Question

The ratio of MKS units and CGS units of coefficient of viscosity is
A. $\dfrac{{{\eta }_{MKS}}}{{{\eta }_{CGS}}}=10$
B. $\dfrac{{{\eta }_{MKS}}}{{{\eta }_{CGS}}}=9.8$
C. $\dfrac{{{\eta }_{MKS}}}{{{\eta }_{CGS}}}=0.1$
D. $\dfrac{{{\eta }_{MKS}}}{{{\eta }_{CGS}}}=\dfrac{1}{9.8}$

Answer 1

As the very first step, you could find the unit of the given quantity in the respective unit systems. Then you could think of the conversion of these units into the other. Keeping this conversion in mind you could take the ratio as per requirement and thus get the answer accordingly.

Complete step by step solution:
In the question, we are asked to find the ratio of MKS that is the SI units and CGS units of coefficient of viscosity. We could simply find the units of this quantity in respective unit systems and then recall the conversion between the two and then take the ratio to get the answer. Before doing all that, let us define this quantity.
The tangential force that is acting per unit area that is required so as to maintain unit velocity gradient is called coefficient of viscosity. Mathematically, it can be expressed as,
$\eta =\dfrac{Fdx}{Adv}$
You may recall that the unit of the coefficient of viscosity in MKS and CGS unit systems respectively are $kg/m/s$ and $poise$ respectively. Also, we know that,
$1poise=0.1kg/m/s$
Therefore, by taking the ratio of units of coefficient of viscosity in MKS and CGS units we will get,
$\dfrac{{{\eta }_{MKS}}}{{{\eta }_{CGS}}}=\dfrac{1kg/m/s}{0.1kg/m/s}$
$\therefore \dfrac{{{\eta }_{MKS}}}{{{\eta }_{CGS}}}=10:1$
Therefore, we got the required ratio as 10:1. Hence, option A is the right answer.

Note: The same unit systems may have different units for a derived quantity. The quantity under discussion falls under this category and it’s another unit in CGS system is$dynes/c{{m}^{2}}$ and in MKS it is$Ns/{{m}^{2}}$ . We have the following conversions.
$1N={{10}^{5}}dyne$
$1{{m}^{2}}={{10}^{4}}c{{m}^{2}}$
We will end up in the same answer as above if we take the ratio this way.