WB JEE Medical WB JEE Medical Solved Paper-2013

  • question_answer
    In an experiment four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4% respectively. Quantity P is calculated as follows \[P=\frac{{{a}^{3}}{{b}^{2}}}{cd}%,\]Error in P is

    A)  14%

    B)  10%

    C)  7%

    D)  4%

    Correct Answer: A

    Solution :

     \[P=\frac{{{a}^{3}}{{b}^{2}}}{cd}\] \[\therefore \] \[\frac{\Delta P}{P}\times 100\] \[=\left( \frac{3\Delta a}{a}+\frac{2\Delta b}{b}+\frac{\Delta c}{c}+\frac{\Delta d}{d} \right)\times 100\] \[=3\frac{\Delta a}{a}\times 100+2\frac{\Delta b}{b}\times 100+\frac{\Delta c}{c}\] \[\times 100+\frac{\Delta d}{d}\times 100\] \[=3\times 1+2\times 2+3+4\] \[=3+4+3+4=14%\]


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