JEE Main & Advanced Sample Paper JEE Main Sample Paper-5

  • question_answer
    Poise is the CGS unit of coefficient of viscosity. Suppose we employ a system of units in which unit of mass is a kg, the unit of length is P metre and unit of time is y s. In this new system, 1Poise is equal to

    A)  \[1000\alpha {{\beta }^{-1}}{{\gamma }^{-1}}\]                

    B)  \[10\alpha {{\beta }^{-1}}{{\gamma }^{-1}}\]

    C) \[0.1{{\alpha }^{-1}}\beta \gamma \]                     

    D)  \[0.01{{\alpha }^{-1}}\beta \gamma \]

    Correct Answer: C

    Solution :

     Idea Dimensions of coefficient of viscosity is \[[\eta ]=[{{M}^{1}}{{L}^{-1}}{{T}^{-1}}]\] also\[{{n}_{2}}{{u}_{2}}={{n}_{1}}{{u}_{1}}\] \[\Rightarrow \]\[{{n}_{2}}=\frac{{{n}_{1}}{{u}_{1}}}{{{u}_{2}}}\] \[={{n}_{1}}{{\left[ \frac{{{M}_{1}}}{{{M}_{2}}} \right]}^{1}}{{\left[ \frac{{{L}_{1}}}{{{L}_{2}}} \right]}^{-1}}{{\left[ \frac{{{T}_{1}}}{{{T}_{2}}} \right]}^{-1}}\] \[{{u}_{1}}\]and\[{{u}_{2}}\]are two units of measurement and\[{{n}_{1}}\]and\[{{n}_{2}}\]are their respective numerical values. Coefficient of viscosity\[\eta =[{{M}^{1}}{{L}^{-1}}{{T}^{-1}}]\] In new system, we have unit of mass \[=\alpha \times {{10}^{3}}g\] Unit of length\[=\beta \times 100\]cm and unit of time = \[\gamma \] s So, new system unit is \[={{\left[ \frac{1g}{\alpha \times {{10}^{3}}g} \right]}^{1}}{{\left[ \frac{1cm}{100\beta cm} \right]}^{-1}}\left[ \frac{1s}{\gamma s} \right]\] \[=0.1{{\alpha }^{-1}}{{\beta }^{-1}}\gamma \] or \[=0.1{{\alpha }^{-1}}\beta \gamma \] TEST Edge Every year about 1 to 2 questions are asked from dimension analysis. Besides conversion of one system of units into another, checking of accuracy of formulae and derivation of formulae can also be asked.

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