] Also, \[DO=\frac{DB}{2}=\frac{\sqrt{2a}}{2}=\frac{a}{\sqrt{2}}\] According to parallel axis theorem \[{{I}_{mm'}}={{I}_{nn'}}+M{{\left( \frac{a}{\sqrt{2}} \right)}^{2}}=\frac{M{{a}^{2}}}{6}+\frac{M{{a}^{2}}}{2}\] \[=\frac{M{{a}^{2}}+3M{{a}^{2}}}{6}=\frac{2}{3}M{{a}^{2}}\]