AMU Medical AMU Solved Paper-2005

  • question_answer
    A machine which is 75% efficient, uses 12 J of energy in lifting 1 kg mass through a certain distance. The mass is then allowed to fall through the same distance. The velocity at the end of its fall is

    A) \[\sqrt{\text{l2}}\text{m}/\text{s}\]                      

    B)  \[\sqrt{18}\text{m}/\text{s}\]

    C)  \[\sqrt{24}\text{m}/\text{s}\]                                 

    D)  \[\sqrt{32}\text{m}/\text{s}\]

    Correct Answer: B

    Solution :

                     From law of conservation of energy Potential energy = kinetic energy Given, potential energy                \[=\frac{75}{100}\times 12=9\,J\]... (i) Kinetic energy                   \[=\frac{1}{2}m{{v}^{2}}\]            ... (ii) Equating Eqs. (i) and (ii), we get                 \[\frac{1}{2}m{{v}^{2}}=9\] \[\Rightarrow \]               \[v=\sqrt{\frac{9\times 2}{1}}=\sqrt{18}\,m/s\]


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