Check dimensional consistency of given equation.

Explanation

For example
`x=x_(0)=v_(0)t+(1)/(2)at^(2)`
Here, x is distance covered by object in time t.
`x_(0)`= initial position of object during motion
x= final position
`v_(0)`= initial velocity
a= acceleration
`LHS=x=M^(0)L^(1)T^(0)`
In RHS tere are three terms,
`x_(0)=M^(0)L^(1)T^(0)`
`v_(0)t=[L^(1)T^(-1)T^(1)]=M^(0)L^(1)T^(0)`
In `(1)/(2)at^(2)` is constant term which is dimensionless.
`:.at^(2)=[M^(0)L^(1)T^(-2)][T^(2)]`
`=M^(0)L^(1)T^(0)`
Here all terms in given equation have same dimension. Hence, given equation,
`x=x_(0)+v_(0)t+(1)/(2)at^(2)` is dimensionally consistence. (Dimensionally valid).
Test of consistency of dimension tells us no more no less than test of consistency of units. If equation fails consistency test it is proved wrong, but if it passes it is not proved right Thus, a dimensionally correct equation need not be actually an exact (correct) equation but al dimensionally wrong (incorrect) equation must be wrong.