A) \[\left( \left( n+\frac{1}{2} \right)\,\pi ,\,m\pi ,\,r\pi \right)\]
B) \[\left( (n-1)\,\frac{\pi }{2},\,\,m\pi ,\,\,r\pi \right)\]
C) \[\left( (2n-1)\,\frac{\pi }{3},\,\,m\pi ,\,\,\frac{r\pi }{2} \right)\]
D) \[\left( (n+1)\,\frac{\pi }{2},\,\,m\pi ,\,r\pi \right)\]
Correct Answer: A
Solution :
\[n\] and \[{{P}_{1}}=n\times \frac{hc}{\lambda }\] So, \[\Rightarrow \] \[\Rightarrow \] and \[1\times \cos \,{{60}^{o}}+2\,\cos \,{{60}^{o}}+x-4\,\cos \,{{60}^{o}}=0\] Or \[\therefore \]\[x=0.5\,\,N\] and \[\frac{M}{\pi {{R}^{2}}}\times \pi {{\left\{ \frac{R}{2} \right\}}^{2}}=\frac{M}{4}\] \[=M-\frac{M}{4}=\frac{3M}{4}\] and \[{{X}_{CM}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}}{{{m}_{1}}+{{m}_{2}}}\]
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