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JEE Main & Advanced Sample Paper JEE Main Sample Paper-21

  • question_answer
    A ring A of mass m is attached to a spring of constant K, which is fixed at C on a smooth circulate track of radius R. Points A and C are diametrically opposite. When the ring slips from rest on the track to point B, making an angle of \[{{30}^{o}}\] with \[(\angle ACB={{30}^{o}})\]spring becomes unstretched. Find the velocity of the ring at B.

    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, \[\frac{1}{2}\,K{{R}^{2}}{{(2-\sqrt{3})}^{2}}\,+mg\,\frac{R}{2}\,=\frac{1}{2}\,m{{v}^{2}}\] \[v={{\left[ \frac{K{{R}^{2}}}{m}{{(2-\sqrt{3})}^{2}}+gR \right]}^{1/2}}\]


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