A) 0.125\[\Upsilon \]
B) 0.25\[\Upsilon \]
C) 0.5\[\Upsilon \]
D) 0.75\[\Upsilon \]
Correct Answer: A
Solution :
Moment of inertia of circular disc is \[I=\frac{m{{r}^{2}}}{2}\]about the centre of mass axis and by perpendicular axis theorem is \[\frac{m{{R}^{2}}}{4}\] about diameter. \[\therefore \] \[K.E.=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\times \frac{1}{4}m{{R}^{2}}\times {{\omega }^{2}}\] \[=\frac{1}{8}m{{R}^{2}}{{\omega }^{2}}\](as\[m{{R}^{2}}{{\omega }^{2}}=Y\]) \[=\frac{1}{8}Y\]
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