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Question 1
ప్రశ్న 1
For a fluid of mass density ρ under gravitational acceleration g, which set of relationships is simultaneously correct for specific weight γ, specific volume v, and specific gravity SG relative to a reference fluid of density ρref?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Specific weight is the weight per unit volume, so γ = ρg. Specific volume is the volume per unit mass and is therefore the reciprocal of mass density, v = 1/ρ. Specific gravity is a dimensionless ratio of the density of the fluid to the density of the stated reference fluid, so SG = ρ/ρref. These three relationships involve different physical quantities even though they are linked through density. Option B is therefore the only set in which all three relations are correct.
వివరణ:
Question 2
ప్రశ్న 2
An oil has specific gravity 0.82 relative to water of density 1000 kg/m³. Taking g = 9.81 m/s², which set gives its density, specific weight, and specific volume most nearly correctly?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The density is ρ = SG × ρwater = 0.82 × 1000 = 820 kg/m³. The specific weight is γ = ρg = 820 × 9.81 = 8044 N/m³, which is about 8.04 kN/m³. The specific volume is v = 1/ρ = 1/820 = 1.22 × 10^-3 m³/kg approximately. Each quantity retains its own SI unit, so density, specific weight, and specific volume must not be interchanged. Option A contains all three correctly calculated values.
వివరణ:
Question 3
ప్రశ్న 3
Match the fluid property with its dimensional formula: (I) mass density ρ, (II) specific weight γ, (III) dynamic viscosity μ, (IV) surface tension σ. Which sequence is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Mass density is mass per unit volume, giving dimensions ML^-3. Specific weight is force per unit volume, so its dimensions are MLT^-2 divided by L³, or ML^-2T^-2. Dynamic viscosity follows from τ = μ(du/dy), which gives μ the dimensions ML^-1T^-1. Surface tension is force per unit length, so its dimensions are MT^-2. Hence every association in Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
A liquid of fixed mass has its density increased from 800 kg/m³ to 840 kg/m³. What is the percentage decrease in its specific volume?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Specific volume is the reciprocal of density, v = 1/ρ. Initially, v1 = 1/800 = 0.00125 m³/kg. Finally, v2 = 1/840 = 0.0011905 m³/kg approximately. The percentage decrease is [(v1 - v2)/v1] × 100, which is about 4.76%. The density increase of 5% does not produce exactly a 5% decrease in its reciprocal, so Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
A liquid has dynamic viscosity μ = 0.085 Pa·s and density ρ = 850 kg/m³. Its kinematic viscosity is:
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Kinematic viscosity is defined by ν = μ/ρ. Substituting the given data gives ν = 0.085/850 m²/s. This evaluates to 1.0 × 10^-4 m²/s. The density appears in the denominator because kinematic viscosity represents dynamic viscosity normalized by mass density. Therefore Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
A Newtonian liquid fills a 6 mm gap between two large parallel plates. The lower plate is stationary and the upper plate moves at 1.8 m/s, producing a linear velocity profile. If μ = 0.45 Pa·s, what is the magnitude of shear stress in the liquid?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a Newtonian fluid, the shear-stress magnitude is τ = μ(du/dy). A linear profile across a 6 mm gap gives du/dy = 1.8/0.006 = 300 s^-1. Therefore τ = 0.45 × 300 = 135 Pa. The stated linear profile makes the velocity gradient uniform, so the shear-stress magnitude is also uniform across the gap. Option C is the correct result.
వివరణ:
Question 7
ప్రశ్న 7
Which statement best characterizes a Newtonian fluid over the range of conditions being considered?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A Newtonian fluid obeys a linear constitutive relation between shear stress and rate of shear strain. In simple shear this is written as τ = μ(du/dy), subject to the chosen sign convention for stress direction. At a specified thermodynamic state, μ is the proportionality coefficient and does not depend on the shear rate for an ideal Newtonian model. Non-Newtonian fluids depart from this linear-through-the-origin relationship or may require additional constitutive parameters. Option B therefore gives the defining behavior.
