JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Self Evaluation Test - Work, Energy and Power

  • question_answer
    A ring of mass m can slide over a smooth vertical rod as shown in figure. The ring is connected to a spring of force constant\[k=4mg\text{ }I\text{ }R\], where 2 R is the natural length of the spring. The other end of spring is fixed to the ground at a horizontal distance 2 R from the base of the rod. If the mass is released at a height 1.5 R, then the velocity of the ring as it reaches the ground is Critical Thinking            

    A)  \[\sqrt{gR}\]

    B) \[2\sqrt{gR}\]  

    C) \[\sqrt{2gR}\]

    D) \[\sqrt{3gR}\]

    Correct Answer: B

    Solution :

    [b] Initial length of the spring, \[{{\ell }_{i}},=\sqrt{{{(2R)}^{2}}+{{(1.5R)}^{2}}} =2.5R\] \[\therefore  {{x}_{i}}=2.5R-2R=0.5R.\] \[Now\,\frac{1}{2}k{{x}_{i}}^{2}+mg\left( 1.5R \right)=\frac{1}{2}m{{v}^{2}}\] or  \[\frac{1}{2}k{{\left( 0.5R \right)}^{2}}+mg\left( 1.5R \right)\]


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