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.
Show Hint
- \( ML^{1}T^{-4} \)
- \( ML^{-1}T^{-4} \)
- \( ML^{2}T^{-4} \)
- \( ML^{-1}T^{-2} \)
The Correct Option is B
Solution and Explanation
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 \( \boxed{ML^{-1}T^{-4}} \).
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