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Question 1
ప్రశ్న 1
In the classical Reynolds dye experiment with flow through a transparent circular pipe, which observation most clearly distinguishes laminar flow from turbulent flow?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Reynolds introduced a thin dye filament into water flowing through a pipe to visualize the nature of the motion. At sufficiently low flow rates, the dye remained as a relatively distinct filament with little transverse mixing. As the velocity and Reynolds number increased, disturbances developed and the filament became irregular. In turbulent flow, the dye dispersed extensively because of fluctuating transverse motion and mixing. Therefore Option A correctly represents the essential experimental observation.
వివరణ:
Question 2
ప్రశ్న 2
What was the principal role of the injected dye in Reynolds' pipe-flow experiment?
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Explanation:
The dye was used primarily as a flow-visualization tracer. Its behaviour made the difference between orderly laminar motion and strongly mixed turbulent motion visible. The dye was not intended to alter the fluid properties significantly. The onset of waviness, breakdown and dispersion of the filament provided qualitative evidence of transition. Hence Option B correctly describes the experimental purpose of the dye.
వివరణ:
Question 3
ప్రశ్న 3
For flow through a circular pipe, which expression correctly defines the Reynolds number based on the mean pipe velocity V and pipe diameter D?
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Explanation:
For circular-pipe flow, the characteristic velocity is conventionally the cross-sectional mean velocity V. The characteristic length is the internal pipe diameter D. Reynolds number is therefore Re = ρVD/μ. Since kinematic viscosity is ν = μ/ρ, the same expression can be written as Re = VD/ν. Hence Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
What is the primary physical interpretation of Reynolds number in viscous-flow classification?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Reynolds number compares the characteristic strength of inertial effects with viscous effects in a flow. A low Reynolds number indicates that viscous action is relatively important. A high Reynolds number indicates that inertia is relatively more dominant. This competition strongly influences the stability and regime of internal pipe flow. Therefore Option A gives the correct physical interpretation.
వివరణ:
Question 5
ప్రశ్న 5
Which statement correctly relates dynamic viscosity μ, kinematic viscosity ν and Reynolds number for pipe flow?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Kinematic viscosity is defined as dynamic viscosity divided by density, so ν = μ/ρ. Substituting μ = ρν into Re = ρVD/μ gives Re = VD/ν. Both forms are dimensionless and are equivalent when the properties are expressed consistently. Kinematic viscosity therefore incorporates the combined influence of dynamic viscosity and density. Hence Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
For conventional engineering classification of flow in a circular pipe, suppose the stated convention is Re < 2000 as laminar, 2000 ≤ Re ≤ 4000 as transitional, and Re > 4000 as turbulent. Which accompanying statement is most accurate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Values near 2000 and 4000 are commonly used as convenient engineering limits for circular-pipe classification. They should not be treated as universal critical values for every flow geometry. Transition in pipe flow is sensitive to disturbances, inlet conditions, wall condition and experimental environment. Laminar motion can sometimes persist above a nominal critical value when disturbances are carefully suppressed. Therefore Option B gives the technically appropriate interpretation.
వివరణ:
Question 7
ప్రశ్న 7
Air of density 1.2 kg/m³ and dynamic viscosity 1.8 × 10^-5 Pa·s flows through a 50 mm diameter pipe at a mean velocity of 3.0 m/s. What is the Reynolds number?
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Explanation:
Reynolds number is Re = ρVD/μ. The diameter is 50 mm = 0.050 m. Substitution gives Re = (1.2 × 3.0 × 0.050)/(1.8 × 10^-5). The result is 10000. Therefore Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
An oil of density 800 kg/m³ and dynamic viscosity 0.080 Pa·s flows through a 40 mm diameter pipe at a mean velocity of 1.5 m/s. What is its Reynolds number and regime using the stated circular-pipe convention Re < 2000 for laminar flow?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Reynolds number is Re = ρVD/μ. Substitution gives Re = 800 × 1.5 × 0.040/0.080. This evaluates to 600. The value is well below the stated laminar limit of 2000. Therefore Option B correctly gives both the Reynolds number and regime.
వివరణ:
Question 9
ప్రశ్న 9
Water flows through a 20 mm diameter pipe at a mean velocity of 0.080 m/s. If ν = 1.0 × 10^-6 m²/s, what is the Reynolds number and regime using the stated pipe-flow convention?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Using kinematic viscosity, Reynolds number is Re = VD/ν. Substituting V = 0.080 m/s and D = 0.020 m gives VD = 0.0016 m²/s. Dividing by 1.0 × 10^-6 m²/s gives Re = 1600. This is below the stated value of 2000 for laminar circular-pipe flow. Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
Water flows through a 25 mm diameter circular pipe at 0.12 m/s. If ν = 1.0 × 10^-6 m²/s, how should the flow be classified using the stated convention Re < 2000 laminar, 2000 ≤ Re ≤ 4000 transitional, and Re > 4000 turbulent?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Reynolds number is Re = VD/ν. Here Re = 0.12 × 0.025/(1.0 × 10^-6). This gives Re = 3000. Under the classification explicitly stated in the question, 3000 lies within the transition range. Therefore Option C is correct.
