Question

One centipoise is equal to
\[\begin{align}
  & A)1kg-{{m}^{-1}}{{s}^{-1}} \\
 & B)1000kg-{{m}^{-1}}{{s}^{-1}} \\
 & C)0.1kg-{{m}^{-1}}{{s}^{-1}} \\
 & D)0.001kg-{{m}^{-1}}{{s}^{-1}} \\
\end{align}\]

Answer 1

Hint: Poise, is the unit of viscosity of a fluid. Students must know what is 1 poise equal to in terms of MKS unit system. Students must also know how to convert poise to centipoise, using mathematical expressions.

Formula used:
$\begin{align}
  & 1poise=1kg-{{m}^{-1}}{{s}^{-1}} \\
 & 1centipoise={{10}^{-2}}poise \\
\end{align}$

Complete step by step answer:
Students must be familiar that viscosity appears in the following formula for a liquid,
$F=\eta A\dfrac{u}{y}$…(1), where F is the frictional force fluid, A is the area, u is the velocity of topmost layers of fluid, y is the distance between the topmost and the lowermost layers (which has zero velocity) and finally, $\eta $is the coefficient of viscosity.
 Rearranging equation (1), we get.
$\eta =\dfrac{F}{A}\dfrac{y}{u}$, now putting unit of force as $kg-m{{s}^{-2}}$, distance as m, velocity as $m{{s}^{-1}}$and area as ${{m}^{2}}$, we get the unit of coefficient of viscosity i.e. 1 poise.
$\begin{align}
  & \Rightarrow 1poise=\dfrac{kg-m{{s}^{-2}}}{{{m}^{2}}}\times \dfrac{m}{m{{s}^{-1}}} \\
 & \Rightarrow 1poise=kg-{{m}^{-1}}{{s}^{-1}} \\
\end{align}$
Now, we know that one centipoise is equal to ${{10}^{-2}}$poise.
$\begin{align}
  & \Rightarrow 1centipoise={{10}^{-2}}poise \\
 & \Rightarrow 1centipoise={{10}^{-2}}kg-{{m}^{-1}}{{s}^{-1}} \\
 & \Rightarrow 1centipoise=0.001kg-{{m}^{-1}}{{s}^{-1}} \\
\end{align}$

So, the correct answer is “Option D”.

Additional Information:
Viscosity of fluid is the measurement of its resistance to deformation at a specific rate.
A liquid of higher viscosity is called a thick liquid.
Viscosity is not the measure of density of a liquid, for instance mustard oil has higher viscosity compared to water but water has higher density.
An ideal fluid has no viscosity.
There are only two elements in the periodic table which exist in liquid state at the room temperature, namely mercury (a metal) and bromine (a nonmetal).
For a long time, it was believed that glasses are supercooled liquids, but now they are considered amorphous solids.
A fluid with extremely high viscosity (like pitch) can be confused with a solid.
At very low temperatures, superfluids have almost zero viscosity.

Note:
Students can also solve the question by the following formula,
$\tau =\eta \dfrac{\partial u}{\partial y}$, where, $\tau $is the pressure and $\dfrac{\partial u}{\partial y}$ is the shear velocity.