• # question_answer A circular arc AB of thin wire frame of radius R and mass M makes an angle of $90{}^\circ$ at the origin. The centre of mass of the arc lies at A)  $\left[ 0,\,\left( \frac{2}{\pi } \right)R \right]$             B)  $\left[ 0,\,\left( \frac{\sqrt{2}}{\pi } \right)R \right]$ C)  $\left[ 0,\,\left( \frac{2\sqrt{2}}{\pi } \right)R \right]$ D)  $\left[ 0,\,\left( \frac{4}{\pi } \right)R \right]$

By symmetry, ${{X}_{CM}}=0$ As,       $dm=\lambda R\,\,d\theta$ $\therefore$    ${{y}_{CM}}=\frac{\int{y\,\,dm}}{\int{dm}}=\frac{1}{m}\,\int{R\,\cos \,\theta \lambda R\,\,d\theta }$ $=\frac{{{R}^{2}}\lambda }{m}\int_{-\pi 4}^{\pi /4}{\cos \,\theta \,d\theta }$ $=\frac{{{R}^{2}}\lambda }{R\times \lambda \times \left( \frac{\pi }{2} \right)}\left[ 2\,\sin \,\frac{\pi }{4} \right]$ $=\frac{2R}{\pi }\,2\frac{1}{\sqrt{2}}=\frac{2\sqrt{2}\,R}{\pi }$