A) \[[L\,\,{{T}^{-1}}{{Q}^{-1}}]\]
B) \[[M{{L}^{2}}\,\,{{T}^{-1}}{{Q}^{-2}}]\]
C) \[\left[ \text{LT}{{\text{Q}}^{-\text{1}}} \right]\]
D) \[{{O}_{1}}\]
Correct Answer: A
Solution :
The equation of trajectory. \[\left[ \frac{{{V}^{2}}}{2} \right]_{{{v}_{0}}}^{v}=-{{\omega }^{2}}\left[ \frac{{{X}^{2}}}{2} \right]_{0}^{x}\] \[\Rightarrow \] ???. (i) And \[{{V}^{2}}-V_{0}^{2}=-{{\omega }^{2}}{{X}^{2}}\] ??.. (ii) On dividing we get, \[\Rightarrow \] \[V=\sqrt{V_{0}^{2}-{{\omega }^{2}}{{X}^{2}}}\] \[2\,mg\,sin{{45}^{\circ }}-{{f}_{2}}-T=2ma\] Now, \[\Rightarrow \] \[\frac{2mg}{\sqrt{2}}-{{\mu }_{1}}{{R}_{1}}-{{T}_{2}}=2ma\] \[\Rightarrow \] \[\frac{2mg}{\sqrt{2}}-\frac{1}{3}.2\,mg\,\cos {{45}^{\circ }}-T=2ma\] \[({{m}_{B}}-{{m}_{A}})g\,\sin \theta =\frac{mg}{\sqrt{2}}is\,\,lesser\] \[({{\mu }_{B}}{{m}_{B}}+{{\mu }_{A}}{{m}_{A}})g\,\cos \theta \] \[=\frac{4mg}{3\sqrt{2}}\]
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