2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    Air of density 1.2 \[\operatorname{Kg}{{\operatorname{m}}^{-3}}\] is blowing across the horizontal wings of on aero plane in such a way that its speeds above and below the wings are \[150\,\,m{{s}^{-1}}\]and \[100\,\,m{{s}^{-1}}\],  respectively. The pressure difference between the upper and lower     JEE Main  Online Paper (Held On 22 April 2013)

    A)  \[60\operatorname{N}{{\operatorname{m}}^{-2}}\]                   

    B)  \[180\operatorname{N}{{\operatorname{m}}^{-2}}\]

    C)  \[7500\operatorname{N}{{\operatorname{m}}^{-2}}\]                              

    D)  \[12500\operatorname{N}{{\operatorname{m}}^{-2}}\]

    Correct Answer: C

    Solution :

     Pressure difference \[{{P}_{2}}-{{P}_{1}}=\frac{1}{2}\rho \left( v_{2}^{2}-v_{1}^{2} \right)\] \[=\frac{1}{2}\times 1.2\left( {{(150)}^{2}}-{{(100)}^{2}} \right)\] \[=\frac{1}{2}\times 1.2(22500-10000)\] \[=7500\,N{{m}^{-2}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner