A) \[M{{R}^{2}}\]
B) \[\frac{2}{5}M{{R}^{2}}\]
C) \[\frac{3}{2}M{{R}^{2}}\]
D) \[\frac{1}{2}M{{R}^{2}}\]
Correct Answer: C
Solution :
The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is \[{{I}_{CM}}=\frac{1}{2}M{{R}^{2}}\] Where M is the mass of disc and R its radius. According to theorem of parallel axis, MI of circular disc about an axis touching the disc at its diameter and normal to the disc is \[I={{I}_{CM}}+M{{R}^{2}}\] \[=\frac{1}{2}M{{R}^{2}}+M{{R}^{2}}=\frac{3}{2}M{{R}^{2}}\]
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