A) \[{{\left[ \frac{K{{R}^{2}}}{2m}{{(2-\sqrt{3})}^{2}}+gR\sqrt{3} \right]}^{1/2}}\]
B) \[{{\left[ \frac{K{{R}^{2}}}{m}{{(2-\sqrt{3})}^{2}}+gR\sqrt{3} \right]}^{1/2}}\]
C) \[{{\left[ \frac{K{{R}^{2}}}{m}{{(\sqrt{2}-1)}^{2}}+gR\sqrt{3} \right]}^{1/2}}\]
D) \[{{\left[ \frac{K{{R}^{2}}}{2m}{{(\sqrt{2}-1)}^{2}}+gR \right]}^{1/2}}\]
Correct Answer: B
Solution :
Decreases in elastic PE + decrease in gravitational PE = increase in KE \[=\frac{1}{2}\,K{{x}^{2}}\,+mg(AD)\,=\frac{1}{2}\,m{{v}^{2}}\] \[x=AC-CB=2R-2R\,\cos \,{{30}^{0}}\] \[=R(2-\sqrt{3})\,(As.\,\angle CBA\,={{90}^{0}})\] \[AD=AB\,\cos \,{{60}^{0}}\] \[=(AC\,\sin \,{{30}^{0}}\,)\,\cos {{60}^{0}}\,=\frac{R}{2}\] So, \[\frac{1}{2}\,K{{R}^{2}}{{(2-\sqrt{3})}^{2}}\,+mg\,\frac{R}{2}\,=\frac{1}{2}\,m{{v}^{2}}\] \[v={{\left[ \frac{K{{R}^{2}}}{m}{{(2-\sqrt{3})}^{2}}+gR \right]}^{1/2}}\]You need to login to perform this action.
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