Topic: Fluid Mechanics
The tank shown in the figure is closed at top and contains air at a pressure of $P_{A}.$ The value of $P_{A}$ for the manometer readings shown will be (Oil RD=0.75, 150cm; Water 60cm; Mercury 10cm open to atmosphere)
- A. -3.573 kPa
- B. -4.573 kPa
- C. -6.573 kPa
- D. -7.573 kPa
Answer: A. -3.573 kPa
Explanation: Using jumping of fluid technique (hydrostatic equation): $P_{A} + (0.75\times1000\times9.81\times1.5) + (1000\times9.81\times0.60) - (13.6\times1000\times9.81\times0.10) = 0$. Solving this gives $P_{A}=-3.577~kPa$.