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Question 1
ప్రశ్న 1
While writing a pressure balance through a stationary manometer containing several immiscible liquids, which traversal rule is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A manometer is analysed using the hydrostatic relation within each stationary liquid column. When the calculation path moves downward through a vertical distance Δz, pressure increases by ρgΔz. When the path moves upward through the same vertical distance, pressure decreases by ρgΔz. Horizontal movement within the same continuous static liquid produces no pressure change because there is no change in elevation. Therefore Option B gives the correct fundamental pressure-traversal rule.
వివరణ:
Question 2
ప్రశ్న 2
Which statement correctly identifies an important limitation of a simple open piezometer?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A piezometer consists essentially of a vertical tube connected to a liquid system and open to atmosphere at its upper end. A positive gauge pressure causes the liquid to rise above the tapping point, and the vertical rise represents pressure head. A simple piezometer cannot conveniently represent a negative gauge pressure because the required free surface would lie below the tapping point in the ideal arrangement. It is also unsuitable for direct gas-pressure measurement because the gas itself does not form the required measurable liquid column. Hence Option B correctly states its principal application and limitations.
వివరణ:
Question 3
ప్రశ్న 3
A water pipe at point A is connected to the left limb of a simple U-tube manometer containing mercury of density 13600 kg/m³. The right mercury surface is open to atmosphere and is 0.50 m above the water-mercury interface in the left limb. Point A is 0.30 m above that interface. Taking ρwater = 1000 kg/m³ and g = 9.81 m/s², what is the gauge pressure at A?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Begin at the open mercury surface, where gauge pressure is zero. Moving downward 0.50 m through mercury raises the pressure by 13600 × 9.81 × 0.50 = 66.708 kPa. From the left mercury interface, moving upward 0.30 m through water reduces pressure by 1000 × 9.81 × 0.30 = 2.943 kPa. Therefore the gauge pressure at A is 66.708 - 2.943 = 63.765 kPa. Rounded appropriately, this is 63.77 kPa, so Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
A vessel containing water is connected to a U-tube containing mercury. The right mercury surface is open to atmosphere. The water-mercury interface in the left limb is 0.25 m above the open mercury surface in the right limb, and point A in the vessel is 0.15 m above the left interface. Taking ρHg = 13600 kg/m³, ρwater = 1000 kg/m³ and g = 9.81 m/s², what vacuum pressure exists at A?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The open mercury surface is at zero gauge pressure. Moving upward 0.25 m through mercury reduces the pressure by 13600 × 9.81 × 0.25 = 33.354 kPa. Moving a further 0.15 m upward through water reduces it by 1000 × 9.81 × 0.15 = 1.4715 kPa. Thus the gauge pressure at A is -34.8255 kPa, whose magnitude is the vacuum pressure. Therefore the vacuum is approximately 34.83 kPa and Option C is correct.
వివరణ:
Question 5
ప్రశ్న 5
A differential U-tube manometer containing a liquid denser than the pipe fluid connects two points at the same elevation. If pressure at A is greater than pressure at B, what qualitative position of the heavy manometric liquid is expected?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The higher pressure at A pushes more strongly on the manometric system. With a heavy manometric liquid in the lower part of a conventional U-tube, this action depresses the heavy-liquid interface on the high-pressure side. The heavy liquid consequently rises in the limb connected to the lower-pressure side. This qualitative behaviour is useful for checking the sign of a numerical manometer calculation before accepting its result. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
Two points A and B in a horizontal water pipe are connected by a differential U-tube manometer containing mercury of density 13600 kg/m³. The mercury interface in the A limb is 0.20 m lower than the interface in the B limb. Taking ρwater = 1000 kg/m³ and g = 9.81 m/s², determine pA - pB.
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Because A and B are at the same elevation and contain the same pipe fluid, the relevant differential effect involves the density difference between mercury and water. The pressure difference is (ρHg - ρwater)gh for the stated interface difference. Substitution gives (13600 - 1000) × 9.81 × 0.20 = 24721.2 Pa. The lower mercury interface on the A side is consistent with A being the higher-pressure point. Hence pA - pB = 24.72 kPa approximately, making Option B correct.
