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JIPMER Jipmer Medical Solved Paper-2001

  • question_answer
    According to Newton, the viscous force acting between liquid layers of area A and velocity gradient\[\frac{\Delta \upsilon }{\Delta z}\]is given by \[F=\eta A\frac{\Delta \upsilon }{\Delta z}\] where \[\eta \]  is a constant called coefficient of viscosity. The dimension of \[\eta \] are:

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

    B)         \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{\text{-1}}}{{\text{T}}^{\text{-1}}}\text{ }\!\!]\!\!\text{ }\]           

    C)         \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{-2}}{{\text{T}}^{-2}}\text{ }\!\!]\!\!\text{ }\]       

    D)         \[\text{ }\!\![\!\!\text{ }{{\text{M}}^{0}}{{\text{L}}^{0}}{{\text{T}}^{0}}\text{ }\!\!]\!\!\text{ }\]

    Correct Answer: B

    Solution :

    Coefficient of viscosity \[\eta =\frac{\text{force}\,\text{ }\!\!\times\!\!\text{ }\,\text{length}}{\text{area}\,\text{ }\!\!\times\!\!\text{ velocity}}\] Dimensions of \[\eta \] \[=\frac{\text{Dimension of}\,\text{force}\,\,\text{ }\!\!\times\!\!\text{ }\,\,\text{dimension of length}}{\text{Dimension of area}\,\,\text{ }\!\!\times\!\!\text{ }\,\,\text{dimension of velocity}}\] \[=\frac{ML{{T}^{-2}}\times L}{{{L}^{2}}\times L{{T}^{-1}}}=[M{{L}^{-1}}{{T}^{-1}}]\]


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