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EAMCET Medical EAMCET Medical Solved Paper-2000

  • question_answer
    A stone is projected with a velocity u at angle \[\theta \] with the horizontal reaches maximum height H1, when it is projected with a velocity u at an angle \[\left( \frac{\pi }{2}-\theta  \right)\] with the horizontal, it reaches to a maximum height H2. The relation between the horizontal range R of the projectile, H1 and H2 is:

    A)  \[R=4\sqrt{{{H}_{1}}{{H}_{2}}}\]                              

    B)  \[R=4({{H}_{1}}-{{H}_{2}})\]

    C)  \[R=4({{H}_{1}}+{{H}_{2}})\]     

    D)  \[R={{H}_{1}}^{2}/{{H}_{2}}^{2}\]

    Correct Answer: A

    Solution :

                     \[{{H}_{1}}=\frac{{{u}^{2}}si{{n}^{2}}\theta }{2g}\] \[{{H}_{2}}=\frac{{{u}^{2}}{{\sin }^{2}}(90-\theta )}{2g}=\frac{{{u}^{2}}{{\cos }^{2}}2\theta }{2g}\]            From \[\frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{T}_{2}}{{V}_{2}}}{{{T}_{2}}}\] \[\frac{{{P}_{1}}{{m}_{1}}}{\rho {{T}_{1}}}=\frac{{{P}_{2}}{{m}_{2}}}{\rho {{T}_{2}}}\] \[\frac{720\times m}{313}=\frac{{{P}_{2}}\left( m-\frac{1}{4}m \right)}{626}\] \[\frac{720\times m}{313}=\frac{{{P}_{2}}\times \frac{3}{4}m}{626}\] \[{{P}_{2}}=\frac{720\times 626\times 4}{313\times 3}\] \[=1920\,kPa.\]


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