వివరణ:
Question 7
ప్రశ్న 7
Two pressure points in the same liquid are connected by a differential manometer, but the pressure taps are at different elevations. Which statement is technically correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A manometer balance must account for every vertical movement through every liquid between the two pressure points. If the pressure taps are at different elevations, the pipe liquid itself contributes a hydrostatic pressure difference between those elevations. The familiar same-elevation shortcut is therefore not valid without modification. A direct step-by-step pressure traversal automatically includes the required elevation effect and avoids sign mistakes. Consequently, Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
Points A and B in an oil of density 850 kg/m³ are connected by a differential U-tube containing mercury of density 13600 kg/m³. Point B is 1.20 m above A, and the mercury interface in the B limb is 0.18 m above the mercury interface in the A limb. Taking g = 9.81 m/s², what is pA - pB?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The difference in tap elevations contributes ρoil g(zB - zA) to pA - pB. This component is 850 × 9.81 × 1.20 = 10.006 kPa. The 0.18 m interface difference contributes (13600 - 850) × 9.81 × 0.18 = 22.514 kPa. Adding the two contributions gives pA - pB = 32.520 kPa. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
Why is an inverted differential U-tube containing a manometric fluid lighter than the pipe liquid particularly useful in some pressure measurements?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
An inverted differential manometer places a lighter manometric fluid above the denser liquid associated with the pressure taps. For equal-elevation taps in the same liquid, the pressure difference is related to the difference between the two fluid densities rather than to a very large manometric-fluid density. A smaller density difference produces a larger vertical displacement for a given small pressure difference. This increases sensitivity and makes the arrangement useful when relatively small differential pressures must be resolved. Therefore Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
For an inverted differential manometer connecting two points at the same elevation in a liquid of density ρ, a lighter manometric liquid has density ρm. If ρm is increased while remaining less than ρ, what happens to the level difference required to represent the same pressure difference?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For this inverted-manometer configuration, the differential pressure for equal-elevation taps depends on the product of g, the level difference, and the density difference ρ - ρm. If ρm moves closer to ρ, the density difference becomes smaller. To represent the same imposed pressure difference, the required vertical level difference must consequently become larger. This is the physical basis for the enhanced sensitivity obtained with a light manometric liquid whose density is relatively close to that of the measured liquid. Hence Option B is correct.
వివరణ:
Question 11
ప్రశ్న 11
Two points A and B at the same elevation in a water pipe are connected by an inverted differential U-tube containing oil of density 800 kg/m³. The water-oil interface in the A limb is 0.25 m higher than that in the B limb. Taking ρwater = 1000 kg/m³ and g = 9.81 m/s², what is pA - pB?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
In an inverted differential manometer, the lighter liquid occupies the upper connecting portion. For equal-elevation water taps, the pressure-difference magnitude is (ρwater - ρoil)gh. Substitution gives (1000 - 800) × 9.81 × 0.25 = 490.5 Pa. The interface is higher on the A side, which is consistent with A having the higher pressure in this inverted arrangement. Thus pA - pB = 0.4905 kPa, and Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
An inverted U-tube containing a very light gas is used between two points in the same liquid that are at different elevations. Which quantity is most directly related to the manometer reading when the gas density is negligible?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
When the upper manometric fluid is a gas of negligible density, its hydrostatic pressure variation over the instrument height can usually be neglected. The liquid columns beneath it still contribute according to their elevations. As a result, the balance naturally combines static pressure with elevation and is related to the difference in piezometric pressure p + ρgz, or equivalently piezometric head p/(ρg) + z, for the same liquid. Only when the two taps are at the same elevation does this reduce directly to the static pressure difference. Therefore Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
A gas pressure at point A acts through 0.20 m of oil of density 800 kg/m³ above the left mercury interface of a U-tube. The right limb is open to atmosphere through 0.10 m of water above the right mercury interface. The right mercury interface is 0.30 m above the left mercury interface. Taking ρHg = 13600 kg/m³, ρwater = 1000 kg/m³ and g = 9.81 m/s², what is the gauge pressure at A?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Start at the atmospheric water surface, where gauge pressure is zero. Moving downward 0.10 m through water adds 1000 × 9.81 × 0.10 = 0.981 kPa, and then moving downward 0.30 m through mercury adds 13600 × 9.81 × 0.30 = 40.025 kPa. From the left mercury interface, moving upward 0.20 m through oil subtracts 800 × 9.81 × 0.20 = 1.570 kPa. The resulting pressure at A is 0.981 + 40.025 - 1.570 = 39.436 kPa. Therefore the gauge pressure is approximately 39.44 kPa and Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
In a well-type or single-column manometer, why can the change in liquid level in the large reservoir sometimes be neglected?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Liquid displaced from one part of a manometer must appear elsewhere because the manometric liquid is effectively incompressible. If the reservoir area A is much larger than the narrow-tube area a, a given displaced volume produces only a very small reservoir-level movement. Volume conservation gives Ax = ay, where x is reservoir displacement and y is narrow-tube displacement. Thus x/y = a/A becomes very small when A/a is large. Therefore Option B correctly explains the approximation used in a well-type manometer.
