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Question 1
ప్రశ్న 1
Pressure intensity at a point in a fluid is most appropriately interpreted as which quantity?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Pressure intensity is the normal compressive force per unit area acting at a point in a fluid. More precisely, it is defined as the limiting value of normal force divided by area as the area becomes very small around the point. Its SI unit is the pascal, where 1 Pa = 1 N/m². The dimensional formula of pressure is ML^-1T^-2. Therefore Option A gives both the correct physical interpretation and the correct SI unit.
వివరణ:
Question 2
ప్రశ్న 2
For a fluid at rest, which statement about pressure at a point is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A fluid at rest cannot sustain a finite shear stress. The stress acting on any small plane through a point is therefore normal to that plane. Static equilibrium of an infinitesimal fluid element requires the pressure magnitude at the point to be independent of the plane orientation. Gravity creates a spatial pressure gradient with elevation, but it does not make pressure at one point direction-dependent. Hence Option B correctly describes pressure in a static fluid.
వివరణ:
Question 3
ప్రశ్న 3
Let z be measured vertically upward in a static fluid of density ρ under uniform gravitational acceleration g. Which hydrostatic relation is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Hydrostatic equilibrium balances the pressure gradient against the body force due to gravity. With z measured positive upward, pressure decreases as elevation increases. The differential hydrostatic equation is therefore dp/dz = -ρg. For a constant-density liquid, integration gives a linear pressure variation with vertical elevation. Option B has the correct sign and dimensions.
వివరణ:
Question 4
ప్రశ్న 4
Points A and B lie at the same elevation in the same connected, homogeneous liquid that is at rest. Which statement is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
In ordinary hydrostatics under gravity, pressure in a homogeneous fluid varies only with vertical elevation. Two points at the same elevation in the same connected static liquid therefore have equal pressure. The shape or width of the containing vessel does not alter this equality. This principle is fundamental when equating pressures at common levels in manometer calculations. Therefore Option C is correct under the stated hydrostatic conditions.
వివరణ:
Question 5
ప్రశ్న 5
Two points in water are separated vertically by 3.0 m, with the lower point beneath the upper point. Taking ρ = 1000 kg/m³ and g = 9.81 m/s², by how much does pressure increase from the upper point to the lower point?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a constant-density liquid at rest, the pressure increase with downward depth is Δp = ρgΔh. Substituting ρ = 1000 kg/m³, g = 9.81 m/s², and Δh = 3.0 m gives Δp = 29430 Pa. This is 29.43 kPa. The result is a pressure difference, so it is independent of whether the pressures are expressed on an absolute or gauge basis as long as the same datum is used at both points. Therefore Option C is correct.
వివరణ:
Question 6
ప్రశ్న 6
An open tank contains an oil of specific gravity 0.80 to a depth of 5.0 m. Taking water density as 1000 kg/m³ and g = 9.81 m/s², what is the gauge pressure at the bottom?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Specific gravity 0.80 gives the oil density as ρ = 0.80 × 1000 = 800 kg/m³. Because the tank is open, the free surface is at zero gauge pressure. The bottom gauge pressure is p = ρgh = 800 × 9.81 × 5 = 39240 Pa. This equals 39.24 kPa. Therefore Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
Which expression correctly defines pressure head h_p for a fluid of specific weight γ?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Pressure head expresses pressure as an equivalent height of the same fluid. Since specific weight γ has units N/m³ and pressure p has units N/m², the ratio p/γ has units of metres. Thus h_p = p/γ = p/(ρg). The numerical head corresponding to a given pressure depends on the specific weight of the fluid used to express the head. Option B correctly gives both the formula and its dimension.
వివరణ:
Question 8
ప్రశ్న 8
A water pressure of 147.15 kPa is to be expressed as pressure head. Taking γwater = 9.81 kN/m³, the pressure head is:
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Pressure head is h_p = p/γ. Using p = 147.15 kPa = 147.15 kN/m² and γwater = 9.81 kN/m³ gives h_p = 147.15/9.81 m. The result is exactly 15 m. Using consistent kN-based units makes the conversion direct because kN/m² divided by kN/m³ gives metres. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
A pressure of 100 kPa is expressed first as a head of water and then as a head of oil having specific gravity 0.80. Taking g = 9.81 m/s², which pair is nearest?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The pressure head is h = p/(ρg). For water, h = 100000/(1000 × 9.81) = 10.19 m approximately. The oil density is 0.80 × 1000 = 800 kg/m³, so its head is 100000/(800 × 9.81) = 12.74 m approximately. A lighter liquid requires a taller column to produce the same pressure. Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
What height of mercury of specific gravity 13.6 is equivalent to a pressure of 100 kPa? Take water density as 1000 kg/m³ and g = 9.81 m/s².
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The density of mercury is ρ = 13.6 × 1000 = 13600 kg/m³. The equivalent pressure head is h = p/(ρg). Substitution gives h = 100000/(13600 × 9.81) = 0.7495 m. Rounded suitably, this is about 0.750 m of mercury. Therefore Option B is correct.
