Dimensional formula for pressure head is

[M^{0} L^{0} T^{0}]

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[M L^{- 1} T^{- 2}]

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[M^{0} L^{1} T^{- 2}]

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[M^{0} L^{1} T^{0}]

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Solution

The correct option is D $[M^{0} L^{1} T^{0}]$
$[P r e s u r e h e a d] = \frac{[P r e s s u r e]}{[D e n s i t y] \times [g]}$

$P r e s s u r e = \frac{F o r c e}{A r e a}$

But, $F o r c e = M a s s \times A c c e l a r a t i o n = [M] [L T^{- 2}] = [M L T^{- 2}]$

Then,

$[P r e s s u r e] = \frac{[M L T^{- 2}]}{[L^{2}]} = [M L^{- 1} T^{- 2}]$

$[D e n s i t y] = \frac{[M a s s]}{[V o l u m e]} = \frac{[M]}{[L^{3}]} = M L^{- 3}$

Now,

$[P r e s u r e h e a d] = \frac{[P r e s s u r e]}{[D e n s i t y] \times [g]} = \frac{[M L^{- 1} T^{- 2}]}{[M L^{- 3}] [L T^{- 2}]} = [M^{0} L^{1} T^{0}]$

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