Your Answer : (Not Answered)

Correct Answer : B

The correct relation is \(C_d = C_c \times C_v\).

Actual discharge is the actual area of the jet at the vena contracta times the actual velocity there. The area is \(C_c\) times the orifice area and the velocity \(C_v\) times the theoretical velocity, so

\[ \begin{aligned} Q_{act} &= (C_c a)(C_v \sqrt{2gH}) \\ C_d &= \frac{Q_{act}}{a\sqrt{2gH}} = C_c C_v \end{aligned} \]

The coefficients multiply, so a sum as in the first option is wrong, and \(C_c = C_v/C_d\) inverts the relation (it should be \(C_c = C_d/C_v\)). The relation offered for \(C_r\) is not a standard result.

Your Answer : (Not Answered)

Correct Answer : D

Flow is treated as incompressible when Ma < 0.3.

Below a Mach number of about 0.3, the density of a gas changes by only a few per cent as it flows, so it can be treated as incompressible with little error. Above that, compressibility effects become significant and must be included.

Ma = 0.3 is the boundary rather than the region, and Ma > 0.3 or Ma > 0.4 describe flows where compressibility can no longer be ignored.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : A

\(\overline{MG}\) is the metacentric height and \(G\) should lie below \(M\).

When a floating body tilts slightly, the centre of buoyancy shifts and the buoyant force acts along a line that crosses the body's centre line at the metacentre M. The metacentric height is the distance GM from the centre of gravity to the metacentre.

If M is above G, the weight and buoyancy form a couple that rotates the body back upright: stable equilibrium. If M were below G, the couple would increase the tilt.

MB is the metacentric radius (\(I/V\)), not the metacentric height. Whether B lies above or below M does not decide stability; for most floating bodies B is below G and still stable, because what matters is the position of G relative to M.

Your Answer : (Not Answered)

Correct Answer : D

The match is P-2, Q-3, R-1.

In the FLT system, mass has dimensions \(FL^{-1}T^{2}\) (from \(F = ma\)).

• Dynamic viscosity is shear stress divided by velocity gradient: \(FL^{-2} / T^{-1} = FL^{-2}T\), item 2.
• Surface tension is force per unit length: \(FL^{-1}\), item 3.
• Density is mass per unit volume: \(FL^{-1}T^{2}/L^{3} = FL^{-4}T^{2}\), item 1.

The other options misplace at least one of these; for instance P-1 gives viscosity the dimensions of density.

Your Answer : (Not Answered)

Correct Answer : B

The correct plot is (B), apparent viscosity falling as the shear rate rises.

A pseudoplastic fluid is shear-thinning: in the power-law model \(\tau = K(du/dy)^{n}\) its index is \(n \lt 1\), so the apparent viscosity \(\eta = K(du/dy)^{n-1}\) decreases with increasing shear rate. Polymer solutions, paints and blood behave this way.

Plot (A), a constant viscosity, describes a Newtonian fluid. Plots (C) and (D), viscosity rising with shear rate, describe a dilatant (shear-thickening) fluid with \(n \gt 1\), such as a dense starch suspension.

Your Answer : (Not Answered)

Correct Answer : C

The pressure gradient is \(\mu A(-\hat{j})\).

Incompressibility fixes \(B\):

\[ \begin{aligned} \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} &= Ay + 2By = 0 \\ \Rightarrow B &= -\frac{A}{2} \end{aligned} \]

The steady Navier-Stokes equation gives \(\nabla p = -\rho(\vec{V}\cdot\nabla)\vec{V} + \mu\nabla^{2}\vec{V}\). At (5, 0) both velocity components vanish (\(u = 5A \cdot 0 = 0\), \(v = 0\)), so the convective term is zero there. The viscous term:

\[ \begin{aligned} \nabla^{2}u &= \nabla^{2}(Axy) = 0 \\ \nabla^{2}v &= \nabla^{2}(By^{2}) = 2B = -A \end{aligned} \]

So \(\nabla p = \mu(0\,\hat{i} - A\,\hat{j}) = \mu A(-\hat{j})\).

Options with an \(\hat{i}\) component wrongly take \(\nabla^{2}(Axy)\) as non-zero, or keep \(B\) without applying continuity.

Your Answer : (Not Answered)

Correct Answer : C

The new thickness is \(2\delta_{1}\).

