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NEET AIPMT SOLVED PAPER SCREENING 2011

  • question_answer
    The instantaneous angular position of a point on a rotating wheel is given by the equation \[Q(\tau )=2{{\tau }^{3}}-6{{\tau }^{2}}\] The torque on the wheel becomes zero at

    A) \[\tau =0.5\,s\]

    B)                        \[\tau =0.25\,s\]              

    C)        \[\tau =2\,s\]    

    D)        \[\tau =1\,s\]

    Correct Answer: D

    Solution :

    According to question, torque, \[\tau =0\] Its means that, \[\alpha =0\] \[\alpha =\frac{{{d}^{2}}\theta }{d{{t}^{2}}}\] Given      \[\theta (t)=2{{t}^{3}}-6{{t}^{2}}\] So,          \[\frac{d\theta }{dt}=6{{t}^{2}}-12t\]                 \[\frac{{{d}^{2}}\theta }{d{{t}^{2}}}=12t-12\left( \because \alpha =\frac{{{d}^{2}}\theta }{d{{t}^{2}}}=0 \right)\] \[12t-12=0\] \[t=1\,s\]


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