Fluid Properties Questions and Answers

Answer: D

The answer is None of these, because none of the options, as printed, is the dimension of a flow rate.

The rate of flow (discharge) is a volume per unit time: m³/s, with dimensions L³T⁻¹, that is L³/T.

The first option is printed as L³/T⁻¹. Dividing by T⁻¹ is multiplying by T, so it equals L³T, a volume multiplied by time, which is not a rate. The other two, L²/T⁻¹ and L⁴/T⁻¹, have the wrong power of length as well. So the correct dimension L³T⁻¹ is not among them.

Answer: D

The model answer is All of these.

1 bar is defined as 10⁵ Pa. Converting:

\[ \begin{aligned} \frac{10^{5}}{101325} &= 0.9869 \text{ atm} \\ \frac{10^{5}}{1000 \times 9.81} &= 10.19 \text{ m} \approx 1020 \text{ cm of water} \end{aligned} \]

The 76 cm mercury column is the standard atmosphere, 1.013 bar, about 1.3% more than 1 bar (1 bar is about 75 cm of mercury). The paper treats it as the same, since 1 bar and 1 atm are equal to within that margin, and so takes all three statements together. Strictly, the first two are exact conversions and the third is approximate.

Answer: C

The mass is 51 kg.

Weight is mass times gravitational acceleration, \(W = mg\), so the mass is the weight divided by g.

\[ \begin{aligned} m &= \frac{W}{g} = \frac{500\ \text{N}}{9.8\ \text{m/s}^2} \\ &\approx 51\ \text{kg} \end{aligned} \]

51 N has the right number but the unit of force, not mass. 4900 kg comes from multiplying by g instead of dividing, and \(1.96 \times 10^{-2}\) is the reciprocal, g divided by W.

Answer: D

The CGS unit of dynamic viscosity is the Poise.

Dynamic viscosity has units of force times time per area. In CGS this is dyne·s/cm², named the poise after Poiseuille; 1 poise = 0.1 N·s/m².

The stokes is the CGS unit of kinematic viscosity (cm²/s). The newton and the dyne are units of force, in SI and CGS respectively.

Answer: B

Doubling the diameter halves the capillary rise.

The capillary rise is given by Jurin's law

\[ h = \frac{4\sigma \cos\theta}{\rho g d} \]

so h is inversely proportional to the diameter d. With everything else unchanged, twice the diameter gives half the rise. That is why capillary rise is noticeable only in narrow tubes, and why piezometer tubes are made wide enough to avoid it. Doubling or tripling the rise would need a direct proportion, and the rise is clearly affected.

Answer: C

The unit of kinematic viscosity is the Stoke.

Kinematic viscosity is dynamic viscosity divided by density, so its dimensions are length squared per time. In SI it is m²/s; in the CGS system it is cm²/s, called the stoke, after George Stokes: 1 stoke = 10^-4 m²/s.

The poise is the CGS unit of dynamic viscosity (g/cm·s), the usual trap. The pascal is the unit of pressure or stress, and the faraday is a quantity of electric charge, so neither measures viscosity.

Answer: B

The compressibility is \(4 \times 10^{-11}\ \text{Pa}^{-1}\).

Compressibility is the reciprocal of the bulk modulus:

\[ \begin{aligned} \beta &= \frac{1}{K} = \frac{1}{25 \times 10^{9}} \\ &= 4 \times 10^{-11}\ \text{Pa}^{-1} \end{aligned} \]

\(4 \times 10^{-9}\) slips by treating 25 GPa as \(25 \times 10^{7}\) Pa. The options with 25 copy the bulk modulus figure instead of inverting it.

Answer: B

It is an ideal fluid.

An ideal fluid is an imaginary fluid that is incompressible and has no viscosity, so it offers no resistance to shear. It is a useful model in fluid mechanics, but no actual fluid behaves this way.

A real fluid always has some viscosity. An ideal plastic (Bingham) fluid needs a yield stress before it flows and then flows with a constant viscosity, and a non-Newtonian fluid has a viscosity that changes with the rate of shear. All three are viscous, so none fits the description.

Answer: B

Absolute viscosity is also known as dynamic viscosity.

Dynamic (absolute) viscosity \(\mu\) is the constant in Newton's law of viscosity, \(\tau = \mu \frac{du}{dy}\), and is measured in Pa s (N s/m²) or poise.

Kinematic viscosity is a different quantity, \(\nu = \mu/\rho\), the dynamic viscosity divided by density, measured in m²/s or stokes. 'Parallel viscosity' and 'linear viscosity' are not standard terms.

Answer: A

1 Pascal equals \(1\ N/m^{2}\).

Pressure is force per unit area, and the SI unit, the pascal, is defined as a force of one newton acting on one square metre.

10 N/m² would be 10 Pa. N s/m² is not a unit of pressure at all: it is the pascal-second, the SI unit of dynamic viscosity.