CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2007

  • question_answer
    The cylindrical tube of a spray pump has a cross-section of \[8\text{ }c{{m}^{2}}\], one end of which has 40 fine holes each of area \[{{10}^{-8}}{{m}^{2}}\]. If the liquid flows inside the tube with a speed of\[0.15\text{ }m\text{ }mi{{n}^{-1}}\], the speed with which the liquid is ejected through the holes is

    A)  \[50\text{ }m{{s}^{-1}}\]                             

    B)  \[5\text{ }m{{s}^{-1}}\]

    C)  \[0.05\text{ }m{{s}^{-1}}\]                         

    D)  \[0.5\text{ }m{{s}^{-1}}\]

    Correct Answer: B

    Solution :

    According to equation of continuity,                 \[av=\] constant \[\therefore \] For tube, \[(8\times {{10}^{-4}})\times \left( \frac{0.15}{60} \right)={{a}_{1}}{{v}_{1}}\] For holes,            \[(40\times {{10}^{-8}})\times v={{a}_{2}}{{v}_{2}}\] \[\therefore \]  \[{{a}_{2}}{{v}_{2}}={{a}_{1}}{{v}_{1}}\] \[\therefore \]  \[40\times {{10}^{-8}}\times v=\frac{8\times {{10}^{-4}}\times 0.15}{60}\] \[\Rightarrow \]               \[v=\frac{8\times {{10}^{-4}}\times 0.15}{40\times {{10}^{-8}}\times 60}\]                 \[=5\text{ }m/s\]


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