For a laminar flat-plate boundary layer (Blasius):

\[ \begin{aligned} \delta &\approx \frac{5x}{\sqrt{Re_{x}}} = 5\sqrt{\frac{\nu x}{U_{\infty}}} \\ &\propto \sqrt{\nu} \end{aligned} \]

At the same x and U, multiplying the kinematic viscosity by 4 multiplies the thickness by \(\sqrt{4} = 2\).

\(4\delta_{1}\) assumes thickness proportional to viscosity, \(\delta_{1}/2\) has the dependence inverted, and \(\delta_{1}\) ignores viscosity altogether. A more viscous fluid diffuses momentum further from the wall, so the layer thickens.

Your Answer : (Not Answered)

Correct Answer : A

The product remains always constant.

A doublet is the limit of a source and a sink of strength \(m\) a distance \(a\) apart as \(a \to 0\). If \(m\) stayed fixed, the two would simply cancel. To keep a finite flow pattern, the strength is increased as the distance shrinks so that the product \(m a\), the doublet strength, is held constant throughout the approach.

A decreasing product would make the pattern vanish, an increasing one would make it blow up, and 'first increases and then decreases' does not describe any limiting process used to define the doublet.

Your Answer : (Not Answered)

Correct Answer : C

The slope is \(-\frac{u}{v}\).

Along a potential (equipotential) line \(\phi\) is constant, so

\[ \begin{aligned} d\phi &= \frac{\partial\phi}{\partial x}dx + \frac{\partial\phi}{\partial y}dy \\ &= u\,dx + v\,dy = 0 \end{aligned} \]

giving \(\frac{dy}{dx} = -\frac{u}{v}\).

\(\frac{v}{u}\) is the slope of a streamline, which is tangent to the velocity. The two slopes multiply to \(-1\), the familiar result that streamlines and equipotentials cross at right angles. \(\frac{u}{v}\) and \(-\frac{v}{u}\) have the right pieces in the wrong arrangement.

Your Answer : (Not Answered)

Correct Answer : D

The relation is \(f = 4C_f\).

The Darcy factor appears in the head-loss equation \(h_f = f\frac{L}{D}\frac{V^{2}}{2g}\), while the Fanning factor is the wall shear stress scaled by dynamic pressure, \(C_f = \tau_w/(\tfrac{1}{2}\rho V^{2})\). A force balance on the pipe gives \(\tau_w = \frac{D}{4}\frac{\Delta p}{L}\), from which \(f = 4C_f\).

For laminar flow this reads \(f = 64/Re\) and \(C_f = 16/Re\). The other ratios (0.25, 0.5, 2) do not match these definitions.

Your Answer : (Not Answered)

Correct Answer : A

The ratio is 1.

With no pressure gradient and no body force, the x-momentum equation for fully developed flow reduces to

\[ \mu\frac{d^{2}u}{dy^{2}} = 0 \]

so the velocity is linear in \(y\) (Couette flow). With \(u(0) = -U_{2}\) and \(u(H) = U_{1}\), \(u = -U_{2} + (U_{1} + U_{2})\frac{y}{H}\), and the shear stress

\[ \tau = \mu\frac{du}{dy} = \frac{\mu(U_{1} + U_{2})}{H} \]

is the same at every \(y\), including both walls. Their ratio is therefore 1, whatever the plate speeds.

The options built from \(U_{1}\) and \(U_{2}\) would require a stress that varies across the gap, which happens only when a pressure gradient adds a parabolic part to the profile.

Your Answer : (Not Answered)

Correct Answer : A

The centre-line velocity is 2 m/s.

The pitot tube reads the stagnation pressure and the static tap the static pressure. In the manometer the difference in piezometric head shows up as the 20 mm deflection of the manometric fluid; the 300 mm vertical spacing between the two openings is accounted for by the fluid columns in the connecting lines and does not appear in the result. So

\[ \begin{aligned} \frac{1}{2}\rho V^{2} &= (\rho_{m} - \rho) g h \\ &= (11000 - 1000)(10)(0.02) = 2000\ \text{Pa} \\ V &= \sqrt{\frac{2 \times 2000}{1000}} = 2\ \text{m/s} \end{aligned} \]

Using the full manometric density 11000 instead of the difference gives about 2.1 m/s, and 3, 5 and 7 m/s do not follow from a 20 mm reading with these densities.

