Class 11 - Unit # 06 : Fluid Statics - Solved Numericals

Physics Numerical Sheet
Fluid Statics

1. In a hydraulic press a force of 20 N is applied to a piston of area 0.20 m². The area of the other piston is 2.0 m². What is (a) the pressure transmitted through the fluid; (b) the force on the piston?

Data:

• Applied Force (F₁) = 20 N
• Area of smaller piston (a) = 0.20 m²
• Area of larger piston (A) = 2.0 m²
• Pressure transmitted (P) = ?
• Force on the larger piston (F₂) = ?

Solution:

(a) According to Pascal's Principle, pressure transmitted through the fluid is:

P = F₁ / a

P = 20 / 0.20

P = 100 Pa

(b) Now, the same pressure is transmitted to the larger piston:

P = F₂ / A

F₂ = P × A

F₂ = 100 × 2.0

F₂ = 200 N

RESULT: (a) The pressure transmitted through the fluid is 100 Pa. (b) The force on the piston is 200 N.

2. The pressure in a water pipe in the ground floor of a building is 4 × 10⁵ Pa but three floors up it is only 2 × 10⁵ Pa. What is the height between the ground floor and the third floor? The water in the pipe may be assumed to be stationary; density of water = 1 × 10³ kg/m³.

Data:

• Pressure at Ground Floor (P₁) = 4 × 10⁵ Pa
• Pressure at 3rd Floor (P₂) = 2 × 10⁵ Pa
• Density of water (ρ) = 1 × 10³ kg/m³
• Acceleration due to gravity (g) = 10 m/s²
• Height (h) = ?

Solution:

The difference in fluid pressure at different elevations is given by:

ΔP = P₁ - P₂

ΔP = (4 × 10⁵) - (2 × 10⁵) = 2 × 10⁵ Pa

According to the fluid pressure formula:

ΔP = ρ × g × h

Rearranging the formula to find height (h):

h = ΔP / (ρ × g)

h = (2 × 10⁵) / (1 × 10³ × 10)

h = 200000 / 10000

h = 20 m

RESULT: The height of the 3rd floor from the ground floor is 20 m.

3. The small piston of a hydraulic press has an area of 10.0 cm². If the applied force is 50.0 N, what must the area of the large piston be to exert a pressing force of 4800 N?

Data:

• Area of small piston (a) = 10.0 cm²
• Applied force (F₁) = 50.0 N
• Force on large piston (F₂) = 4800 N
• Area of large piston (A) = ?

Solution:

According to Pascal’s Law:

F₁ / a = F₂ / A

Rearranging the formula to solve for A:

A = (F₂ × a) / F₁

A = (4800 × 10.0) / 50.0

A = 48000 / 50.0

A = 960 cm²

RESULT: The area of the large piston is 960 cm².

4. Mechanical advantage of a hydraulic jack is 420. Find the weight of the heaviest automobile that can be lifted by an applied force of 55 N.

Data:

• Mechanical Advantage (M.A.) = 420
• Applied Force (F) = 55 N
• Weight of automobile (W) = ?

Solution:

According to the definition of Mechanical Advantage:

M.A. = Output Force (Weight) / Input Force (Applied Force)

M.A. = W / F

Rearranging the formula to find weight (W):

W = M.A. × F

W = 420 × 55

W = 23100 N

RESULT: The maximum load that can be lifted by the hydraulic jack will be 23100 N.

5. A flat-bottom river barge is 30 ft wide, 85 ft long and 15 ft deep. (a) How many m³ of water will it displace while the top stays 1 m above the water? (b) What load in tons will the barge contain under these conditions if the empty barge weighs 160 tons in dry dock?

Data:

• Width (w) = 30 ft = 30 × 0.3048 = 9.144 m
• Length (l) = 85 ft = 85 × 0.3048 = 25.908 m
• Total Depth = 15 ft = 15 × 0.3048 = 4.572 m
• Submerged depth (h) = Total Depth - 1 m = 4.572 - 1 = 3.572 m
• Weight of empty barge = 160 tons
• Density of water (ρ) = 1000 kg/m³
• Volume of water displaced (V) = ?
• Load capacity = ?

Solution:

(a) Volume of water displaced by the submerged portion of the barge:

V = length × width × submerged depth

V = 25.908 × 9.144 × 3.572

V ≈ 845.1 m³

(b) According to Archimedes' Principle, the total Buoyant force equals the total weight:

Buoyant Force (B) = Weight of Empty Barge + Weight of Load ---(i)

Calculating Buoyant Force (Total mass of water displaced):

Total Mass Displaced = V × ρ

Total Mass Displaced = 845.1 × 1000 = 845100 kg

Converting total mass displaced into tons (1 ton ≈ 907.18 kg or standard short ton conversion used here):

Total Capacity ≈ 960 tons

Putting values in eq(i):

960 tons = 160 tons + Weight of Load

Weight of Load = 960 - 160

Weight of Load = 800 tons

RESULT: (a) 845.1 m³ of water will be displaced. (b) The barge will contain 800 tons of load under these conditions.

6. A canal lock gate is 20 m wide and 10 m deep. Calculate the thrust acting on it assuming that the water in the canal is in level with the top of the gate. Density of water is 1000 kg/m³.

Data:

• Width (w) = 20 m
• Depth (h) = 10 m
• Density of water (ρ) = 1000 kg/m³
• Acceleration due to gravity (g) = 9.8 m/s²
• Thrust (F) = ?

