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Mechanical Properties of Fluids — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsMechanical Properties of Fluids4 Marks
Question
Find the conversion factor between the SI and CGS units of coefficient of viscosity using dimensional analysis.
✓

Answer


The dimensions of the coefficient of viscosity $\eta$ are
$
[\eta]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]
$
The SI and CGS units of coefficient of viscosity are the pascal-second and poise, respectively.
$1 \mathrm{~Pa} \cdot \mathrm{s}=1 \mathrm{~N} \cdot \mathrm{m}^{-2} \cdot \mathrm{s}=1 \mathrm{~kg}-\mathrm{m}^{-1} \cdot \mathrm{s}^{-1}$
$1 \mathrm{P}=1 \mathrm{dyn} \cdot \mathrm{cm}^{-2} \cdot \mathrm{s}=1 \mathrm{~g}-\mathrm{cm}^{-1} \cdot \mathrm{s}^{-1}$
Let $1 \mathrm{~Pa} \cdot \mathrm{s}=\mathrm{xP}$
$
\therefore 1\left[\mathrm{M}_1 \mathrm{~L}_1^{-1} \mathrm{~T}_1^{-1}\right]=\mathrm{x}\left[\mathrm{M}_2 \mathrm{~L}_2^{-1} \mathrm{~T}_2^{-1}\right]
$
where subscripts 1 and 2 pertain to SI and CGS units.
$
\begin{aligned}
& \therefore x=\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right]\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{-1}\left[\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right]^{-1} \\
&=\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right]\left[\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right]\left[\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right] \\
&=\frac{1 \mathrm{~kg}}{1 \mathrm{~g}} \cdot \frac{1 \mathrm{~cm}}{1 \mathrm{~m}} \cdot \frac{1 \mathrm{~s}}{1 \mathrm{~s}} \\
&=\frac{1000}{1} \times \frac{1}{100} \times \frac{1}{1}=10^3 \times 10^{-2}=10 \\
& \therefore 1 \mathrm{~Pa} \cdot \mathrm{s}=10 \mathrm{P}
\end{aligned}
$

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