The hydrostatic pressure 'P' of a liquid column depends upon the density'D', height'H' of liquid column and also an acceleration'g'due to gravity. Using dimensional analysis, derive formula for pressures
Answers
Steps:
1) We know that,
[tex]dim(D) = \frac{M}{L^{3}} \\ \\ dim( H) = L \\ \\ dim( g) = \frac{L}{T^{2}} [/tex]
2) Since , Hydrostatic Pressure is dependent on above given quantities .
Therefore,
=> [tex]dim(P) = dim(k D^{x}H^{y}g^{z}) \\ \\ =\ \textgreater \ \frac{M}{LT^{2}} = (\frac{M}{L^{3}} )^{x}*(L)^{y} *( \frac{L}{T^{2}} )^{z} \\ \\ =\ \textgreater \ \frac{M}{LT^{2}} = \frac{M^{x}}{L^{(3x-y-z)}*T^{2z}} [/tex]
=> Comparing Powers ,we get
x = 1 , 2z=2 => z= 1
=> 3x-y-z=1
=> 3*1-y-1=1
=> y= 1
3) Therefore, Formula of Hydrosatic Pressure is given by :
By Experiments ,it is observed that value of k = 1 ;
Answer:
Final Answer: P = DHg.
Steps:
1) We know that,
dim(P) = \frac{M}{LT^{2}}dim(P)=
LT
2
M
\begin{gathered}dim(D) = \frac{M}{L^{3}} \\ \\ dim( H) = L \\ \\ dim( g) = \frac{L}{T^{2}} \end{gathered}
dim(D)=
L
3
M
dim(H)=L
dim(g)=
T
2
L
2) Since , Hydrostatic Pressure is dependent on above given quantities .
Therefore,
P = k D^{x}H^{y}g^{z}P=kD
x
H
y
g
z
where k is constant.
=> \begin{gathered}dim(P) = dim(k D^{x}H^{y}g^{z}) \\ \\ =\ \textgreater \ \frac{M}{LT^{2}} = (\frac{M}{L^{3}} )^{x}*(L)^{y} *( \frac{L}{T^{2}} )^{z} \\ \\ =\ \textgreater \ \frac{M}{LT^{2}} = \frac{M^{x}}{L^{(3x-y-z)}*T^{2z}} \end{gathered}
dim(P)=dim(kD
x
H
y
g
z
)
= \textgreater
LT
2
M
=(
L
3
M
)
x
∗(L)
y
∗(
T
2
L
)
z
= \textgreater
LT
2
M
=
L
(3x−y−z)
∗T
2z
M
x
=> Comparing Powers ,we get
x = 1 , 2z=2 => z= 1
=> 3x-y-z=1
=> 3*1-y-1=1
=> y= 1
3) Therefore, Formula of Hydrosatic Pressure is given by :
\boxed{H=kDHg}
H=kDHg
By Experiments ,it is observed that value of k = 1 ;
\boxed{H=DHg}
H=DHg