Question

The unit of stress is:
$\begin{align}
  & A.\text{ kg}{{\text{m}}^{-2}} \\
 & B.\text{ N k}{{\text{g}}^{-1}} \\
 & C.\text{ N}{{\text{m}}^{-2}} \\
 & D.\text{ N} \\
\end{align}$

Answer 1

Hint: As we know elastic properties of the body are mainly described in terms of concepts like stress and strain. Stress is one of the types of pressure which is defined as internal elastic restoring force per unit cross sectional area (A) of the body. Unit of force is N and unit of area is ${{m}^{2}}$. Use units of force and area and calculate units of stress.

Formula used:
$\text{stress=}\dfrac{\text{applied force (restoring force)}}{\text{area}}$

Complete answer:
To express the properties of any elastic body we use the concept of stress. If the body is elastic in nature then internal restoring force per unit cross sectional area of the body is called stress. In simple words it is also defined as applied force per unit cross sectional area of the body. Thus, stress can be written as,
$\text{stress=}\dfrac{\text{applied force}}{\text{area}}$
Mathematically, $stress=\dfrac{F}{A}$
We know that force can be defined as a product of mass and acceleration. Whereas acceleration can be defined as rate of change of velocity with respect to time and velocity and can be expressed as displacement per unit time. Therefore unit of acceleration is $m/{{s}^{2}}$
Therefore unit of force is written as,
$\begin{align}
  & F=ma \\
 & F=kgm/{{s}^{2}} \\
\end{align}$
Consider where the unit of mass is kg.
We know that $kgm/{{s}^{2}}$ can be written as Newton denoted by N. Unit of area is ${{m}^{2}}$.
Let force is 1N and area of cross section is $1{{m}^{2}}$ then stress is written as,
$\text{stress}=N/{{m}^{2}}$
Hence the unit of stress is $N{{m}^{-2}}$.

Therefore option (c) is the correct option.

Additional information:
There are two types of body namely elastic and plastic body. Internal restoring force in the elastic body is large while the internal restoring force is small in the case of the plastic body. We know that stress is directly proportional to internal restoring force. Therefore stress in the elastic body is more than plastic.
Dimensions of stress are $\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-2}} \right]$
There are two types of stress present:
1) Longitudinal stress which is also known as tensile stress the ratio of normal applied force to the area of cross section of longitudinal body is known as longitudinal stress.
2) Volume stress can be expressed as applied normal force produces a charge in volume of body then that normal force acting on body per unit surface area.
3) Shearing stress is a ratio of tangential applied force to the force surface area (basically it is a parallel to force)

Note:
In equilibrium position, the applied force and internal elastic restoring force are numerically equal and however they are oppositely directed. As we has calculate S.I unit is stress is $N/{{m}^{2}}$ and C.G.S unit of stress can be written as $Dyne/c{{m}^{2}}$ if we take ratio of S.I unit to C.G.S. unit then value of ratio is 10.