EAMCET Medical EAMCET Medical Solved Paper-2006

  • question_answer
    The speeds of air-flow on the upper and lower surfaces of a wing of an aeroplane are v1 and v2 respectively. If A is the cross-sectional area of the wing and p is the density of air, then the upward lift is:

    A)  \[\frac{1}{2}\rho A({{v}_{1}}-{{v}_{2}})\]                             

    B)  \[\frac{1}{2}\rho A({{v}_{1}}+{{v}_{2}})\]

    C)  \[\frac{1}{2}\rho A({{v}_{1}}^{2}-{{v}_{2}}^{2})\]                            

    D)  \[\frac{1}{2}\rho A({{v}_{1}}^{2}+{{v}_{2}}^{2})\]

    Correct Answer: C

    Solution :

                     Due to the specific shape of wings when the aeroplane runs, air passes at higher speed over it as compared to its lower surface. This difference of air speeds above and below the wings, in accordance with Bernoullis principle, creates a pressure difference, due to which an upward force called dynamic lift acts on the plate. \[\therefore \] Upward lift = pressure difference \[\times \]area of wing \[=\frac{1}{2}\rho A(v_{1}^{2}-v_{2}^{2})\]


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