Manipal Medical Manipal Medical Solved Paper-2003

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

    B)  \[996\text{ }cm/se{{c}^{2}}\]

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

    D)  \[650\text{ }cm/se{{c}^{2}}\]

    Correct Answer: B

    Solution :

     Time period \[=\frac{No.\text{ }of\text{ }revolutions}{time}=\frac{25}{14}=1.79\,\sec \] 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 \[a={{\omega }^{2}}I={{(3.51)}^{2}}\times 80\] \[=985.6\text{ }cm/se{{c}^{2}}\] \[\approx 996\text{ }cm/se{{c}^{2}}\]


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