Dimensional Analysis Help: Understanding Correct Equations in MKS System

[hthrcru]

I'm having trouble understanding what exactly a dimensionally correct equation is..here's my problem-

In the following expressions:
V-volume
A-area
P-pressure
p(lowercase)-density
t-time
m-mass
v-velocity
g-acceleration due to gravity

I'm supposed to state whether the following equations are dimensionally correct or incorrect, using the above variables in terms [L,M,t], the MKS system..

1. m/t=(3/2)pV/t

2. PV=(1/2)mv²+mg(V/A)

Can ANYONE explain to me if these are dimensionally correct or not and why?

 

[hthrcru]

did I post this wrong or does no one know the answer? I'm new to this website so I am not sure how long it normally takes for a response!

 

[vissh]

All the physical quantities of interest can be derived from the base i.e. fundamental quantities [mass,length and 5 more].By dimension of a quantity[Q let] in a base quantity, we mean "the exponent of a base quantity that enters into the expression of that qunatity[Q]".
Eg:- Force = mass * acceleration = mass *(velocity/time) = mass * [(length/time)/time] =mass * length * (time)^-2 => Dimensions of force are 1 in mass, 1 in length and -2 in time. It is denoted as [Force] = MLT^-2 {M for mass, L for length, T for time}
And MLT^-2 is called dimensional formula for Force.
->For any physical quantity, you can just go on breaking the formula to the base ones{like i did for Force}.
--->Now, for an equation to be dimensionally correct, Dimensional formula for LHS term must be same as Dimensional formula for RHS term.
>Also, if 2 terms on any side are adding or subracting, then the 2 terms must have same dimensional formula also.
>As nos [1,2,5,10.1 etc] are unitless, they don't contribute anything to dimensional formula of a term
Eg:- Let's take up one of the kinematic's equation :- S = ut - (1/2)at²
-- = L
-- [ut] = (LT^-1)(T) = L
-- [(1/2)at²] = (LT^-2)(T)² = L
And hence this equation is dimensionally correct.
Eg:- Let i write :- velocity, v = aS [a for acceleration and S for displacement]
[v] = LT^-1
[aS] = ((LT^-2)(L) = L²T^-2
clearly, both are not same and this equation i wrote is dimensionally incorrect.

Try out those You have asked, I think you will be able to do them now ^.^
Hope it help :)