Question

The SI units of thrust and pressure are respectively given by:
(a). \[N,N{m^2}\]
(b). \[N,N{m^{ - 2}}\]
(c). \[N{m^{ - 1}},N{m^{ - 2}}\]
(d). \[N{m^{ - 2}},N\]

Answer 1

- Hint: Thrust is basically the reaction force to an action force in accordance with Newton's third law of motion.

Complete step-by-step solution -

We will have to find the SI units of thrust and pressure so we need to know the formula for the physical quantity first. Since thrust is a kind of force it has the same dimensions as that of Force. Since the SI unit of force is Newton, thrust would also have the same SI unit which is N.
Pressure is the amount of force applied per unit area. i.e.
\[P = \dfrac{F}{A}\]
Therefore, SI unit of pressure is \[N{m^{ - 2}}\]
The correct answer is option B. \[N,N{m^{ - 2}}\]

Additional Information:

The SI units and Dimensional formula of some important physical quantities to remember are:
Work, Energy of all kinds = \[J,[{{M}^{1}}{{L}^{2}}{{T}^{-2}}]\]
Power =\[W,[{{M}^{1}}{{L}^{2}}{{T}^{-3}}]\]
Planck’s Constant (h) = \[Js,[{{M}^{1}}{{L}^{2}}{{T}^{-1}}]\]
Angular displacement (\[\theta\])=$rad,[{{M}^{0}}{{L}^{0}}{{T}^{0}}]$.
Angular velocity (\[\omega\])=\[rad{{s}^{-1}},[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\]
Force constant ($\dfrac{\text{force}}{\text{displacement}}$) = $N{{m}^{-1}},\left[ {{M}^{1}}{{L}^{0}}{{T}^{-2}} \right]$
Coefficient of elasticity ($\dfrac{\text{stress}}{\text{strain}}$) = $N{{m}^{-2}},\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-2}} \right]$
Angular frequency \[(\omega )=,rad{{s}^{-1}}[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]\]
Angular momentum \[I\omega =kg{{m}^{2}}{{s}^{-1}}[{{M}^{1}}{{L}^{2}}{{T}^{-1}}]\]

Note: The SI unit for similar types of physical quantities remains the same. For example, Force, Thrust and Tension all have the same SI unit Newton. Similarly, all types of energies (Kinetic, potential, thermal etc.) have the same SI unit Joules. Pressure, Stress, Coefficient of elasticity all have the same SI unit that is. \[N,N{m^{ - 2}}\]