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NEET Sample Paper NEET Sample Test Paper-52

  • question_answer
    A car of mass 1500 kg is moving with a speed of \[12.5\text{ }m{{s}^{-}}^{1}\] on a circular path of radius 20 m on a level road. What should be the value of coefficient of friction to attain this centripetal force?

    A) 0.2                               

    B) 0.4

    C) 0.6                               

    D) 0.8

    Correct Answer: D

    Solution :

    Here, \[m=1500\text{ }kg,\text{ }v=12.5\text{ }m{{s}^{-}}^{1}\] \[r\,\,=\,\,20\,m\] Friction force = centripetal force required \[F=\frac{m{{v}^{2}}}{r}\,\,=\,\,\frac{1500{{(12.5)}^{2}}}{20}\,=\,\,1.172\times {{10}^{4}}N\] \[As\,\,\,F=\mu R=\mu \,mg\] \[\mu \,\,=\,\,\frac{F}{mg}\,\,=\,\,\frac{1.172\times {{10}^{4}}}{1500\times 9.8}\,\,=\,\,0.8\,\]


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