Question

Dimensional formula for pressure head is

• A. $\left[M^0{L}^0{T}^0\right]$
• B. $\left[M{L}^{-1}{T}^{-2}\right]$
• C. $\left[M^0{L}^1{T}^{-2}\right]$
• D. $\left[M^0{L}^1{T}^0\right]$

Answer 1

The correct option is D $\left[M^0{L}^1{T}^0\right]$

$\left[Presurehead\right]=\frac{\left[Pressure\right]}{\left[Density\right]\times \left[g\right]}$

$Pressure=\frac{Force}{Area}$

But, $Force=Mass\times Accelaration=\left[M\right]\left[L{T}^{-2}\right]=\left[ML{T}^{-2}\right]$

Then,

$\left[Pressure\right]=\frac{\left[ML{T}^{-2}\right]}{\left[L^2\right]}=\left[M{L}^{-1}{T}^{-2}\right]$

$\left[Density\right]=\frac{\left[Mass\right]}{\left[Volume\right]}=\frac{\left[M\right]}{\left[L^3\right]}=M{L}^{-3}$

Now,

$\left[Presurehead\right]=\frac{\left[Pressure\right]}{\left[Density\right]\times \left[g\right]}=\frac{\left[M{L}^{-1}{T}^{-2}\right]}{\left[M{L}^{-3}\right]\left[L{T}^{-2}\right]}=\left[M^0{L}^1{T}^0\right]$