Fluid Statics Questions and Answers

Answer: A

A tyre gauge reads gauge pressure.

A tyre pressure gauge, like most pressure gauges, measures the pressure above the surrounding atmosphere. A tyre showing 30 psi on the gauge actually holds about 30 + 14.7 = 44.7 psi absolute.

The gauge reads zero, not 14.7 psi, when disconnected, which shows it is not reading absolute or atmospheric pressure. A vacuum reading would mean the tyre is below atmospheric pressure, which an inflated tyre is not.

Answer: D

The force at the plunger is \(1 \times 10^{4}\ \text{N}\).

By Pascal's law the pressure is the same under the plunger and the ram, so the forces are in the ratio of the areas, that is of the squares of the diameters:

\[ \begin{aligned} F &= W \left(\frac{d}{D}\right)^{2} = 16 \times 10^{4} \times \left(\frac{5}{20}\right)^{2} \\ &= \frac{16 \times 10^{4}}{16} = 1 \times 10^{4}\ \text{N} \end{aligned} \]

\(4 \times 10^{4}\) N uses the ratio of diameters instead of areas. \(64 \times 10^{4}\) and \(256 \times 10^{4}\) N multiply instead of divide, which would make the press a force reducer.

Answer: A

The device that does not measure pressure is the Rotameter.

A rotameter is a variable-area flow meter: a float rises in a tapered vertical tube until the flow balances its weight, and its height shows the flow rate.

A piezometer measures gauge pressure by the height of liquid in a vertical tube, a manometer measures pressure with a column of liquid, and a differential manometer measures the difference of pressure between two points.

Answer: B

Pascal's law holds for a fluid at rest.

In a fluid at rest there are no shear stresses, so the pressure at a point acts equally in all directions. That is the content of Pascal's law, and it underlies hydraulic presses and jacks.

In moving fluids, whether laminar, turbulent or transient, viscous shear stresses act and the normal stresses at a point are in general not equal in all directions, so the law is stated for fluids at rest.

Answer: A

The pressure difference is \(12800\ N/m^{2}\).

For a differential manometer with mercury below a lighter liquid, the pressure difference is the head of mercury less the equal head of the pipe liquid it displaces.

\[ \begin{aligned} \Delta p &= g h (\rho_{Hg} - \rho_{oil}) \\ &= 10 \times 0.1 \times (13600 - 800) \\ &= 12800\ N/m^{2} \end{aligned} \]

13600 forgets to subtract the oil column, and 1280 and 1360 are the same figures with a factor of ten slip in the 10 cm head.