Write dimensional formula of pressure.
Answer
642.9k+ views
Hint: To write the dimensional formula for a physical quantity its formula should be known. Pressure is Force upon area.
Complete step by step answer:
Pressure is defined as the amount of physical force exerted on a unit area. It is given by:
$P=\dfrac{Force}{Area}$
$P=\dfrac{F}{A}$
To find the dimensional formula for pressure we need the dimensional formula of force,
$F=ma$
$F=[M][L{{T}^{-2}}]$
$\begin{align}
& \Rightarrow F=[{{M}^{1}}{{L}^{1}}{{T}^{-2}}] \\
& \\
\end{align}$
Now, for pressure,
$P=\dfrac{[{{M}^{1}}{{L}^{1}}{{T}^{-2}}]}{[{{L}^{2}}]}$
$P=[{{M}^{1}}{{L}^{-1}}{{T}^{-2}}]$
Its SI unit is Pascal
$1Pa=\dfrac{1N}{1{{m}^{2}}}$
$1Pa=N{{m}^{-2}}$
The correct answer is SI units of pressure is $P=[{{M}^{1}}{{L}^{-1}}{{T}^{-2}}]$
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: While solving dimensional formula questions students must note that every physical quantity must be expressed in its absolute units only.
Complete step by step answer:
Pressure is defined as the amount of physical force exerted on a unit area. It is given by:
$P=\dfrac{Force}{Area}$
$P=\dfrac{F}{A}$
To find the dimensional formula for pressure we need the dimensional formula of force,
$F=ma$
$F=[M][L{{T}^{-2}}]$
$\begin{align}
& \Rightarrow F=[{{M}^{1}}{{L}^{1}}{{T}^{-2}}] \\
& \\
\end{align}$
Now, for pressure,
$P=\dfrac{[{{M}^{1}}{{L}^{1}}{{T}^{-2}}]}{[{{L}^{2}}]}$
$P=[{{M}^{1}}{{L}^{-1}}{{T}^{-2}}]$
Its SI unit is Pascal
$1Pa=\dfrac{1N}{1{{m}^{2}}}$
$1Pa=N{{m}^{-2}}$
The correct answer is SI units of pressure is $P=[{{M}^{1}}{{L}^{-1}}{{T}^{-2}}]$
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: While solving dimensional formula questions students must note that every physical quantity must be expressed in its absolute units only.
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