Question

Water flows through a horizontal tube as shown. If the difference of heights of water column in the vertical tubes is \(h=2\ \text{cm}\), and the areas of cross section at A and B are \(4\ \text{cm}^{2}\) and \(2\ \text{cm}^{2}\) respectively, the difference in pressure that is \((P_{A}-P_{B})\) is {\(v_{A}=\) velocity of water at point A, \(v_{B}=\) velocity of water at point B, density of water \(\rho =1\ \text{g/cm}^{3}\)}

• A. \(\frac{3}{2}v_{A}^{2}\)
• B. \(\frac{1}{2}v_{A}^{2}\)
• C. \(v_{A}^{2}\)
• D. \(2v_{A}^{2}\)

Hint:

Use the continuity equation to relate the two speeds, then apply Bernoulli's equation along the horizontal tube.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Continuity gives \(A_{A}v_{A}=A_{B}v_{B}\), and Bernoulli's equation for a horizontal tube gives \(P_{A}-P_{B}=\frac{1}{2}\rho\left(v_{B}^{2}-v_{A}^{2}\right)\).

Step 2: Calculation
\(4v_{A}=2v_{B}\Rightarrow v_{B}=2v_{A}\). Then \(P_{A}-P_{B}=\frac{1}{2}(1)\left(4v_{A}^{2}-v_{A}^{2}\right)=\frac{3}{2}v_{A}^{2}\) in CGS units, with \(\rho=1\ \text{g/cm}^{3}\).

Step 3: Compare the options
The result \(\frac{3}{2}v_{A}^{2}\) matches the first option.

Final Answer:
The pressure difference is \(\frac{3}{2}v_{A}^{2}\). \[ \boxed{\text{(A) }\dfrac{3}{2}v_{A}^{2}} \]