10 FE practice problems: fluid statics, with solutions
These ten original problems practice fluid statics, a topic from the FE exam specification, using the Fluid Mechanics, Manometers part of the FE Reference Handbook 10.6. Each problem gives the situation and the values with units. Work it with the handbook PDF open and commit to an answer before you open the solution. Every solution shows the handbook page, the equation, the substitution with units, a size check, and the mistake behind each wrong option, so a wrong pick tells you exactly what to fix.
How to use these problems
Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.
The problems
Problem 1 · FE, Fluid statics
A mercury manometer (SG = 13.6) connects two points in a water pipeline, and the mercury levels differ by 2.5 in. Use γ = 62.4 lbf/ft³ for water. The pressure difference between the two points is most nearly:
Handbook: Fluid Mechanics, Manometers, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (1.14 psi)
- Given
- h = 2.5 in., sg = 13.6, γw = 62.4 lbf/ft³, si =
- Find
- pressure difference ΔP (psi)
- Handbook
- Fluid Mechanics, Manometers, page 183
- Equation
P1 - P2 = (γm - γw)h for a mercury-water manometer- Substitute
h = 2.5 in. = 0.2083 ftΔP = (γm - γw)h = (13.6 - 1)(62.4 lbf/ft³)(0.2083 ft) = 163.8 lbf/ft² ÷ 144 in.²/ft² = 1.14 psi
- Result
- 1.14 psi, 3 significant figures
- Check
- the water in both legs above the mercury cancels except for the height h, so only (γm - γw) acts.
- Why the others are wrong
- A: added the water column instead of subtracting it
- B: used γm·h and ignored the water above the mercury
- C: used the unit weight of water alone
Problem 2 · FE, Fluid statics
A mercury manometer (SG = 13.6) connects two points in a water pipeline, and the mercury levels differ by 4 in. Use γ = 62.4 lbf/ft³ for water. The pressure difference between the two points is most nearly:
Handbook: Fluid Mechanics, Manometers, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (1.82 psi)
- Given
- h = 4 in., sg = 13.6, γw = 62.4 lbf/ft³, si =
- Find
- pressure difference ΔP (psi)
- Handbook
- Fluid Mechanics, Manometers, page 183
- Equation
P1 - P2 = (γm - γw)h for a mercury-water manometer- Substitute
h = 4 in. = 0.3333 ftΔP = (γm - γw)h = (13.6 - 1)(62.4 lbf/ft³)(0.3333 ft) = 262.1 lbf/ft² ÷ 144 in.²/ft² = 1.82 psi
- Result
- 1.82 psi, 3 significant figures
- Check
- the water in both legs above the mercury cancels except for the height h, so only (γm - γw) acts.
- Why the others are wrong
- A: used the unit weight of water alone
- B: added the water column instead of subtracting it
- C: used γm·h and ignored the water above the mercury
Problem 3 · FE, Fluid statics
A mercury manometer (SG = 13.6) connects two points in a water pipeline, and the mercury levels differ by 330 mm. Use γ = 9.81 kN/m³ for water. The pressure difference between the two points is most nearly:
Handbook: Fluid Mechanics, Manometers, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (40.8 kPa)
- Given
- h = 330 mm, sg = 13.6, γw = 9.81 kN/m³, si =
- Find
- pressure difference ΔP (kPa)
- Handbook
- Fluid Mechanics, Manometers, page 183
- Equation
P1 - P2 = (γm - γw)h for a mercury-water manometer- Substitute
h = 330 mm = 0.330 mΔP = (γm - γw)h = (13.6 - 1)(9.81 kN/m³)(0.330 m) = 40.8 kPa
- Result
- 40.8 kPa, 3 significant figures
- Check
- the water in both legs above the mercury cancels except for the height h, so only (γm - γw) acts.
