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) \[\left[ 0,\,\left( \frac{4}{\pi } \right)R \right]\]
Correct Answer: C
Solution :
By symmetry, \[{{X}_{CM}}=0\] As, \[dm=\lambda R\,\,d\theta \] \[\therefore \] \[{{y}_{CM}}=\frac{\int{y\,\,dm}}{\int{dm}}=\frac{1}{m}\,\int{R\,\cos \,\theta \lambda R\,\,d\theta }\] \[=\frac{{{R}^{2}}\lambda }{m}\int_{-\pi 4}^{\pi /4}{\cos \,\theta \,d\theta }\] \[=\frac{{{R}^{2}}\lambda }{R\times \lambda \times \left( \frac{\pi }{2} \right)}\left[ 2\,\sin \,\frac{\pi }{4} \right]\] \[=\frac{2R}{\pi }\,2\frac{1}{\sqrt{2}}=\frac{2\sqrt{2}\,R}{\pi }\]You need to login to perform this action.
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