వివరణ:
Question 11
ప్రశ్న 11
Which relationship among absolute pressure p_abs, gauge pressure p_g, and local atmospheric pressure p_atm is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Absolute pressure is referenced to a perfect vacuum. Gauge pressure is referenced to the local atmospheric pressure. Consequently, p_g = p_abs - p_atm, which may be rearranged as p_abs = p_atm + p_g. A positive gauge pressure means the absolute pressure exceeds atmospheric pressure, while a negative gauge pressure indicates a sub-atmospheric condition. Therefore Option C is correct.
వివరణ:
Question 12
ప్రశ్న 12
A pressure gauge on a closed water line reads 250 kPa while the local atmospheric pressure is 98 kPa absolute. What is the absolute pressure in the line?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The gauge reading represents pressure above the local atmospheric pressure. The absolute pressure is therefore p_abs = p_atm + p_g. Substitution gives p_abs = 98 + 250 = 348 kPa. Subtracting atmospheric pressure would be appropriate when converting from absolute pressure to gauge pressure, not in this direction. Hence Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
A vessel contains gas at an absolute pressure of 80.0 kPa when the local atmospheric pressure is 101.3 kPa absolute. Which description is correct?
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Explanation:
Gauge pressure is p_g = p_abs - p_atm. Here p_g = 80.0 - 101.3 = -21.3 kPa. A negative gauge pressure means the vessel is below atmospheric pressure. The magnitude of this deficiency is commonly called the vacuum pressure, so the vacuum pressure is 21.3 kPa. Therefore Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
A vacuum gauge indicates a vacuum of 35 kPa in a vessel when local atmospheric pressure is 100 kPa absolute. What is the vessel absolute pressure?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A vacuum pressure is the amount by which atmospheric pressure exceeds the absolute pressure in the vessel. Thus p_vac = p_atm - p_abs. Rearranging gives p_abs = p_atm - p_vac = 100 - 35 = 65 kPa. The corresponding gauge pressure would be -35 kPa. Therefore Option B is correct.
వివరణ:
Question 15
ప్రశ్న 15
A vacuum is specified as 0.40 m of mercury. Taking ρHg = 13600 kg/m³ and g = 9.81 m/s², what pressure deficiency below atmosphere does this represent?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A pressure head h of a liquid corresponds to pressure magnitude p = ρgh. Using mercury gives p = 13600 × 9.81 × 0.40 Pa. This equals 53366.4 Pa, or about 53.37 kPa. Because the value is described as a vacuum, it represents a pressure deficiency below atmospheric pressure rather than a positive gauge pressure. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
In the ideal pressure-reference sense, what is the lowest possible gauge pressure when the local atmospheric pressure is p_atm and absolute pressure is not permitted to fall below zero?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Gauge pressure is related to absolute pressure by p_g = p_abs - p_atm. The limiting ideal vacuum corresponds to p_abs = 0. Substituting this limiting value gives p_g = -p_atm. A more negative gauge pressure would imply a negative absolute pressure under this classical pressure convention. Therefore Option B is correct.
వివరణ:
Question 17
ప్రశ్న 17
An open vessel contains 2.0 m of oil of specific gravity 0.80 over 3.0 m of water. Taking ρwater = 1000 kg/m³ and g = 9.81 m/s², what is the gauge pressure at the bottom?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The pressure increase through each liquid layer equals its specific weight multiplied by its vertical depth. The oil contributes 800 × 9.81 × 2 = 15696 Pa. The water contributes 1000 × 9.81 × 3 = 29430 Pa. Adding the layer contributions gives 45126 Pa, or about 45.13 kPa gauge because the top surface is open to atmosphere. Therefore Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
Two open vessels of different shapes contain the same liquid to the same vertical depth h. Neglecting capillary effects, how do the gauge pressures at their bottom points compare if both bottom points are at depth h below their respective free surfaces?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a constant-density liquid at rest with an atmospheric free surface, bottom gauge pressure is p = ρgh. This relation contains density, gravitational acceleration, and vertical depth but not the vessel shape or total liquid volume. Consequently, equal depths of the same liquid produce equal bottom gauge pressures. Vessel geometry can change the total force on a boundary because area may change, but it does not alter pressure at a specified depth. Therefore Option C is correct.
వివరణ:
Question 19
ప్రశ్న 19
For the same pressure difference p, which statement correctly compares the corresponding pressure heads in two liquids having specific weights γ1 and γ2, where γ2 > γ1?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Pressure head in a specified liquid is h = p/γ. For the same pressure p, head is inversely proportional to the liquid specific weight. Since γ2 is greater than γ1, p/γ2 is smaller than p/γ1. This is why dense manometric liquids such as mercury require a smaller height difference for a given pressure difference than lighter liquids. Therefore Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
When traversing vertically through a static liquid of constant density during a manometer calculation, which pressure-tracking rule is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Hydrostatic pressure increases with depth in a static liquid. Therefore moving downward through a vertical distance Δh increases pressure by ρgΔh. Conversely, moving upward through the same liquid decreases pressure by ρgΔh. This rule is applied successively across each liquid column when writing a manometer pressure balance. Option B correctly states the fundamental traversal rule.
