A) proportional to \[51{{m}^{2}}/{{s}^{2}}\]
B) independent of \[521{{m}^{2}}/{{s}^{2}}\]
C) proportional to \[\frac{1}{\sqrt{a}}\]
D) proportional to \[a\]
Correct Answer: A
Solution :
\[\Rightarrow \] \[\frac{1}{T}=\frac{1}{{{T}_{1}}}+\frac{1}{{{T}_{2}}}\] \[\therefore \] \[T=\frac{{{T}_{1}}\times {{T}_{2}}}{{{T}_{1}}+{{T}_{2}}}=\frac{810\times 1620}{810+1620}=540yr\] ?.(i) Also, for SHM \[\frac{1}{4}th\] and \[hv-{{W}_{0}}=\frac{1}{2}mv_{\max }^{2}\] \[\Rightarrow \] acceleration, \[\frac{hc}{\lambda }-\frac{hc}{{{\lambda }_{0}}}=\frac{1}{2}mv_{\max }^{2}\] \[\Rightarrow \] \[hc\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda {{\lambda }_{0}}} \right)=\frac{1}{2}mv_{\max }^{2}\] \[{{v}_{\max }}=\sqrt{\frac{2hc}{m}\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda .{{\lambda }_{0}}} \right)}\] \[\lambda \] ?.(ii) From Eqs. (i) and (ii), we get \[v=\sqrt{\frac{2hc}{m}\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda {{\lambda }_{0}}} \right)}\] \[\frac{3\lambda }{4}\] \[v,\] \[=2\pi \,\sqrt{\frac{m}{3k\,(a\,\sin \omega t)}}\] \[\frac{v}{v}=\sqrt{\frac{[{{\lambda }_{0}}-(3\lambda /4)]}{\frac{3}{4}\lambda {{\lambda }_{0}}}\times \frac{\lambda {{\lambda }_{0}}}{{{\lambda }_{0}}-\lambda }}\] \[v=v{{\left( \frac{4}{3} \right)}^{1/2}}\sqrt{\frac{[{{\lambda }_{0}}-(3\lambda /4)]}{{{\lambda }_{0}}-\lambda }}\]
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