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NEET Sample Paper NEET Sample Test Paper-83

  • question_answer
    A syringe containing water is held horizontally with its nozzle at a height h above the ground as shown in figure. The cross-sectional areas of the piston and the nozzle are A and a respectively. The piston is pushed with a constant speed v. The horizontal range R of the stream of water on the ground is

    A)  \[R=v\sqrt{\frac{2h}{g}}\]        

    B)  \[R=v\sqrt{\frac{h}{2g}}\]

    C)  \[R=\frac{av}{A}\sqrt{\frac{2h}{g}}\]               

    D)  \[R=\frac{Av}{a}\sqrt{\frac{2h}{g}}\]

    Correct Answer: D

    Solution :

    Let v be the horizontal speed of water when it emerges from the nozzle. From the equation of continuity, we have \[\operatorname{AV}=\,\,av\,\,\,or\,\,v=\,\,\frac{AV}{a}\]             ... (i) Let t be the time taken by the stream of water to strike the ground. The horizontal and vertical distances covered in time t are \[\operatorname{R}=vt\]                                     ... (ii) \[h=\frac{1}{2}\]                                   ... (iii) From Eq. (iii), we have \[t=\sqrt{\frac{2h}{g}}\] Using this value in Eq. (ii), we get \[R=v\sqrt{\frac{2h}{g}}\]                                  ... (iv) Using Eqs. (i) and (iv), we have \[R=\frac{AV}{a}\sqrt{\frac{2h}{g}}\]  


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