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BVP Medical BVP Medical Solved Paper-2011

  • question_answer
    On hanging a mass of 1 kg from a spring of50 cm length, its length increases by 2 cm. The mass is pulled down until the length of the spring becomes 60 cm. Then energy stored in the spring in this condition is

    A)  1.5 J                                     

    B)  2.0 J

    C)  2.5 J                                     

    D)  3 J

    Correct Answer: C

    Solution :

                    Force constant of a spring \[\frac{aT}{2}\]                 \[\frac{aT}{4}\] \[aT\] Increase in the length \[{{s}^{2}}\] \[y=8t-5{{t}^{2}}\]                 \[\frac{2}{5}M{{R}^{2}}\]


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