How to show an equation is dimensionally consistent t=2π√(l/g)

To see whether an equation is dimensionally consistent, first look for all the
places in the equation where there's anything that has units, and just write
the units in place of the quantity. 
If there's something that doesn't have units, you can just ignore it.

In this equation . . .

--  The  2π  is just a number.  It has no units, so we'll just leave it out.

--  The ' l ' is 'length', so write 'meter' in its place.

--  The ' g ' is 'gravity'.  That's an acceleration, so write  'm/sec² ' in its place.

Now the equation says:        T  =  √ meter/(m/s²) .

Look at the fraction inside the square root:

                                (meter) / (meter/sec²)

Cancel the 'meter'
in the numerator
and denominator:          (1)  /  (1/sec²)

                               =       sec² .

So the fraction is  sec² ,  and the square root is √sec²  =  seconds.

Now the equation says       T  =  seconds .

The ' T ' is the period of a pendulum with length of ' l '.
' T ' is a time, and the equation says it's equal to 'seconds'.
The equation is dimensionally consistent.            yay !

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