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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer
    According to Newton viscus force is given by \[F=-\eta A\frac{d\upsilon }{dx}\]where\[\eta =\]coefficient of viscosity, so dimensions of \[\eta \] will be:

    A)  \[[M{{L}^{-1}}{{T}^{-2}}]\]         

    B)         \[[ML{{T}^{-2}}]\]

    C)  \[[M{{L}^{-1}}{{T}^{-1}}]\]         

    D)         \[[{{M}^{-1}}{{L}^{2}}{{T}^{-2}}]\]

    Correct Answer: C

    Solution :

                 \[F=-\eta A\frac{d\upsilon }{dc}\] \[\therefore \]  \[\eta =-\frac{F}{A}\frac{dx}{d\upsilon }\] Writing the dimensions \[[\eta ]=\frac{[ML{{T}^{-2}}]}{[{{L}^{2}}]}\frac{[L]}{[L{{T}^{-1}}]}\] \[=[M{{L}^{-1}}{{T}^{-1}}]\]


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