వివరణ:
Question 8
ప్రశ్న 8
For a power-law fluid described by τ = K(du/dy)^n in simple shear, which classification is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
In the power-law model, the exponent n controls how apparent viscosity changes with shear rate. When n is less than one, the apparent viscosity decreases as shear rate increases, which is shear-thinning behavior. When n equals one, the equation becomes linear in shear rate and reduces to a Newtonian-type relation with K equal to the dynamic viscosity. When n exceeds one, apparent viscosity increases with shear rate, giving shear-thickening behavior. Option A correctly states all three cases.
వివరణ:
Question 9
ప్రశ్న 9
For many common fluids over ordinary engineering temperature ranges, which trend is generally correct when temperature increases?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The dynamic viscosity of most liquids decreases as temperature rises because intermolecular cohesive effects become less influential. For gases, viscosity generally increases with temperature because molecular momentum transfer becomes more effective. Newtonian behavior describes the relation between shear stress and shear rate; it does not imply temperature-independent viscosity. This opposite temperature trend for liquids and gases is a standard fluid-mechanics distinction. Option C is therefore correct.
వివరణ:
Question 10
ప్రశ్న 10
Which conversion pair is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The poise is a non-SI unit of dynamic viscosity. One poise equals 0.1 Pa·s in SI units. The stoke is a unit of kinematic viscosity, with 1 St = 10^-4 m²/s. A centistoke is one hundredth of a stoke, so 1 cSt = 10^-6 m²/s. Option B correctly converts both quantities and also keeps dynamic and kinematic viscosity units distinct.
వివరణ:
Question 11
ప్రశ్న 11
If pressure on a fluid element increases by dp while its volume decreases by dV, which expression correctly represents a positive bulk modulus K?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Bulk modulus measures resistance to volumetric compression. An increase in pressure normally causes a decrease in volume, so dp and dV have opposite signs during compression. The definition K = -V(dp/dV) includes a minus sign so that K is positive for a stable compressible fluid. The reciprocal quantity, with appropriate differential form, is the compressibility rather than the bulk modulus. Therefore Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
A liquid occupies 0.250 m³ and has a bulk modulus of 1.6 GPa. If its pressure is increased by 8 MPa and K is treated as constant, what decrease in volume is predicted by the small-change relation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a small pressure change with constant bulk modulus, K = -Δp/(ΔV/V). The magnitude of the fractional volume decrease is Δp/K = 8 × 10^6/(1.6 × 10^9) = 0.005. Multiplying by the initial volume gives 0.005 × 0.250 = 0.00125 m³. The negative sign in ΔV would indicate a reduction, while the question asks for the magnitude of the decrease. Hence Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
A liquid initially has density 1000 kg/m³ and bulk modulus 2.0 GPa. For a small isothermal pressure increase of 10 MPa, assuming the bulk modulus remains constant, what is the approximate percentage increase in density?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a fixed mass, a small decrease in volume produces approximately an equal fractional increase in density. The bulk-modulus relation gives Δρ/ρ ≈ Δp/K for a small change. Here Δp/K = 10 × 10^6/(2.0 × 10^9) = 0.005. Expressed as a percentage, this is 0.5%. Therefore Option B is correct under the stated small-change and constant-bulk-modulus assumptions.
వివరణ:
Question 14
ప్రశ్న 14
Which statement about vapour pressure and cavitation is technically correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Vapour pressure is the saturation pressure associated with a liquid at a specified temperature. Cavitation can begin when the local absolute pressure falls to the vapour pressure or below it, allowing vapour bubbles or cavities to form. Gauge pressure alone is not the correct comparison because vapour pressure is an absolute thermodynamic pressure. Vapour pressure generally varies strongly with temperature and rises as temperature increases for ordinary liquids. Option B correctly states the relevant physical condition.