Your Answer : (Not Answered)

Correct Answer : B

The pressure difference is about 1.76 mm Hg.

A Pitot-static tube reads the dynamic pressure, the difference between stagnation and static pressure:

\[ \begin{aligned} \Delta p &= \tfrac{1}{2}\rho_{air}V^{2} = \tfrac{1}{2}(1.2)(20)^{2} = 240\ \text{Pa} \\ h &= \frac{\Delta p}{\rho_{Hg}g} = \frac{240}{13600 \times 10} \\ &= 1.76 \times 10^{-3}\ \text{m} = 1.76\ \text{mm} \end{aligned} \]

1760 is the same column expressed in micrometres, a units slip. 0.57 is the reciprocal of 1.76, and \(0.57 \times 10^{-5}\) compounds that with a wrong power of ten.

Your Answer : (Not Answered)

Correct Answer : D

The match is P - 1; Q - 3.

An orifice meter is just a thin plate with a hole clamped between flanges, so it is cheap to make and install. But the jet contracts sharply (the vena contracta) and then expands abruptly, so much of the pressure drop is never recovered: high head loss, low cost.

A venturi meter has a smooth converging section, a throat and a long gradual diffuser that recovers most of the pressure, so its permanent head loss is low; but it is long and precisely machined, so it is expensive. Options pairing the orifice with high cost or the venturi with low cost or high loss reverse these features.

Your Answer : (Not Answered)

Correct Answer : B

The velocity in the 0.1 m section is 1.0 m/s.

The constriction acts as a venturi meter. Continuity links the two velocities: the diameter halves, so the area falls to a quarter and \(V_{2} = 4V_{1}\). Bernoulli’s equation between the two sections then gives

\[ \begin{aligned} \Delta p &= \tfrac{1}{2}\rho(V_{2}^{2} - V_{1}^{2}) = \tfrac{1}{2}\rho(16 - 1)V_{1}^{2} \\ 7500 &= 500 \times 15 V_{1}^{2} \\ V_{1} &= 1.0\ \text{m/s} \end{aligned} \]

The throat velocity is 4.0 m/s. A value of 0.5 or 1.5 would give pressure drops of 1.9 kPa or 16.9 kPa, and 2.0 m/s would need 30 kPa.

Your Answer : (Not Answered)

Correct Answer : C

The prototype velocity is 10 m \(\text{s}^{-1}\).

Froude similarity requires equal \(V/\sqrt{gL}\) in model and prototype. With the same \(g\):

\[ \begin{aligned} V_{p} &= V_{m}\sqrt{L_{p}/L_{m}} \\ &= 1\times\sqrt{100} = 10\ \text{m/s} \end{aligned} \]

100 scales velocity with length directly, as Reynolds or geometric reasoning might suggest, and 0.1 inverts the ratio. 1 would hold only for a full-scale model.

Your Answer : (Not Answered)

Correct Answer : B

The ratio of inertial to viscous forces is the Reynolds number, \(Re = \rho VL/\mu\).

It decides whether a flow is laminar (low \(Re\), viscous forces dominate) or turbulent (high \(Re\), inertia dominates), as in pipe flow, where the change happens at about \(Re = 2300\).

The Froude number compares inertial forces with gravity forces and governs open-channel and free-surface flow. The power number relates the power drawn by an impeller or agitator to the fluid density, rotational speed and impeller diameter. The Biot number is a heat-transfer group comparing conduction resistance inside a body with convection resistance at its surface. None of these involves viscosity in the way the question describes.

Your Answer : (Not Answered)

Correct Answer : D

The statement that the upward buoyant force on a body immersed in a fluid equals the weight of the fluid displaced is Archimedes' principle. Newton's third law is about action and reaction, Bernoulli's principle is about pressure and speed in flow, and Pascal's principle is about transmission of pressure in a confined fluid, so only Archimedes' principle fits the statement.

The same principle gives the condition for floating: a body floats when the weight of fluid it displaces equals its own weight, so the buoyant force balances its weight.

Your Answer : (Not Answered)

Correct Answer : C

Viscosity is the property of a fluid that resists motion or deformation because of internal friction between its layers. Surface tension is the force at the surface of a liquid, elasticity is the tendency to return to the original shape, and density is mass per unit volume, so none of these measures the resistance to flow of the fluid.