TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

The situation given can be shown as. Here, $R$ be the radius of circular disc $=2\,\text{cm}$ The radius of removed portion, $r=1\,\text{cm}$ Area of whole disc, $A_{1}=\pi R^{2}=4\pi\,\text{cm}^{2}$ Area of removed portion, $A_{2}=\pi r^{2}=\pi\,\text{cm}^{2}$ As, $\;\frac{1}{4}$ th area of the disc is removed, so remaining area of the disc is $\;\frac{3}{4}$ th initial area. Similarly, remaining mass, $m_{1}=\;\frac{3}{4} m$ Mass of removed portion, $m_{2}=\;\frac{1}{4} m$ Let $x$ be the maximum shift in centre of mass. So, initially centre of mass along $X$ -axis of complete disc, $x_{\text{CM}}=\;\frac{m_{1}x+m_{2}r}{m_{1}+m_{2}}$ $0=\;\frac{\;\frac{3}{4} m x + \frac{1}{4} m r}{\frac{3}{4} m + \;\frac{1}{4} m}$ $\Rightarrow \;\; \;\frac{3}{4} m x = -\;\frac{1}{4} m r$ or $x=-\;\frac{1}{3} r = -\;\frac{1}{3}\,\text{cm}$ i.e., the centre of mass of remaining portion will shift to the left of origin at $\;\frac{1}{3}\,\text{cm}$. Now, the disc is rotated by angle $\theta$, so the centre of mass will also shift by angle $\theta$ as shown $\triangle OPQ$ is isoscales. So, a perpendicular drawn from 0, on $P Q$ divide the angle $\theta$ and length $P Q$ in equal parts i.e., $RQ = \;\frac{1}{2} PQ = \;\frac{1}{2}\left(\;\frac{1}{\sqrt{3}}\right)\,\text{cm} \;\; ($ given, $PQ = \;\frac{1}{\sqrt{3}})$ From $\triangle ORQ$. $\sin \;\frac{\theta}{2} = \;\frac{RQ}{OQ} = \;\frac{ \;\frac{1}{2} \times \frac{1}{\sqrt{3}} }{ \frac{1}{3} }$ $\Rightarrow \;\; \sin \;\frac{\theta}{2} = \;\frac{\sqrt{3}}{2} = \sin 60^{\circ}$ $\Rightarrow \;\; \theta = 120^{\circ}$