వివరణ:
Question 21
ప్రశ్న 21
In a connected U-tube manometer containing a single continuous manometric liquid at rest, points X and Y are selected in that same liquid at exactly the same elevation. Which relation is valid?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Pressure in a connected homogeneous liquid at rest is the same at all points having the same elevation under ordinary gravity. Therefore points X and Y at the same level in the same continuous manometric liquid have equal pressures. The limb in which a point lies does not affect this hydrostatic equality. Tube diameter is also irrelevant to the equality when capillary effects are neglected. Hence Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
Which statement best describes the use of a simple open piezometer tube connected to a point in a liquid system?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A simple piezometer is a vertical transparent tube connected to the liquid at the pressure-tapping point and open to atmosphere at its top. The rise of the liquid above the tapping point directly represents positive gauge pressure head p/γ under hydrostatic conditions in the tube. A simple open piezometer is not suitable for sub-atmospheric pressure because the required free surface would lie below the tapping elevation in the idealized arrangement. It is also generally unsuitable for gas-pressure measurement because a gas does not provide the required measurable liquid column in the same way. Therefore Option A correctly states the principal use and limitation.
వివరణ:
Question 23
ప్రశ్న 23
Standard atmospheric pressure of 101.325 kPa is equivalent to approximately what head of water if ρwater = 1000 kg/m³ and g = 9.81 m/s²?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Equivalent water head is obtained from h = p/(ρg). Substituting p = 101325 Pa gives h = 101325/(1000 × 9.81). The result is approximately 10.33 m of water. The familiar mercury barometer height near 0.76 m corresponds to the same atmospheric pressure because mercury is much denser than water. Therefore Option C is correct.
వివరణ:
Question 24
ప్రశ్న 24
Point A is 2.0 m below the free surface of water, and point B is 2.0 m below the free surface of an oil of specific gravity 0.75. Both free surfaces are exposed to the same atmosphere. Taking g = 9.81 m/s², by how much does the gauge pressure at A exceed that at B?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At the same depth, hydrostatic gauge pressure is proportional to liquid density. For water, pA = 1000 × 9.81 × 2 = 19.62 kPa. For the oil, ρ = 0.75 × 1000 = 750 kg/m³, giving pB = 750 × 9.81 × 2 = 14.715 kPa. Their difference is 19.62 - 14.715 = 4.905 kPa, which rounds to 4.91 kPa. Therefore Option B is correct.
వివరణ:
Question 25
ప్రశ్న 25
A water vessel is connected at point A to the left limb of a U-tube manometer containing mercury of density 13600 kg/m³. The right limb is open to atmosphere. The water-mercury interface in the left limb is 0.25 m below A, and the mercury free surface in the right limb is 0.40 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:
Start at the open mercury surface, where the gauge pressure is zero. Moving downward 0.40 m through mercury increases pressure by 13600 × 9.81 × 0.40 = 53.366 kPa at the left mercury interface. Moving upward from that interface to point A through 0.25 m of water decreases pressure by 1000 × 9.81 × 0.25 = 2.453 kPa. Hence pA = 53.366 - 2.453 = 50.914 kPa gauge, which rounds to 50.91 kPa. Therefore Option B is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. Normal force per unit area in the limiting sense as the area tends to zero, with SI unit N/m² సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: B. Pressure at a point is a scalar thermodynamic quantity, and the normal compressive stress has the same magnitude in all directions in a static fluid సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: B. dp/dz = -ρg సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: C. Pressure at A equals pressure at B, provided no other body-force field produces a horizontal pressure gradient సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: C. 29.43 kPa సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: B. 39.24 kPa సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: B. h_p = p/γ, with dimensions of length సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: C. 15 m of water సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. Water: 10.19 m; oil: 12.74 m సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: B. 0.750 m సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: C. p_abs = p_atm + p_g సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: C. 348 kPa సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. Gauge pressure = -21.3 kPa and vacuum pressure = 21.3 kPa సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: B. 65 kPa సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: C. 53.37 kPa సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: B. -p_atm సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: C. 45.13 kPa సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: C. The bottom pressures are equal because hydrostatic pressure depends on ρ, g, and vertical depth rather than vessel shape సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: C. The head in liquid 1 is larger because h = p/γ సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: B. Moving downward by Δh increases pressure by ρgΔh, while moving upward decreases it సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: A. pX = pY సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: A. It directly measures positive gauge pressure as a liquid-column height, but it is unsuitable for measuring a vacuum pressure because the liquid level would have to fall below the tapping point in the simple arrangement సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: C. 10.33 m సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: B. 4.91 kPa సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: B. 50.91 kPa సమాధానం: B.