వివరణ:
Question 11
ప్రశ్న 11
Water with kinematic viscosity 1.0 × 10^-6 m²/s flows at a mean velocity of 1.5 m/s through a 50 mm diameter pipe. What is the Reynolds number and corresponding regime under the stated pipe-flow convention?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Reynolds number based on pipe diameter is Re = VD/ν. Substitution gives Re = 1.5 × 0.050/(1.0 × 10^-6). The result is 75000. This is far above the stated turbulent threshold of 4000. Therefore Option C is correct.
వివరణ:
Question 12
ప్రశ్న 12
For a particular circular-pipe calculation, take the chosen critical Reynolds number for the onset of transition as 2000. A liquid has ν = 1.2 × 10^-6 m²/s and flows through a 30 mm pipe. What mean velocity corresponds to Re = 2000?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Reynolds number is Re = VD/ν. Solving for velocity gives V = Reν/D. Substituting Re = 2000, ν = 1.2 × 10^-6 m²/s and D = 0.030 m gives V = 0.080 m/s. This is a calculated velocity based on the explicitly selected critical Reynolds number rather than a universal physical threshold. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
An oil of density 900 kg/m³ flows at 1.2 m/s through a 40 mm diameter pipe. If its Reynolds number is 2400, what is its dynamic viscosity?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
From Re = ρVD/μ, the viscosity is μ = ρVD/Re. Substitution gives μ = 900 × 1.2 × 0.040/2400. The result is 0.018 Pa·s. The stated Reynolds number would lie within the stated transition range, but only viscosity is requested here. Therefore Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
A liquid of kinematic viscosity 1.5 × 10^-6 m²/s flows at 0.20 m/s in a circular pipe. What pipe diameter would correspond to Re = 4000?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Starting with Re = VD/ν, solve for diameter as D = Reν/V. Substituting the data gives D = 4000 × 1.5 × 10^-6/0.20. This equals 0.030 m. Converting to millimetres gives 30 mm. Therefore Option B is correct.
వివరణ:
Question 15
ప్రశ్న 15
For water with ν = 1.0 × 10^-6 m²/s flowing through a 25 mm diameter circular pipe, what discharge corresponds to Re = 2000?
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Explanation:
For circular-pipe flow, V = 4Q/(πD²). Substituting this into Re = VD/ν gives Re = 4Q/(πDν). Therefore Q = ReπDν/4. With Re = 2000, D = 0.025 m and ν = 1.0 × 10^-6 m²/s, Q is approximately 3.93 × 10^-5 m³/s. Therefore Option B is correct.
వివరణ:
Question 16
ప్రశ్న 16
Oil of kinematic viscosity 2.0 × 10^-5 m²/s flows at a discharge of 0.0020 m³/s through a 40 mm diameter circular pipe. Using Re = 4Q/(πDν), what is the Reynolds number and regime under the stated convention?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a circular pipe, combining Q = AV with Re = VD/ν gives Re = 4Q/(πDν). Substitution gives Re = 4(0.0020)/[π(0.040)(2.0 × 10^-5)]. The result is approximately 3183. Under the stated convention, this lies between 2000 and 4000 and is therefore transitional. Hence Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
For a given fluid flowing through a fixed circular pipe, what happens to Reynolds number if the mean velocity is doubled while all fluid properties remain unchanged?
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Explanation:
For a fixed pipe and fixed fluid properties, Re = ρVD/μ. Density, diameter and dynamic viscosity remain constant under the stated conditions. Reynolds number is therefore directly proportional to mean velocity. Doubling the velocity consequently doubles Reynolds number. Hence Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
The same incompressible liquid discharge Q passes successively through two circular pipes, and the second pipe diameter is one-half of the first. If kinematic viscosity is unchanged, what is the ratio Re2/Re1?
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Explanation:
For fixed discharge in a circular pipe, Re = 4Q/(πDν). Thus Reynolds number is inversely proportional to pipe diameter when Q and ν are fixed. Halving D therefore doubles Re. Although velocity itself increases by a factor of four, Reynolds number contains the additional multiplying factor D. Consequently Re2/Re1 = 2 and Option C is correct.
