2. Solved Papers
  3. Manipal Medical

Manipal Medical Manipal Medical Solved Paper-2012

  • question_answer
    Water from a tap emerges vertically downwards with initial velocity\[4\text{ }m{{s}^{-1}}\]. The cross-sectional area of the tap is A. The flow is steady and pressure is constant throughout the stream of water. The distance h vertically below the tap, where the cross-sectional area of the stream becomes\[\left( \frac{2}{3} \right)A,\]is\[(g=10\text{ }m/{{s}^{2}})\]

    A)  0.5 m            

    B)  1 m

    C)  1.5 m            

    D)  2.2 m

    Correct Answer: B

    Solution :

     The equation of continuity \[{{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}\] \[A\times 4=\frac{2}{3}A\times {{v}_{2}}\] \[{{v}_{2}}=6\,m{{s}^{-1}}\] From Bernoullis theorem, \[p+\rho g{{h}_{1}}+\frac{1}{2}\rho v_{1}^{2}=p+\rho g{{h}_{2}}+\frac{1}{2}\rho v_{2}^{2}\] Or \[g({{h}_{1}}-{{h}_{2}})=\frac{1}{2}(v_{1}^{2}-v_{1}^{2})\] \[g\times h=\frac{1}{2}[{{(6)}^{2}}-{{(4)}^{2}}]\]\[[\because {{h}_{1}}-{{h}_{2}}=h]\] \[10\times h=\frac{1}{2}[36-16]\] Or \[h=\frac{20}{20}\] \[=1\,m\]


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