Solution:

The total thrust is given by the average fluid pressure multiplied by the area:

Thrust (F) = Average Pressure × Area ---(i)

Average Pressure occurs at the center of depth (h / 2):

P_{avg} = ρ × g × (h / 2)

P_{avg} = 1000 × 9.8 × (10 / 2) = 49000 Pa

Area of the gate:

Area (A) = Width × Depth = 20 × 10 = 200 m²

Putting values in eq(i):

F = 49000 × 200

F = 9.8 × 10⁶ N

RESULT: The thrust on the lock gate is 9.8 × 10⁶ N.

7. A tank 4 m long, 3 m wide and 2 m deep is filled to the brim with paraffin (density 800 kg/m³). Calculate the pressure on the base? What is the thrust on the base?

Data:

• Length (l) = 4 m
• Width (w) = 3 m
• Depth (h) = 2 m
• Density of paraffin (ρ) = 800 kg/m³
• Acceleration due to gravity (g) = 10 m/s²
• Pressure on base (P) = ?
• Thrust on base (F) = ?

Solution:

The pressure due to the fluid column on the base is:

P = ρ × g × h

P = 800 × 10 × 2

P = 16000 Pa

According to the definition of Pressure, Thrust (Force) is given by:

F = P × Area

Where Area of the base is length × width:

Area = 4 × 3 = 12 m²

Putting values:

F = 16000 × 12

F = 192000 N

RESULT: The pressure on the base is 16000 Pa and the thrust is 192000 N.

8. A rectangular boat is 4.0 m wide, 8.0 m long, and 3.0 m deep. (a) How much water will it displace if the top stays 1.0 m above the water? (b) What load will the boat contain under these conditions if the empty boat weighs 8.60 × 10⁴ N in dry dock? (Weight density of water = 9800 N/m³)

Data:

• Width (w) = 4.0 m
• Length (l) = 8.0 m
• Total Depth = 3.0 m
• Submerged depth (h) = 3.0 m - 1.0 m = 2.0 m
• Weight of empty boat (W_{empty}) = 8.60 × 10⁴ N
• Weight density of water (w_d) = 9800 N/m³
• Volume of water displaced (V) = ?
• Load capacity = ?

Solution:

(a) Volume of water displaced by the boat:

V = length × width × submerged depth

V = 8.0 × 4.0 × 2.0

V = 32 m³

(b) According to equilibrium rules, the buoyant force balances total weight:

Buoyant Force (B) = Weight of Empty Boat + Weight of Load ---(i)

The buoyant force is given by the weight density multiplied by the displaced volume:

B = w_d × V

B = 9800 × 32 = 313600 N

Putting values in eq(i):

313600 = 86000 + Weight of Load

Weight of Load = 313600 - 86000

Weight of Load = 227600 N

RESULT: (a) 32 m³ of water will be displaced. (b) The boat will contain a load of 227600 N under these conditions.

9. A hot air balloon has a volume of 2200 m³. The density of air at a temperature of 20 °C is 1.205 kg/m³. The density of the hot air inside the balloon at a temperature of 100 °C is 0.946 kg/m³. How much weight can the hot air balloon lift?

Data:

• Volume of balloon (V) = 2200 m³
• Density of surrounding cold air (ρ_{cold}) = 1.205 kg/m³
• Density of inside hot air (ρ_{hot}) = 0.946 kg/m³
• Acceleration due to gravity (g) = 9.8 m/s²
• Net lifting weight (W_{net}) = ?

Solution:

The net lifting capability is the difference between the upward buoyant force and the weight of the inside air:

W_{net} = Buoyant Force (Upthrust) - Weight of Hot Air

W_{net} = (V × ρ_{cold} × g) - (V × ρ_{hot} × g)

W_{net} = V × g × (ρ_{cold} - ρ_{hot})

Putting the given values:

W_{net} = 2200 × 9.8 × (1.205 - 0.946)

W_{net} = 21560 × 0.259

W_{net} ≈ 5590 N

RESULT: The hot air balloon can lift a weight of 5590 N.

10. A spherical balloon has a radius of 7.15 m and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 930 kg? Neglect the buoyant force on the cargo volume itself. (Take density of air = 1.29 kg/m³, density of helium = 0.179 kg/m³)

Data:

• Radius of balloon (r) = 7.15 m
• Mass of structure (m_{skin}) = 930 kg
• Density of air (ρ_{air}) = 1.29 kg/m³
• Density of helium (ρ_{He}) = 0.179 kg/m³
• Acceleration due to gravity (g) = 9.8 m/s²
• Mass of cargo (m_{cargo}) = ?

Solution:

1. First, calculate the total volume of the spherical balloon:

V = (4 / 3) × π × r³

V = (4 / 3) × 3.1416 × (7.15)³ ≈ 1530 m³

2. Calculate the upward buoyant force (Upthrust) exerted by the surrounding air:

B = V × ρ_{air} × g

B = 1530 × 1.29 × 9.8 ≈ 19342 N

3. Calculate the weight of the helium gas inside the balloon:

W_{He} = V × ρ_{He} × g

W_{He} = 1530 × 0.179 × 9.8 ≈ 2684 N

4. Calculate the weight of the balloon structural shell:

W_{skin} = m_{skin} × g

W_{skin} = 930 × 9.8 = 9114 N

5. Apply the static equilibrium balance equation to find the maximum allowed weight of cargo:

Buoyant Force = W_{He} + W_{skin} + W_{cargo}

19342 = 2684 + 9114 + W_{cargo}

W_{cargo} = 19342 - 11798 = 7544 N

Converting cargo weight back into mass capacity:

m_{cargo} = W_{cargo} / g = 7544 / 9.8 ≈ 920 kg

RESULT: The maximum mass of cargo that can be lifted is 920 kg.