J & K CET Engineering J and K - CET Engineering Solved Paper-2005

  • question_answer
    A man does a given amount of work in \[10\text{ }s\]. Another man does the' same amount of work in \[20\text{ }s\]. The ratio of the output power of first man to the second man is

    A)  \[1\]           

    B)  \[1/2\]                                

    C)  \[2/1\]          

    D)  None of these                      

    Correct Answer: C

    Solution :

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