Write the dimensional formula of pressure.
Answer
635.1k+ views
Hint The dimensional formula of the pressure is given by using the formula of the pressure and the terms which are in the pressure formula, then the SI units of the terms which are in the pressure formula are used. Then the SI unit of the terms are converted to the dimensional formula.
Useful formula
The formula for the pressure is given as,
$P = \dfrac{F}{A}$
Where, $P$ is the pressure in the object, $F$ is the force in the object and $A$ is the are of the object.
The formula of the force is given as,
$F = ma$
Where, $F$ is the force of the object, $m$ is the mass of the object and $a$ is the acceleration of the object.
Complete step by step solution
Now,
The formula for the pressure is given as,
$P = \dfrac{F}{A}\,...................\left( 1 \right)$
Now,
The formula of the force is given as,
$F = ma\,..................\left( 2 \right)$
By substituting the equation (2) in the equation (1), then the equation (1) is written as,
$P = \dfrac{{ma}}{A}\,...................\left( 3 \right)$
Now,
The SI unit of the mass is given as, $kg$.
The dimensional unit of the mass is given as, $M$.
The SI unit of the acceleration is given as, $m{s^{ - 2}}$.
The dimensional unit of the acceleration is given as, $L{T^{ - 2}}$.
The SI unit of the area is given as, ${m^2}$.
The dimensional unit of the area is given as, ${L^2}$.
By substituting the dimensional unit of the mass, acceleration and the area in the equation (3), then the equation (3) is written as,
$P = \dfrac{{\left[ M \right]\left[ {L{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}}$
By rearranging the terms in the above equation, then the above equation is written as,
$P = \left[ M \right]\left[ {L{T^{ - 2}}} \right]\left[ {{L^{ - 2}}} \right]$
By multiplying the terms in the above equation, then the above equation is written as,
$P = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$
Thus, the above equation shows the dimensional formula of the pressure.
Note The pressure is directly proportional to the mass and the acceleration and the pressure is inversely proportional to the area. As the mass and acceleration increases, then the pressure also increases. As the area increases, then the pressure is decreasing.
Useful formula
The formula for the pressure is given as,
$P = \dfrac{F}{A}$
Where, $P$ is the pressure in the object, $F$ is the force in the object and $A$ is the are of the object.
The formula of the force is given as,
$F = ma$
Where, $F$ is the force of the object, $m$ is the mass of the object and $a$ is the acceleration of the object.
Complete step by step solution
Now,
The formula for the pressure is given as,
$P = \dfrac{F}{A}\,...................\left( 1 \right)$
Now,
The formula of the force is given as,
$F = ma\,..................\left( 2 \right)$
By substituting the equation (2) in the equation (1), then the equation (1) is written as,
$P = \dfrac{{ma}}{A}\,...................\left( 3 \right)$
Now,
The SI unit of the mass is given as, $kg$.
The dimensional unit of the mass is given as, $M$.
The SI unit of the acceleration is given as, $m{s^{ - 2}}$.
The dimensional unit of the acceleration is given as, $L{T^{ - 2}}$.
The SI unit of the area is given as, ${m^2}$.
The dimensional unit of the area is given as, ${L^2}$.
By substituting the dimensional unit of the mass, acceleration and the area in the equation (3), then the equation (3) is written as,
$P = \dfrac{{\left[ M \right]\left[ {L{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}}$
By rearranging the terms in the above equation, then the above equation is written as,
$P = \left[ M \right]\left[ {L{T^{ - 2}}} \right]\left[ {{L^{ - 2}}} \right]$
By multiplying the terms in the above equation, then the above equation is written as,
$P = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$
Thus, the above equation shows the dimensional formula of the pressure.
Note The pressure is directly proportional to the mass and the acceleration and the pressure is inversely proportional to the area. As the mass and acceleration increases, then the pressure also increases. As the area increases, then the pressure is decreasing.
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