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If momentum [\mathrm{P}], area [\mathrm{A}] and time [\mathrm{T}] are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is :
 

Option: 1

\mathrm{\left[\mathrm{P} \mathrm{A}^1 \mathrm{~T}^{0}\right] }


Option: 2

\mathrm{\left[\mathrm{PA} \mathrm{T}^{-1}\right]}
 


Option: 3

\mathrm{\left[\mathrm{PA}^{-1} \mathrm{~T}\right]}
 


Option: 4

\mathrm{\left[\mathrm{P} \mathrm{A}^{-1} \mathrm{~T}^{-1}\right]}


Answers (1)

best_answer

Let,

\mathrm{\eta \; \alpha \; P^{x}\; A^{y}\; T^{z}}

\mathrm{\eta \; }\rightarrowCoefficient of viscosity

\mathrm{\therefore \left [ \eta \right ]=\left [ P \right ]^{x}\left [ A \right ]^{y}\left [ T \right ]^{z}}

\mathrm{ {\left[M^{\prime} L^{-1} T^{-1}\right]=\left[M^{\prime} L^{\prime} T^{-1}\right]^x\left[L^2\right]^y[T]^z}}

\mathrm{{\left[M^{\prime} L^{-1} T^{-1}\right]=\left[M^x \: L^{x+2 y} \: \: T^{-x+z}\right]} }

Comparing both the sides ,we get,

\mathrm{\begin{aligned} &x=9 \rightarrow(1) \\ \end{aligned}}

\begin{aligned} &x+2 y=-1 \\ \end{aligned}

\begin{aligned} &y=-1 \rightarrow(2) \\ \end{aligned}

\begin{aligned} &-x+z=-1 \\ \end{aligned}

\begin{aligned} &z=0 \rightarrow(3) \\ \end{aligned}

\begin{aligned} &\therefore[\eta]=\left[P^1 A^{-1} T^0\right] \end{aligned}

Hence (1) is correct option.

Posted by

Devendra Khairwa

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