J & K CET Medical J & K - CET Medical Solved Paper-2005

  • question_answer
    A man does a given amount of work in 10 s. Another man does the same amount of work in 20 s. The ratio of the output power of first man to the second man is :

    A)  1                                            

    B)  ½

    C)  2/1                                       

    D)  none of these

    Correct Answer: C

    Solution :

                     The rate of doing work is defined as power \[Power=\frac{work\text{ }done}{time}\]           \[\therefore \]  \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{W}_{1}}/{{t}_{1}}}{{{W}_{2}}/{{t}_{2}}}\] Given, \[{{t}_{1}}=10s,{{t}_{2}}=20s,{{W}_{1}}={{W}_{2}}\] \[\therefore \]  \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{t}_{2}}}{{{t}_{1}}}=\frac{2}{1}\]


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