Question

check the correctness of the relation. P=1/2pv²+pgh​

Answer 1

Given :

check the correctness of the relation:

$\rm{p=\dfrac{1}{2}pv^2+pgh}$

Dimensions :

$\\ (1) \bf \: pressure \: = \frac{force}{area}$

$\\ \to\rm{pressure= ML^{-1}T{^-2}}$

$\\ (2) \bf\: velocity = \frac{meter}{second}$

$\\ \to\rm{Velocity=LT^{-1}}$

$\\ (3) \bf \: g \: = \frac{m}{ {s}^{2} }$

$\\ \to\rm{acceleration(g)=LT^{-2}}$

$(4) \bf \: height \: = \: (in \: meter)$

$\\ \to\rm{height=L}$

Solution :

LHS:

$\\ \implies \large \sf \: pressure \: = \frac{force}{area} = \frac{n}{ {m}^{2} }$

$\sf{\bullet \: Force =Newton \:and \: area =m²}$

➡ $\\ \sf{Pressure=ML^{-1}T^{-2}}$

RHS:

$\\ \implies\large\sf{ \dfrac{1}{2}pv^{2}+phg}$

➡ here in dimensions 1/2 is neglicted because it is constant.

$\\ \implies\large\sf{pv^2 +pgh}$

$\\ \implies\large\sf{\dfrac{kg}{m^{2}} \times \dfrac{m}{s^{2}} \times m}$

$\\ \implies\large\sf{kgm^{-1} s^{-2}}$

$\\ \implies\large\sf{ML^{-1} T^{-2}}$

As we see that ,LHS=RHS

Therefore both LHS and RHS has a dimensions of pressure hence both are equal to each other

so we can say,

$\therefore \boxed{ \large \sf{p = \frac{1}{2} p {v}^{2} + pgh}}$