9th Class Science Work and energy Question Bank Work, Energy and Power

  • question_answer
    A motor of power P is employed to deliver water at a certain rate from a given pipe. In order to obtain thrice as much water from the same pipe in the same time, another motor of power P' is required. Given:\[\text{g}=\text{9}.\text{3489 m}{{\text{s}}^{\text{-2}}}\]Now, the ratio \[P'/P\] equals

    A)  27                           

    B)         81

    C)  561                       

    D)         \[{{\left( \text{9}.\text{3489} \right)}^{\text{3}}}\]

    Correct Answer: A

    Solution :

     If the volume of water obtained becomes \[n\] times, the velocity with which the water is ejected from the same pipe must also become \[n\] times because, volume \[=\] area \[\times \] velocity, and area remains same. If volume becomes \[n\] times, the mass of the water ejected also becomes \[n\] times. Now, to think about the power of motor, the most suitable formula for the given variables may be taken as rate of change kinetic energy. \[P=\frac{1}{2}m{{v}^{2}}\]or\[P\propto \frac{m{{v}^{2}}}{t}\] As decided above, \[\frac{m}{t}\] has become n times, so power becomes \[{{n}^{3}}\] times. Hence in the given problem \[\frac{P'}{P}\] becomes\[{{3}^{3}}\]\[=27\,\,times\] Note that, in such problems force becomes\[{{n}^{3}}\]times


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