BHU PMT BHU PMT (Screening) Solved Paper-2011

  • question_answer
    A pressure of\[x\]dyne\[c{{m}^{-2}}\]is equivalent to

    A)  \[\left( \frac{x}{10} \right)N{{m}^{-2}}\]                              

    B)  \[\left( \frac{x}{100} \right)N{{m}^{-2}}\]

    C)  \[10x\,N{{m}^{-2}}\]                    

    D)  \[\left( \frac{x}{{{10}^{4}}} \right)\,N{{m}^{-2}}\]

    Correct Answer: A

    Solution :

                     Dyne\[c{{m}^{-2}}\]can be converted into\[N{{m}^{-2}}\]either using dimensions or remembering 1 dyne\[={{10}^{-5}}m\] \[\frac{1\,dyne}{c{{m}^{2}}}=\frac{{{10}^{-5}}}{{{10}^{-4}}}\frac{N}{{{m}^{2}}}={{10}^{-1}}N/{{m}^{2}}\] \[x\frac{dyne}{c{{m}^{2}}}=\frac{x}{10}\frac{N}{{{m}^{2}}}\]


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