వివరణ:
Question 15
ప్రశ్న 15
A pump handles the same pure liquid at two different temperatures, with all other local absolute-pressure conditions unchanged. If the higher temperature causes a higher vapour pressure, what is the direct implication for the margin against cavitation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a given local absolute pressure, cavitation margin depends on how far that pressure remains above the liquid vapour pressure. Raising the liquid temperature generally raises its vapour pressure. The difference between the unchanged local absolute pressure and the vapour pressure therefore becomes smaller. This makes vapour formation easier if other hydraulic conditions are unchanged. Option B is consequently correct.
వివరణ:
Question 16
ప్రశ్న 16
Which pair correctly gives the SI unit and dimensional formula of surface tension σ?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Surface tension is defined as force per unit length acting tangentially at an interface. Its SI unit is therefore N/m. Since force has dimensions MLT^-2, division by length leaves MT^-2. Surface tension can also be interpreted as surface energy per unit area under appropriate reversible conditions, which has the equivalent SI unit J/m² = N/m. Option A gives the correct unit and dimensions.
వివరణ:
Question 17
ప్రశ్న 17
For the same surface tension σ and the same diameter d, which ordering of pressure excess Δp inside the interface is correct for an ideal cylindrical liquid jet, a spherical liquid droplet, and a spherical soap bubble?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A cylindrical liquid jet has one principal curved direction, giving a pressure excess of 2σ/d. A spherical liquid droplet has curvature in two principal directions and one interface, giving Δp = 4σ/d. A soap bubble has two liquid-gas interfaces, so its pressure excess is twice that of a single spherical interface, giving Δp = 8σ/d. These expressions assume ideal curvature and uniform surface tension. Option B gives the correct ordering and formulas.
వివరణ:
Question 18
ప్రశ్న 18
A spherical water droplet of diameter 2.0 mm has surface tension 0.072 N/m. Neglecting other effects, what is the pressure excess inside the droplet?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a spherical liquid droplet with one interface, the pressure excess is Δp = 4σ/d. Here σ = 0.072 N/m and d = 2.0 mm = 0.002 m. Substitution gives Δp = 4 × 0.072/0.002 = 144 Pa. Using 8σ/d would incorrectly apply the two-interface soap-bubble expression. Therefore Option C is correct.
వివరణ:
Question 19
ప్రశ్న 19
For a circular capillary tube of diameter d containing a liquid with contact angle θ measured through the liquid, which expression and sign interpretation are correct for capillary displacement h relative to the outside free surface?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Vertical force balance in a circular capillary gives h = 4σ cosθ/(ρgd) when the sign of h is referenced upward from the outside free surface. If θ is less than 90 degrees, cosθ is positive and the liquid rises. If θ is greater than 90 degrees, cosθ is negative and the liquid level is depressed. The magnitude increases with surface tension and decreases with density, gravity, and tube diameter. Option A correctly states both the formula and its sign interpretation.
వివరణ:
Question 20
ప్రశ్న 20
Water in a clean glass capillary may be idealized with contact angle θ = 0°. If σ = 0.073 N/m, ρ = 1000 kg/m³, d = 1.0 mm, and g = 9.81 m/s², the capillary rise is nearest to:
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a circular capillary, h = 4σ cosθ/(ρgd). With θ = 0 degrees, cosθ = 1. Substituting the given values gives h = 4 × 0.073/(1000 × 9.81 × 0.001) = 0.0298 m approximately. This is 29.8 mm after converting metres to millimetres. Therefore Option C is the nearest value.
