RAJASTHAN PMT Rajasthan - PMT 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 makes25 revolutions in 14 sec. the magnitude of acceleration of stone is:

    A)  \[850\,cm/{{s}^{2}}\]   

    B)                         \[996\,cm/{{s}^{2}}\]

    C)  \[720\,cm/{{s}^{2}}\]   

    D)         \[650\,cm/{{s}^{2}}\]

    Correct Answer: B

    Solution :

    Time period \[\text{=}\frac{\text{No}\text{.of}\,\text{revolutions}}{\text{time}}\text{=}\frac{\text{25}}{\text{14}}\text{=1}\text{.79}\,\text{sec}\] Now angular speed\[\omega =\frac{2\pi }{T}=\frac{2\times 314}{1.79}=3.51\,\text{rad/sec}\] Now magnitude of acceleration is given by \[a={{\omega }^{2}}l={{(3.51)}^{2}}\times 80\] \[=985.6\,cm/{{\sec }^{2}}\] \[\approx 996\,cm/{{\sec }^{2}}\]


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