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

Given equation:

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

• \( P \) is pressure,
• \( \rho \) is density,
• \( V \) is speed,
• \( A \) and \( B \) are quantities for which we need to find the dimensions of \( \frac{A}{B} \).

 

Step-by-Step Solution:

1. Dimensions of Pressure \( P \):

Pressure is force per unit area. The dimensions of force are \( MLT^{-2} \), and area has dimensions \( L^2 \). Therefore, the dimensions of pressure are:

\[ [P] = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}. \]

2. Dimensions of \( \frac{A^2}{B} \):

Since the sum \( \left( P + \frac{A^2}{B} \right) \) is equal to a constant, the dimensions of both terms must be the same. Therefore, the dimensions of \( \frac{A^2}{B} \) must be \( ML^{-1}T^{-2} \).

Let’s assume the dimensions of \( A \) and \( B \) are \( [A] = M^xL^yT^z \) and \( [B] = M^pL^qT^r \), respectively. Then, the dimensions of \( \frac{A^2}{B} \) are:

\[ \left[\frac{A^2}{B}\right] = \frac{(M^x L^y T^z)^2}{M^p L^q T^r} = M^{2x-p} L^{2y-q} T^{2z-r}. \]

For the dimensions to be consistent with \( ML^{-1}T^{-2} \), we must have:

• \( 2x - p = 1 \) (for mass),
• \( 2y - q = -1 \) (for length),
• \( 2z - r = -2 \) (for time).

3. Dimensions of \( A/B \):

From the above, we can solve for the dimensions of \( A \) and \( B \). After solving, we find that:

\[ \left[ \frac{A}{B} \right] = ML^{-1}T^{-4}. \]

Final Answer: The dimensions of \( \frac{A}{B} \) are \( \boxed{ML^{-1}T^{-4}} \).