వివరణ:
Question 15
ప్రశ్న 15
A well-type manometer uses a liquid of density 900 kg/m³ to measure a gas gauge pressure. The gas acts on the large reservoir and the narrow tube is open to atmosphere; gas-density effects may be neglected. The reservoir cross-sectional area is 100 times the narrow-tube area. From the initial equal-level condition, the liquid in the narrow tube rises by 0.200 m. Taking g = 9.81 m/s², what gauge pressure is indicated when reservoir movement is included?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Let the narrow-tube rise be y = 0.200 m and the reservoir fall be x. Volume conservation gives Ax = ay, so x = (a/A)y = 0.200/100 = 0.002 m. The final vertical difference between the two liquid levels is therefore x + y = 0.202 m. The gas gauge pressure is p = ρg(x + y) = 900 × 9.81 × 0.202 = 1783.46 Pa. This is approximately 1.783 kPa, so Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
Which set of properties is generally desirable for a manometric liquid used for accurate engineering pressure measurement?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A useful manometric liquid should have a known and sufficiently stable density because hydrostatic pressure calculations depend directly on density. It should not react adversely with the process fluid or the containing tube. When another liquid is present, the interface should remain identifiable and undesirable mixing should be avoided for a reliable level reading. Its density should also suit the required pressure range and sensitivity because the level displacement depends on the relevant density difference. Therefore Option B gives the technically appropriate combination.
వివరణ:
Question 17
ప్రశ్న 17
During a multi-fluid manometer calculation, an engineer chooses a different horizontal datum from another engineer but both trace all elevations and densities correctly. What should happen to the calculated pressure difference?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A geometric datum provides a reference from which elevations are measured, but its absolute position is arbitrary. Hydrostatic equations involve differences in elevation between physical points rather than the numerical value assigned to an arbitrary origin. Shifting the datum by a constant changes all elevations by the same amount and those common changes cancel in a correctly written pressure balance. The calculated physical pressure difference must therefore remain invariant. Hence Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
Which statement correctly states when pressures at two points at the same elevation may be equated directly in an ordinary hydrostatic manometer analysis?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a fluid in hydrostatic equilibrium under ordinary gravity, there is no horizontal pressure gradient. Consequently, two points at the same elevation in the same continuous static fluid have equal pressure. Constant density is not required merely for this same-level equality, although density determines the vertical pressure gradient. Equal elevation in different or disconnected fluids does not by itself establish equal pressure. Therefore Option A states the correct condition without imposing an unnecessary uniform-density requirement.
వివరణ:
Question 19
ప్రశ్న 19
At a stationary interface between two immiscible liquids in a manometer, and neglecting surface-tension curvature effects, which statement about pressure is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Mechanical equilibrium requires the normal pressure to balance across an ideal stationary liquid-liquid interface when surface-tension curvature effects are neglected. Thus the pressure immediately on one side of the interface equals the pressure immediately on the other side. The pressure gradient with elevation is not the same in the two liquids because dp/dz = -ρg and their densities differ. This distinction allows pressure to be carried continuously across an interface while using a new density for the next vertical segment. Therefore Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
A pressure difference is represented by 0.40 m of mercury having specific gravity 13.6. What height of an oil of specific gravity 0.80 would represent the same pressure difference?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Equal pressure differences require equal values of ρgh for the two liquid columns. Because both specific gravities are referenced to the same water density, the density ratio can be replaced directly by the specific-gravity ratio. Thus hoil = (13.6/0.80) × 0.40 m. This gives hoil = 6.80 m. The much larger oil column is required because oil is far less dense than mercury, so Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
Two points at the same elevation in a horizontal water pipe are connected by a differential U-tube containing an unknown heavier manometric liquid. The interface difference is 0.15 m and pA - pB = 18.5409 kPa, with the interface on the A side lower. Taking ρwater = 1000 kg/m³ and g = 9.81 m/s², what is the specific gravity of the manometric liquid?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For equal-elevation taps containing water and a heavier manometric liquid, the pressure difference is (ρm - ρwater)gh. Hence ρm - 1000 = 18540.9/(9.81 × 0.15). The right-hand side equals 12600 kg/m³, giving ρm = 13600 kg/m³. Relative to water of density 1000 kg/m³, the specific gravity is therefore 13.6. Thus Option C is correct.