వివరణ:
Question 19
ప్రశ్న 19
A liquid flows through a 40 mm pipe at 0.090 m/s. Its kinematic viscosity decreases from 2.0 × 10^-6 m²/s to 1.0 × 10^-6 m²/s while velocity and diameter remain unchanged. Using the stated pipe-flow convention, how does the classification change?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Initially Re = VD/ν = 0.090 × 0.040/(2.0 × 10^-6) = 1800. After the kinematic viscosity is halved, Re becomes 0.090 × 0.040/(1.0 × 10^-6) = 3600. Lower viscosity therefore produces a larger Reynolds number when velocity and diameter are unchanged. Under the stated convention, 1800 is laminar while 3600 lies in the transition range. Therefore Option A is correct.
వివరణ:
Question 20
ప్రశ్న 20
If density is doubled while dynamic viscosity, pipe diameter and mean velocity remain unchanged, what happens to Reynolds number?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Using dynamic viscosity, Reynolds number is Re = ρVD/μ. Under the stated conditions, V, D and μ remain constant. Reynolds number is therefore directly proportional to density. Doubling density doubles the Reynolds number. Hence Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
If the dynamic viscosity of a fluid increases while density, mean velocity and pipe diameter remain unchanged, what is the resulting trend in Reynolds number?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Reynolds number is Re = ρVD/μ. Dynamic viscosity appears in the denominator of this expression. Increasing μ while holding the other quantities constant therefore decreases Reynolds number. A larger viscosity strengthens the relative importance of viscous effects compared with inertia. Consequently Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
For the same fluid and the same selected critical Reynolds number in circular pipes, how does the corresponding critical mean velocity vary with pipe diameter?
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Explanation:
For a selected critical Reynolds number, Vc = Recν/D. The fluid kinematic viscosity and the selected critical Reynolds number are constant in this comparison. Critical velocity is therefore inversely proportional to pipe diameter. A smaller pipe requires a larger mean velocity to attain the same Reynolds number. Hence Option B is correct.
వివరణ:
Question 23
ప్రశ్న 23
When Reynolds number is formed for ordinary flow through a completely filled circular pipe, which characteristic quantities are conventionally used?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The conventional pipe Reynolds number is based on the cross-sectional mean velocity. For a circular pipe, the characteristic length is the internal diameter. This produces Re = ρVD/μ or VD/ν. Using the centreline velocity instead would define a different numerical Reynolds number and would not match the usual pipe-flow transition criteria. Therefore Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
Why should a Reynolds-number transition limit obtained for circular-pipe flow not be applied automatically to flow over a flat plate, around a cylinder or through every other geometry?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Reynolds number itself is a general dimensionless parameter and is not restricted to pipe flow. However, its characteristic length and sometimes its characteristic velocity are selected according to the particular flow geometry. More importantly, hydrodynamic instability and transition depend strongly on the physical configuration and disturbance environment. Critical Reynolds numbers for pipe flow, boundary layers and flow around bodies therefore differ greatly. Hence Option B is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements about Reynolds number and circular-pipe flow: I. Reynolds number is dimensionless. II. For a circular pipe it is commonly based on mean velocity and pipe diameter. III. At fixed V, D and ρ, increasing dynamic viscosity decreases Reynolds number. IV. The nominal transition range used for circular-pipe flow should not be treated as a universal transition criterion for every flow geometry. Which option is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement I is correct because the dimensions in ρVD/μ cancel completely. Statement II is correct because mean velocity and internal diameter are the standard characteristic quantities for a circular pipe. Statement III is correct because dynamic viscosity occurs in the denominator of Reynolds number. Statement IV is correct because transition behaviour and characteristic quantities depend on the physical flow configuration. Therefore all four statements are valid and Option D is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. In laminar flow the dye filament remains relatively straight and distinct, whereas in turbulent flow it spreads and mixes across the flowing fluid సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: B. To act as a visible tracer revealing the degree of orderly motion and transverse mixing సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: A. Re = ρVD/μ = VD/ν సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: A. Ratio of inertial effects to viscous effects సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: A. ν = μ/ρ and Re = VD/ν సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: B. These are conventional circular-pipe guidelines, while actual transition can depend on disturbances, inlet conditions, roughness and geometry సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: C. 10000 సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: B. Re = 600; laminar సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: B. Re = 1600; laminar సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: C. Re = 3000; transitional సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: C. Re = 75000; turbulent సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: C. 0.080 m/s సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. 0.018 Pa·s సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: B. 30 mm సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: B. 3.93 × 10^-5 m³/s సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: C. Re ≈ 3183; transitional సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: C. It doubles సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: C. 2 సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: A. Re changes from 1800 to 3600, so the classification changes from laminar to transitional సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: B. It doubles సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: B. Reynolds number decreases because viscosity appears in the denominator సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: B. It is inversely proportional to diameter సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: B. Cross-sectional mean velocity and internal pipe diameter సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: B. Because the appropriate characteristic length and the stability behaviour depend on the flow geometry and disturbance environment సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: D. I, II, III and IV సమాధానం: D.