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NEET Sample Paper NEET Sample Test Paper-58

  • question_answer
    A large tank is filled water\[(density={{10}^{3}}Kg/{{m}^{3}})\]. A small hole is mode at a depth 10m below water surface. The range of water oozing out of the hole is R on ground. What extra pressure must be applied on the water surface, so that the range becomes 2R: \[(use1\,\,atm\,={{10}^{5}}Pa\,\,and\ g=10/{{s}^{2}})\]

    A) 9 atm                           

    B) 4 atm   

    C) 5 atm                           

    D) 3 atm

    Correct Answer: D

    Solution :

    [d] \[\Delta \,P=\frac{1}{2}\rho {{v}^{2}}\Rightarrow \,v=\sqrt{\frac{2\,\Delta \,p}{p}}\] \[Range\,={{V}_{(Horizantal)}}\,\times t\] \[R\propto \,\sqrt{p}\] We have to attain range two times of initial valve \[\Delta \,P={{\rho }_{w}}gh\] \[\Delta \,P={{10}^{3}}\times 10\times 10={{10}^{5}}\frac{N}{{{m}^{2}}}=1\,atm\] \[In\,{{2}^{nd}}\,case\] \[{{\rho }_{w}}gh+{{P}_{1\,(extra\,pressure)}}=4\Delta P=4\,atm\] \[1\,atm\,+{{P}_{1}}=\,4\,atm\] \[Extra\,pressure\,{{P}_{1}}=3\,atm\]


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