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NEET AIPMT SOLVED PAPER SCREENING 2009

  • question_answer
    A simple pendulum performs simple harmonic motion about x = 0 with an amplitude \[a\] and time period T. The speed of the pendulum at \[\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}\]will be

    A) \[\overrightarrow{r}\]                  

    B) \[\overrightarrow{F}\]

    C) \[\therefore \]                 

    D) \[\overrightarrow{\tau }.\overrightarrow{r}=0\]

    Correct Answer: D

    Solution :

    \[=-B\pi 2r\frac{dr}{dt}\] \[e=-0.04\times \pi \times 2\times 2\times {{10}^{-2}}\times 2\times {{10}^{-3}}\] Here,\[=3.2\pi \mu V\] \[=[{{M}^{0}}{{L}^{1}}{{T}^{-1}}]\]\[=[{{M}^{0}}{{L}^{1}}{{T}^{-2}}]\]


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