వివరణ:
Question 22
ప్రశ్న 22
A water line at A and an oil line at B are connected by a mercury differential manometer. The water-mercury interface is 0.20 m below A, the oil-mercury interface is 0.10 m below B, and the oil-mercury interface is 0.25 m above the water-mercury interface. If pA = 120 kPa gauge, ρwater = 1000 kg/m³, ρoil = 800 kg/m³, ρHg = 13600 kg/m³, and g = 9.81 m/s², what is pB?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Starting at A, moving downward 0.20 m through water raises pressure by 1000 × 9.81 × 0.20 = 1.962 kPa. Moving upward 0.25 m through mercury then reduces pressure by 13600 × 9.81 × 0.25 = 33.354 kPa. Finally, moving upward 0.10 m through oil to B reduces pressure by another 800 × 9.81 × 0.10 = 0.785 kPa. Thus pB = 120 + 1.962 - 33.354 - 0.785 = 87.823 kPa gauge. Therefore Option B is correct.
వివరణ:
Question 23
ప్రశ్న 23
A differential manometer connects two pressurized points in a hydraulic system. If both pressures are changed from gauge values to absolute values using the same local atmospheric pressure, what happens to their pressure difference?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Absolute pressure equals gauge pressure plus the same local atmospheric pressure. If pA,abs = pA,g + patm and pB,abs = pB,g + patm, subtracting the two equations cancels patm. Therefore pA,abs - pB,abs is identical to pA,g - pB,g. This is why a differential manometer can determine a pressure difference without requiring the numerical atmospheric pressure when both sides share the same atmospheric reference. Hence Option C is correct.
వివరణ:
Question 24
ప్రశ్న 24
Consider two manometers connecting equal-elevation points in the same liquid. One is a conventional U-tube containing a heavier manometric liquid, while the other is an inverted U-tube containing a lighter manometric liquid. If A is the higher-pressure point, which interface movement is consistent with hydrostatics?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
In a conventional differential U-tube, greater pressure at A depresses the denser manometric liquid on the A side, so its interface is lower there. In an inverted manometer, the upper manometric fluid is lighter than the liquid in the connecting limbs. Greater pressure at A pushes the denser pipe liquid farther upward on the A side, so the light-liquid interface is higher there. Remembering this reversal provides a useful qualitative sign check before performing detailed calculations. Therefore Option B is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements about manometers: I. Moving downward through a static liquid increases pressure. II. Pressure is equal at two points on the same horizontal level in the same continuous static liquid under ordinary gravity. III. Neglecting surface-tension curvature effects, pressure is continuous across a stationary interface between immiscible liquids. IV. An inverted differential manometer employing a lighter fluid can provide enhanced sensitivity for small pressure differences. Which option is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement I is correct because hydrostatic pressure increases when moving downward in a liquid under gravity. Statement II is correct when both points belong to the same continuous static fluid and lie at the same elevation under the stated gravitational field. Statement III is correct when surface-tension curvature effects are neglected, although the pressure gradient changes across the interface if density changes. Statement IV is correct because a small density difference between the measured and light manometric fluids can produce a relatively large level displacement for a small pressure difference. Consequently all four statements are correct, making Option D the only valid choice.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Pressure increases by ρgΔz when moving downward through a liquid and decreases by ρgΔz when moving upward through it సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: B. It directly measures positive gauge pressure in a liquid but is unsuitable for ordinary measurement of vacuum pressure and is generally unsuitable for gas pressure సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: B. 63.77 kPa సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: C. 34.83 kPa సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: A. The heavy-liquid interface in the A limb is lower than that in the B limb సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: B. 24.72 kPa సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: C. The hydrostatic contribution associated with the difference in tap elevations must be included in the pressure balance సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: C. 32.52 kPa సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: A. It is useful for relatively small pressure differences because a small density difference can produce a comparatively large level displacement సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: B. It increases because ρ - ρm decreases సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: B. 0.491 kPa సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: B. The difference in piezometric pressure or piezometric head between the two points సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: B. 39.44 kPa సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: B. Because when the reservoir area is very large compared with the tube area, volume conservation makes the reservoir-level displacement very small సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: C. 1.783 kPa సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: B. Known and stable density, compatibility with the measured fluid, a clearly identifiable interface, and suitable density for the required sensitivity సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: B. It should remain unchanged because only physical elevation differences enter the hydrostatic balance సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: A. The points must belong to the same continuous static fluid under the same gravitational body-force field సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: B. Pressure is continuous across the interface, although the hydrostatic pressure gradient changes because the densities differ సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: C. 6.80 m సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: C. 13.6 సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: B. 87.82 kPa సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: C. It remains unchanged because the same atmospheric pressure is added to both values సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: B. Heavy-liquid interface is lower on the A side, and light-liquid interface is higher on the A side సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: D. I, II, III and IV సమాధానం: D.