Question

What are the S.I units of surface tension and coefficient of viscosity respectively?
A) ${\text{N }}{{\text{m}}^{ - 1}}{\text{ and kg }}{{\text{m}}^{ - 1}}{\text{ }}{{\text{s}}^{ - 1}}$
B) ${\text{N }}{{\text{m}}^{ - 1}}{\text{ and kg }}{{\text{m}}^{ - 2}}{\text{ }}{{\text{s}}^{ - 1}}$
C) ${{\text{N}}^2}{\text{ }}{{\text{m}}^{ - 1}}{\text{ and kg }}{{\text{m}}^{ - 1}}{\text{ }}{{\text{s}}^{ - 2}}$
D) ${\text{N m and kg }}{{\text{m}}^{ - 1}}{\text{ }}{{\text{s}}^{ - 1}}$

Answer 1

To solve this we must know the definitions of surface tension and coefficient of viscosity. From the definitions, derive the expressions for surface tension and coefficient of viscosity. From the expressions derive the S.I units of the surface tension and coefficient of viscosity.

Complete step-by-step answer:
We know that the tendency of the fluid surface to shrink to the minimum surface area possible is known as surface tension. The surface tension can be determined by the difference in the interactions between the molecules of the fluid with the wall of the container.
The expression for surface tension is as follows:
$\gamma = \dfrac{1}{2}\dfrac{F}{L}$
Where $\gamma$ is the surface tension,
$F$ is the force,
$L$ is the length.
The S.I unit of force is Newton i.e. ${\text{N}}$ and the S.I unit of length is meter i.e. ${\text{m}}$. Thus, the S.I unit of surface tension is,
$\gamma = \dfrac{{\text{N}}}{{\text{m}}}$
$\gamma = {\text{N }}{{\text{m}}^{ - 1}}$
Thus, the S.I unit of surface tension is ${\text{N }}{{\text{m}}^{ - 1}}$.
We know that the force of friction between the two layers of a fluid having the area in square meter and are separated by a distance has a velocity which is given by the equation as follows:
$F = \eta A\dfrac{{dV}}{{dx}}$
Where $F$ is the force of friction,
$\eta$ is the coefficient of viscosity,
$A$ is the area,
$\dfrac{{dV}}{{dx}}$ is the velocity gradient.
The S.I unit of force is newton i.e. ${\text{N}}$, the S.I unit of area is square meter i.e. ${{\text{m}}^2}$, the S.I unit of velocity gradient is per second i.e. ${\text{se}}{{\text{c}}^{ - 1}}$. Thus, the S.I unit of coefficient of viscosity is,
${\text{N}} = \eta \times {{\text{m}}^2} \times {\text{se}}{{\text{c}}^{ - 1}}$
$\eta = \dfrac{{\text{N}}}{{{{\text{m}}^2} \times {\text{se}}{{\text{c}}^{ - 1}}}}$
But the S.I unit of Newton is ${\text{ m kg }}{{\text{s}}^{ - 2}}$. Thus,
$\eta = \dfrac{{{\text{ m kg }}{{\text{s}}^{ - 2}}}}{{{{\text{m}}^2} \times {{\text{s}}^{ - 1}}}}$
$\eta = {\text{kg }}{{\text{m}}^{ - 1}}{\text{ }}{{\text{s}}^{ - 1}}$
Thus, the S.I unit of coefficient of viscosity is ${\text{kg }}{{\text{m}}^{ - 1}}{\text{ }}{{\text{s}}^{ - 1}}$.
Thus, the S.I units of surface tension and coefficient of viscosity respectively are ${\text{N }}{{\text{m}}^{ - 1}}{\text{ and kg }}{{\text{m}}^{ - 1}}{\text{ }}{{\text{s}}^{ - 1}}$.
Thus, the correct option is (A) ${\text{N }}{{\text{m}}^{ - 1}}{\text{ and kg }}{{\text{m}}^{ - 1}}{\text{ }}{{\text{s}}^{ - 1}}$.

Note: The S.I units are the standard units of the measurement. S.I units are defined by the International System of Units. S.I stands for System International which is the set of physical units agreed upon by international convention. The International System of Units is the modern form of the metric system.