Fluid MechanicsHard

Question

Consider a modified Bernoulli equation.

$$\left( P + \frac{A}{Bt^{2}} \right) + \rho g(h + Bt) + \frac{1}{2}\rho V^{2} = \text{~constant~}$$

If $t$ has the dimension of time then the dimensions of A and B are $\_\_\_\_$ , $\_\_\_\_$ respectively.

Options

A.$\left\lbrack {ML}^{0}{\text{ }T}^{- 1} \right\rbrack$ and $\left\lbrack M^{0}LT \right\rbrack$
B.$\left\lbrack {ML}^{0}{\text{ }T}^{- 1} \right\rbrack$ and $\left\lbrack M^{0}{LT}^{- 1} \right\rbrack$
C.$\left\lbrack {ML}^{0}{\text{ }T}^{- 2} \right\rbrack$ and $\left\lbrack M^{0}{LT}^{- 2} \right\rbrack$
D.$\left\lbrack {ML}^{0}{\text{ }T}^{- 2} \right\rbrack$ and $\left\lbrack M^{0}{LT}^{- 1} \right\rbrack$

Solution

⇒ $\lbrack P\rbrack = \left\lbrack \frac{A}{Bt^{2}} \right\rbrack$

$$\begin{array}{r} \Rightarrow \lbrack h\rbrack = \lbrack Bt\rbrack\#(2) \end{array}$$

$$\Rightarrow \lbrack B\rbrack = \left\lbrack \frac{h}{t} \right\rbrack = \left\lbrack \frac{L}{T} \right\rbrack = \left\lbrack {LT}^{- 1} \right\rbrack $$Putting B is equation (1)

$${\left\lbrack {ML}^{- 1}{\text{ }T}^{- 2} \right\rbrack = \left\lbrack \frac{A}{{LT}^{- 1} \times T^{2}} \right\rbrack }{\lbrack A\rbrack = \left\lbrack {ML}^{0}{\text{ }T}^{- 1} \right\rbrack}$$

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