- Why the others are wrong
- B: used the unit weight of water alone
- C: added the water column instead of subtracting it
- D: used γm·h and ignored the water above the mercury
Problem 4 · FE, Fluid statics
A vertical rectangular gate 3.9 m wide and 2.6 m high holds back water, with its top edge 2.1 m below the water surface. Use γ = 9.81 kN/m³ for water. The resultant water force on the gate is most nearly:
Handbook: Fluid Mechanics, Forces on Submerged Surfaces, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (338 kN)
- Given
- d = 2.1 m, H = 2.6 m, b = 3.9 m, γw = 9.81 kN/m³
- Find
- resultant force F (kN)
- Handbook
- Fluid Mechanics, Forces on Submerged Surfaces and the Center of Pressure, page 184
- Equation
F = PC·A = γ·hc·A (pressure at the centroid times the area)- Substitute
Depth to the centroid: hc = d + H/2 = 2.1 + 1.3 = 3.400 mF = γ hc A = (9.81 kN/m³)(3.400 m)(3.9 m × 2.6 m) = 338 kN
- Result
- 338 kN, 3 significant figures
- Check
- F is the pressure at the centroid, γhc = 33.35 kN/m², times the area 10.14 m².
- Why the others are wrong
- A: used H/2 as the depth, ignoring the water above the gate
- C: used half the gate area
- D: left out the gate width
Problem 5 · FE, Fluid statics
A solid rectangular block 4.0 m tall with a specific gravity of 0.55 floats upright in fresh water (SG = 1.0). The depth of the block below the water line is most nearly:
Handbook: Fluid Mechanics, Archimedes Principle and Buoyancy, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (2.20 m)
- Given
- H = 4.0 m, sgb = 0.55, sgf = 1.0
- Find
- submerged depth y (m)
- Handbook
- Fluid Mechanics, Archimedes Principle and Buoyancy, page 184
- Equation
Buoyant force = weight of displaced fluid; γf·Vsub = γb·V- Substitute
Floating: buoyant force = weight, so γf·A·y = γb·A·Hy = H·SGb/SGf = 4.0 m × 0.55/1.0 = 2.20 m
- Result
- 2.20 m, 3 significant figures
- Check
- the block floats because SG 0.55 < 1.0; the part above the water is 1.80 m.
- Why the others are wrong
- A: divided the height by the block's specific gravity instead of multiplying
- B: assumed the block floats half submerged
- C: is the height above the water line, not the submerged depth
Problem 6 · FE, Fluid statics
A mercury manometer (SG = 13.6) connects two points in a water pipeline, and the mercury levels differ by 7 in. Use γ = 62.4 lbf/ft³ for water. The pressure difference between the two points is most nearly:
Handbook: Fluid Mechanics, Manometers, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (3.19 psi)
- Given
- h = 7 in., sg = 13.6, γw = 62.4 lbf/ft³, si =
- Find
- pressure difference ΔP (psi)
- Handbook
- Fluid Mechanics, Manometers, page 183
- Equation
P1 - P2 = (γm - γw)h for a mercury-water manometer- Substitute
h = 7 in. = 0.5833 ftΔP = (γm - γw)h = (13.6 - 1)(62.4 lbf/ft³)(0.5833 ft) = 458.6 lbf/ft² ÷ 144 in.²/ft² = 3.19 psi
- Result
- 3.19 psi, 3 significant figures
- Check
- the water in both legs above the mercury cancels except for the height h, so only (γm - γw) acts.
- Why the others are wrong
- A: added the water column instead of subtracting it
- B: used γm·h and ignored the water above the mercury
- D: used the unit weight of water alone
Problem 7 · FE, Fluid statics
A solid rectangular block 3.6 m tall with a specific gravity of 0.55 floats upright in fresh water (SG = 1.0). The depth of the block below the water line is most nearly:
Handbook: Fluid Mechanics, Archimedes Principle and Buoyancy, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (1.98 m)
- Given
- H = 3.6 m, sgb = 0.55, sgf = 1.0
- Find
- submerged depth y (m)
- Handbook
- Fluid Mechanics, Archimedes Principle and Buoyancy, page 184
- Equation
Buoyant force = weight of displaced fluid; γf·Vsub = γb·V- Substitute
Floating: buoyant force = weight, so γf·A·y = γb·A·Hy = H·SGb/SGf = 3.6 m × 0.55/1.0 = 1.98 m
- Result
- 1.98 m, 3 significant figures
- Check
- the block floats because SG 0.55 < 1.0; the part above the water is 1.62 m.
