NEET AIPMT SOLVED PAPER SCREENING 2012

  • question_answer
    A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless road. If the banking angle is 45°, the speed of the car is

    A) \[\text{20}\,\text{m}{{\text{s}}^{\text{-1}}}\]         

    B)        \[\text{30}\,\text{m}{{\text{s}}^{\text{-1}}}\]  

    C)        \[5\,\text{m}{{\text{s}}^{\text{-1}}}\]                  

    D)        \[10\,\text{m}{{\text{s}}^{\text{-1}}}\]

    Correct Answer: B

    Solution :

    The angle of banking \[\tan \theta =\frac{{{v}^{2}}}{rg}\] Given, \[\theta ={{45}^{o}},r=90\,m\]and\[g=10\,m/{{s}^{2}}\] \[\tan {{45}^{o}}=\frac{{{v}^{2}}}{90\times 10}\] \[v=\sqrt{90\times 10\times \tan {{45}^{o}}}\] Speed of car v = 30 m/s


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