BVP Medical BVP Medical Solved Paper-2005

  • question_answer
    A stone tied to one end of spring 80 cm long is whirled in a horizontal circle with a constant speed. If stone makes 25 revolutions in 14 sec, the magnitude of acceleration of stone is :

    A)  850 cm/\[{{s}^{2}}\]                      

    B)  996 cm/\[{{\sec }^{2}}\]

    C)  720 cm/\[{{\operatorname{s}}^{2}}\]                   

    D)  650 cm/\[{{\sec }^{2}}\]

    Correct Answer: A

    Solution :

                    Time period \[=\frac{No.\,\,of\,\,revolutions}{time}=\frac{25}{14}=1.79\sec \] Now angular speed \[\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{1.79}=3.51\,rad/\sec \] Now magnitude of acceleration is given by                 \[a={{\omega }^{2}}l={{(3.51)}^{2}}\times 80\]                 \[=985.6cm/{{\sec }^{2}}\]                 \[=996cm/{{\sec }^{2}}\]


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