వివరణ:
Question 21
ప్రశ్న 21
Mercury in a glass capillary has σ = 0.48 N/m, ρ = 13600 kg/m³, diameter d = 2.0 mm, and contact angle θ = 130° measured through the mercury, for which cos130° = -0.643. Taking upward h as positive and g = 9.81 m/s², what is h approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The capillary displacement is h = 4σ cosθ/(ρgd). Substituting the data gives h = 4 × 0.48 × (-0.643)/(13600 × 9.81 × 0.002). The result is approximately -0.00463 m, or -4.63 mm. The negative sign represents capillary depression because the contact angle exceeds 90 degrees and cosθ is negative. Therefore Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
Two clean circular capillary tubes of diameters 0.8 mm and 1.6 mm are placed in the same wetting liquid under identical conditions and have the same contact angle. What is the ratio of capillary rise in the 0.8 mm tube to that in the 1.6 mm tube?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the same liquid and contact angle, capillary rise is h = 4σ cosθ/(ρgd). All factors other than diameter are identical in the two tubes. Consequently, h is inversely proportional to d. The rise ratio is h0.8/h1.6 = 1.6/0.8 = 2. Therefore Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
A soap film has surface tension 0.040 N/m and two liquid-gas surfaces. If the area of one face is increased reversibly by 0.015 m² at constant surface tension, what work is required solely to create the additional surface area?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Surface tension is equivalent to reversible surface energy per unit area for a single interface. A soap film has two interfaces, so increasing the area of one face by ΔA creates a total new interfacial area of 2ΔA. The required work is therefore W = 2σΔA = 2 × 0.040 × 0.015 J. This gives W = 0.0012 J. Option C is correct and the factor of two is essential for a soap film.
వివరణ:
Question 24
ప్రశ్న 24
Which statement correctly relates compressibility β to bulk modulus K for a fluid?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Compressibility measures fractional volume change per unit pressure change, while bulk modulus measures resistance to compression. Their differential definitions make them reciprocals under the same thermodynamic condition, so β = 1/K. Compressibility therefore has units of Pa^-1, whereas bulk modulus has units of Pa. A fluid with a large bulk modulus undergoes only a small fractional volume change for a given pressure change. Option B is thus correct.
వివరణ:
Question 25
ప్రశ్న 25
An oil has specific gravity 0.90, kinematic viscosity 2.5 × 10^-4 m²/s, and is subjected to a velocity gradient of 120 s^-1. Taking water density as 1000 kg/m³ and assuming Newtonian behavior, what shear-stress magnitude is developed?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The oil density is ρ = 0.90 × 1000 = 900 kg/m³. Dynamic viscosity follows from μ = ρν, giving μ = 900 × 2.5 × 10^-4 = 0.225 Pa·s. For Newtonian simple shear, the stress magnitude is τ = μ(du/dy). Therefore τ = 0.225 × 120 = 27 Pa. Option B is correct and this calculation links specific gravity, kinematic viscosity, dynamic viscosity, and Newton's law of viscosity.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. γ = ρg, v = 1/ρ, SG = ρ/ρref సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: A. 820 kg/m³, 8.04 kN/m³, 1.22 × 10^-3 m³/kg సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. I: ML^-3, II: ML^-2T^-2, III: ML^-1T^-1, IV: MT^-2 సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: B. 4.76% సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: B. 1.0 × 10^-4 m²/s సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. 135 Pa సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: B. Its shear stress is proportional to rate of shear strain, with viscosity as the proportionality coefficient సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: A. n < 1 indicates shear-thinning behavior, n = 1 gives the Newtonian special case, and n > 1 indicates shear-thickening behavior సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: C. Dynamic viscosity decreases for liquids but increases for gases సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: B. 1 poise = 0.1 Pa·s and 1 centistoke = 10^-6 m²/s సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: B. K = -V(dp/dV) సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: B. 1.25 × 10^-3 m³ సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: B. 0.5% సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: B. If local absolute pressure in a liquid falls to or below the liquid's vapour pressure at that temperature, vapour cavities can form సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: B. The margin against cavitation decreases because the local pressure is closer to the higher vapour pressure సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: A. N/m and MT^-2 సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: B. Jet: 2σ/d, droplet: 4σ/d, soap bubble: 8σ/d సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: C. 144 Pa సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: A. h = 4σ cosθ/(ρgd); h is positive for θ < 90° and negative for θ > 90° సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. 29.8 mm సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. -4.63 mm సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: C. 2.0 సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: C. 0.0012 J సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: B. β = 1/K, so a large K means a relatively incompressible fluid సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: B. 27 Pa సమాధానం: B.