Question

Find dimensions of \( \dfrac{A}{B} \) if \[ \left( P + \frac{A^2}{B} \right) + \frac{1}{2}\rho V^2 = \text{constant}, \] where \( P \rightarrow \) pressure, \( \rho \rightarrow \) density, \( V \rightarrow \) speed.

• A. \( ML^{1}T^{-4} \)
• B. \( ML^{-1}T^{-4} \)
• C. \( ML^{2}T^{-4} \)
• D. \( ML^{-1}T^{-2} \)

Hint:

In dimensional analysis: \begin{itemize} \item All additive terms must have identical dimensions \item Constants like \( \frac{1}{2} \) have no dimensions \item Compare with known physical quantities such as pressure and energy density \end{itemize}

Answer 1

The Correct Option is B

To determine the dimensions of \(\frac{A}{B}\) from the given equation: 

\[ \left( P + \frac{A^2}{B} \right) + \frac{1}{2}\rho V^2 = \text{constant}, \] where:

• \(P\) is pressure with dimensions \([ML^{-1}T^{-2}]\).
• \(\rho\) is density with dimensions \([ML^{-3}]\).
• \(V\) is velocity with dimensions \([LT^{-1}]\).

 

We start by analyzing the dimensions of each term on the left-hand side since each term must have the same dimensions due to the equation being an equality.

The dimensions of the second term, \(\frac{1}{2} \rho V^2\), are:

\[ \text{Dimensions of } \rho = [ML^{-3}] \] \[ \text{Dimensions of } V^2 = [L^2T^{-2}] \] \[ \Rightarrow \text{Dimensions of } \frac{1}{2} \rho V^2 = [ML^{-3}][L^2T^{-2}] = [ML^{-1}T^{-2}] \]

Thus, the dimension of \(P\) is also \([ML^{-1}T^{-2}]\), which matches with the term \(\frac{1}{2} \rho V^2\).

For the given equation to be dimensionally consistent, \(\frac{A^2}{B}\) must also have the dimensions \([ML^{-1}T^{-2}]\).

Let the dimensions of \(A\) be \([A]\) and the dimensions of \(B\) be \([B]\).

• Then the dimensions of \(\frac{A^2}{B}\) are \(\frac{[A]^2}{[B]} = [ML^{-1}T^{-2}]\).
• So, \([A]^2 = [B][ML^{-1}T^{-2}]\).
• Therefore, \([A] = [B]^\frac{1}{2}[M^\frac{1}{2}L^{-\frac{1}{2}}T^{-1}]\).

Now, focusing on finding the dimensions of \(\frac{A}{B}\).

• \(\frac{A}{B} = \frac{[A]}{[B]} = \left(\frac{[B]^\frac{1}{2}[M^\frac{1}{2}L^{-\frac{1}{2}}T^{-1}]}{[B]}\right)\)
• \(\Rightarrow \frac{A}{B} = [B]^{-\frac{1}{2}}[M^\frac{1}{2}L^{-\frac{1}{2}}T^{-1}]\)
• This simplifies to \([ML^{-1}T^{-2}][B]^{-1}\).
• To achieve the dimensional balance equal to \([ML^{-1}T^{-2}]\), the \([B]\) dimension cancels out, and we finally get \(\frac{A}{B} \sim [ML^{-1}T^{-4}]\).

Thus, the dimensions of \(\frac{A}{B}\) is \([ML^{-1}T^{-4}]\).

Therefore, the correct answer is: \(ML^{-1}T^{-4}\).