- Why the others are wrong
- A: divided the height by the block's specific gravity instead of multiplying
- B: assumed the block floats half submerged
- D: is the height above the water line, not the submerged depth
Problem 8 · FE, Fluid statics
A vertical rectangular gate 9 ft wide and 10 ft high holds back water, with its top edge 1 ft below the water surface. Use γ = 62.4 lbf/ft³ for water. The resultant water force on the gate is most nearly:
Handbook: Fluid Mechanics, Forces on Submerged Surfaces, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (33,700 lbf)
- Given
- d = 1 ft, H = 10 ft, b = 9 ft, γw = 62.4 lbf/ft³
- Find
- resultant force F (lbf)
- Handbook
- Fluid Mechanics, Forces on Submerged Surfaces and the Center of Pressure, page 184
- Equation
F = PC·A = γ·hc·A (pressure at the centroid times the area)- Substitute
Depth to the centroid: hc = d + H/2 = 1 + 5 = 6.000 ftF = γ hc A = (62.4 lbf/ft³)(6.000 ft)(9 ft × 10 ft) = 33,700 lbf
- Result
- 33,700 lbf, 3 significant figures
- Check
- F is the pressure at the centroid, γhc = 374.4 lbf/ft², times the area 90.00 ft².
- Why the others are wrong
- A: used half the gate area
- C: used the depth of the top edge instead of the centroid
- D: left out the gate width
Problem 9 · FE, Fluid statics
A mercury manometer (SG = 13.6) connects two points in a water pipeline, and the mercury levels differ by 50 mm. Use γ = 9.81 kN/m³ for water. The pressure difference between the two points is most nearly:
Handbook: Fluid Mechanics, Manometers, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (6.18 kPa)
- Given
- h = 50 mm, sg = 13.6, γw = 9.81 kN/m³, si =
- Find
- pressure difference ΔP (kPa)
- Handbook
- Fluid Mechanics, Manometers, page 183
- Equation
P1 - P2 = (γm - γw)h for a mercury-water manometer- Substitute
h = 50 mm = 0.050 mΔP = (γm - γw)h = (13.6 - 1)(9.81 kN/m³)(0.050 m) = 6.18 kPa
- Result
- 6.18 kPa, 3 significant figures
- Check
- the water in both legs above the mercury cancels except for the height h, so only (γm - γw) acts.
- Why the others are wrong
- B: added the water column instead of subtracting it
- C: used γm·h and ignored the water above the mercury
- D: used the unit weight of water alone
Problem 10 · FE, Fluid statics
A vertical rectangular gate 1.3 m wide and 3.1 m high holds back water, with its top edge 1.9 m below the water surface. Use γ = 9.81 kN/m³ for water. The resultant water force on the gate is most nearly:
Handbook: Fluid Mechanics, Forces on Submerged Surfaces, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (136 kN)
- Given
- d = 1.9 m, H = 3.1 m, b = 1.3 m, γw = 9.81 kN/m³
- Find
- resultant force F (kN)
- Handbook
- Fluid Mechanics, Forces on Submerged Surfaces and the Center of Pressure, page 184
- Equation
F = PC·A = γ·hc·A (pressure at the centroid times the area)- Substitute
Depth to the centroid: hc = d + H/2 = 1.9 + 1.55 = 3.450 mF = γ hc A = (9.81 kN/m³)(3.450 m)(1.3 m × 3.1 m) = 136 kN
- Result
- 136 kN, 3 significant figures
- Check
- F is the pressure at the centroid, γhc = 33.84 kN/m², times the area 4.030 m².
- Why the others are wrong
- A: used the depth of the top edge instead of the centroid
- C: left out the gate width
- D: used H/2 as the depth, ignoring the water above the gate
A new problem posts every day inside the lab, with the full worked solution the same evening.
Frequently asked questions
Where is fluid statics in the FE Reference Handbook?
Look in the Fluid Mechanics, Manometers part of FE Reference Handbook 10.6. Each solution gives the exact page, so you can practice finding it in the PDF the way you will on exam day.
Are these real FE exam questions?
No. They are original problems generated from the handbook formulas and checked by code. Real exam questions are confidential, and sharing them breaks the NCEES agreement every examinee accepts.
How do I check my answer before opening the solution?
Check the units of your result and whether its size makes sense for the situation. Each solution ends with the same kind of check, so you can compare your habit with ours.
Sources
- NCEES FE Civil CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Environmental CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Mechanical CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.