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Haryana PMT Haryana PMT Solved Paper-2009

  • question_answer
    The formula for horizontal range of a projectile is \[S=\frac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}\] where v is initial speed, \[\theta \] angle of inclination and g is acceleration due to gravity. The Wheatstone bridge shown in figure can be used to determine the range if the following arrangement is made

    A)  Q proportional to \[{{v}^{2}}\]  P proportional to g R proportional to \[si{{n}^{2}}\] \[\theta \]

    B)  Q proportional to g P proportional to \[{{v}^{2}}\] R proportional to \[si{{n}^{2}}\]\[\theta \]

    C)  Q proportional to g P proportional to sin2 \[\theta \] R proportional to \[{{v}^{2}}\]

    D)  Q proportional to \[si{{n}^{2}}\] 6 P proportional to \[{{v}^{2}}\] R proportional to g

    Correct Answer: A

    Solution :

                    For the given Wheatstone bridge \[\frac{P}{R}=\frac{Q}{S}\] \[S=\frac{Q\times R}{P}\] \[S=\frac{{{v}^{2}}\,{{\sin }^{2}}\,\theta }{2g}\] Hence option (c) is correct.


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