JEE Main & Advanced Physics Two Dimensional Motion Question Bank Self Evaluation Test - Motion in a Plane

  • question_answer
    The length of second's hand in a watch is 1 cm. The change in velocity of its tip in 15 seconds is:

    A) zero                              

    B) \[\frac{\pi }{30\sqrt{2}}\text{ cm/s}\]

    C) \[\frac{\pi }{30}\text{ cm/s}\]     

    D)        \[\frac{\pi \sqrt{2}}{30}\text{ cm/s}\]

    Correct Answer: D

    Solution :

    [d] \[\Delta \text{v}=\sqrt{2}v=\sqrt{2}\omega \,\text{r}=\sqrt{2}\left( \frac{2\pi }{60} \right)\times 1=\frac{\pi \sqrt{2}}{30}\,cm\text{/}s\]


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