Punjab Medical Punjab - MET Solved Paper-1999

  • question_answer
    A hydraulic lift is designed to lift cars of maximum mass of\[3000\,\,kg\]. The area of cross section of the piston carrying the load is\[4.25\times {{10}^{-2}}{{m}^{2}}\]. Then, the maximum pressure would the smaller piston have to bear, is

    A) \[7.82\times {{10}^{7}}\,\,N/{{m}^{2}}\]

    B) \[6.92\times {{10}^{5}}\,\,N/{{m}^{2}}\]

    C) \[13.76\times {{10}^{5}}\,\,N/{{m}^{2}}\]

    D) \[9.63\,\,N/{{m}^{2}}\]

    Correct Answer: B

    Solution :

    Maximum mass of the car\[=3000\,\,kg\] Area of cross-section of the piston \[=4.25\times {{10}^{-2}}{{m}^{2}}\] The maximum pressure beared by the smaller piston is \[P=\frac{F}{A}=\frac{3000\times 9.8}{4.25\times {{10}^{-2}}}\]     \[=6.92\times {{10}^{5}}\,\,N/{{m}^{2}}\]


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