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.


You need to login to perform this action.
You will be redirected in 3 sec spinner