CET Karnataka Medical CET - Karnataka Medical Solved Paper-2014

  • question_answer
    A person is driving a vehicle at uniform speed of  \[5m\text{ }{{s}^{-1}}\]on a level curved track of radius 5 m. The coefficient of static friction between tyres and road is 0.1. Will the person slip while taking the turn with the same speed? Take \[g=10m{{s}^{-2}}\] Choose the correct statement.

    A)  A person will slip if \[{{v}^{2}}<5m\,{{s}^{-1}}\].

    B)  A person will slip if  \[{{v}^{2}}=5m\,{{s}^{-1}}\].

    C)  A person will not slip if  \[{{v}^{2}}>10\,m\,{{s}^{-1}}\].

    D)  A person will slip if \[{{v}^{2}}>5\text{ }m\text{ }{{s}^{-1}}\].

    Correct Answer: D

    Solution :

    : The person will not slip if \[{{v}^{2}}\le 5{{m}^{2}}{{s}^{-2}}\] Here, \[{{\mu }_{s}}=0.1,\,R=5m,\,g=10m\,{{s}^{-2}}\] \[\therefore \]  \[{{\mu }_{s}}Rg=0.1\times 5\times m\times 10m\,\,{{s}^{-2}}=5{{m}^{2}}\,{{s}^{-2}}\] or   \[{{v}^{2}}\le 5\,{{m}^{2}}\,\,{{s}^{-2}}\] Therefore the person will slip if \[{{v}^{2}}>5\,{{m}^{2}}\,\,{{s}^{-2}}\]. Note: In the question paper the unit of \[{{v}^{2}}\] is given to be \[m\text{ }{{s}^{-1}}\]. But it is wrong. It is a unit of v not \[{{v}^{2}}\]. The correct unit of \[{{v}^{2}}\] is \[{{m}^{2}}\text{ }{{s}^{-2}}\].


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