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BHU PMT BHU PMT (Mains) Solved Paper-2008

  • question_answer
    A ball of mass 0.6 kg attached to a light inextensible string rotates in a vertical circle of radius 0.75 m such that it has a speed of \[5\text{ }m{{s}^{-1}}\]when the string is horizontal. Tension in string when it is horizontal on other side is \[(g=10\,m{{s}^{-2}})\]

    A)  30 N                                     

    B)  26 N

    C)  20 N                                     

    D)  6N

    Correct Answer: C

    Solution :

                     Tension in the string when it makes angle\[\theta \]with the vertical, \[T=\frac{m{{v}^{2}}}{r}+mg\,\cos \theta \] When the string is horizontal,\[\theta =90{}^\circ \] Here,\[m=0.6\text{ }kg,\text{ }r=0.75\text{ }m,\text{ }v=5\text{ }m/s\] Hence, \[T=\frac{m{{v}^{2}}}{r}+mg\times 0=\frac{m{{v}^{2}}}{r}\] \[\therefore \]  \[T=\frac{0.6\times {{(5)}^{2}}}{0.75}=20\,N\]


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