Fluid Mechanics MCQ Solutions with Detailed Explanations

For the best learning experience, attempt all questions before checking the answers. Reviewing the detailed solutions afterward will sharpen your concepts and ensure complete clarity on Fluid Mechanics.

Q1. A piece of metal sinks in water but floats in mercury. Which statement about its density is correct?

(A) Its density is less than that of water

(B) Its density is greater than that of mercury

(C) Its density lies between the densities of water and mercury

(D) It has the same density as mercury

Answer: (C) An object that sinks in water but floats in mercury must have a density greater than that of water but less than that of mercury.

Q2. A difference in pressure of 1 Pa corresponds to what difference in height in a water column?

(A) Approximately 1 mm

(B) Approximately 1 cm

(C) Approximately 10 cm

(D) Approximately 0.1 mm

Answer: (D) P = ρh g. Therefore a pressure difference of 1 Pa corresponds to a head of h = P/(ρ g) ≈ 1/(1000 × 9.8) ≈ 1×10⁻⁴ m, or roughly 0.1 mm.

Q3. When comparing two points on different streamlines, Bernoulli’s equation:

(A) Can still be applied because pressure is the same throughout

(B) Cannot be applied unless the flow is irrotational and the points lie on the same streamline

(C) Is always valid regardless of the streamline

(D) Requires the fluid to be compressible

Answer: (B) Bernoulli’s equation relates energy along a streamline. If two points are on different streamlines, the equation only applies if the flow is irrotational (so that the energy constant is the same across streamlines)

What is a streamline - Imagine a flowing river. A streamline is an imaginary line drawn in the water that shows the exact direction a tiny particle is moving at a single instant. The water flows along these lines, not across them. If the flow is smooth and steady, these lines stay in the same place, and they represent the actual paths the water particles follow. At any given point on a streamline, the velocity vector of the fluid particle at that point is perfectly tangent to the streamline's curve.

The reason streamlines are so important is that Bernoulli's equation is a statement about energy conservation that applies specifically along one of these streamlines. It tells us that if you pick any two points on the same streamline, the sum of the pressure energy, kinetic energy, and potential energy (P + ½ρv² + ρgh) will be constant between those two points. You are essentially tracking the energy of a fluid particle as it moves along its designated path.

Q4. A stagnation point on a body in a flow is where the local velocity goes to zero. At a stagnation point the static pressure is:

(A) Zero

(B) Cannot be defined

(C) Higher than far upstream

(D) Equal to far-upstream static pressure

Answer: (C) At a stagnation point, the fluid velocity decreases all the way to zero, which means that the kinetic energy per unit volume of the moving fluid (represented by the dynamic pressure term ½ρv²) is completely converted into additional static pressure.

In other words, the flow’s energy that was originally carried as motion is redirected into pressure at that point. This makes the static pressure at a stagnation point higher than the static pressure far upstream, where some of the energy is still in the form of fluid motion. This is why stagnation pressure is defined as the sum of static and dynamic pressures.

Q5. A container of fluid is accelerated downward with acceleration g (free fall). What happens to the pressure variation with depth within the fluid?

(A) The pressure still increases linearly with depth

(B) The pressure is uniform throughout the fluid

(C) The pressure decreases with depth

(D) The fluid becomes incompressible

Answer: (B) The pressure variation with depth in a fluid that is accelerating vertically is given by the formula:

P = P₀ + ρ(g − a)h

Where:

P is the pressure at depth h.

P₀ is the pressure at the surface.

ρ (rho) is the density of the fluid.

g is the acceleration due to gravity.

a is the downward acceleration of the container.

h is the depth.

In this specific case, the container is in free fall, which means its downward acceleration is equal to the acceleration due to gravity (so, a = g).

Substituting a = g into the equation:

P = P₀ + ρ(g − g)h P = P₀ + ρ(0)h P = P₀

Q6. A U-tube manometer is connected to a gas container. The other end is open to the atmosphere. The mercury in the open arm is 15 cm higher than in the arm connected to the gas. What is the gauge pressure of the gas? (Use ρHg = 13600 kg/m³ and g = 9.8 m/s²).

(A) 1.99 × 10⁴ Pa

(B) 1.33 × 10⁵ Pa

(C) 2.04 × 10³ Pa

(D) 1.47 × 10³ Pa

Answer: (A) 1.99 × 10⁴ Pa The gauge pressure of the gas is equal to the pressure exerted by the excess height of the mercury column. The height difference is h = 15 cm = 0.15 m. The gauge pressure is Pᵍ = ρ g h = (13600 kg/m³)(9.8 m/s²)(0.15 m) ≈ 1.9992 × 10⁴ Pa ≈ 1.99 × 10⁴ Pa.

Q7. A force exerted by a static fluid on a surface is always directed:

(A) Parallel to the surface

(B) Perpendicular to the surface

(C) In the direction of gravity

(D) Opposite to the direction of gravity

Answer: (B) In a static fluid, there can be no shear forces. If there were a component of force parallel to the surface, the fluid would flow in response to that force, contradicting the condition that it is static. Therefore, the force must be entirely perpendicular (normal) to the surface.

Q8. In an ideal hydraulic system, a force F₁ is applied to an input piston of area A₁, causing it to move down a distance d₁. The output piston of area A₂ moves up a distance d₂. Which of the following statements about the work done is correct?

(A) The work done on the input piston is greater than the work done by the output piston

(B) The work done by the output piston is greater than the work done on the input piston

(C) The work done on the input piston is equal to the work done by the output piston

(D) The work done is independent of the distances moved

Answer: (C) A hydraulic lift multiplies force, but it does not create energy. In an ideal system (no friction), the work input must equal the work output due to the conservation of energy. Work is force times distance, so W_in = F₁ d₁ and W_out = F₂ d₂. Therefore, F₁ d₁ = F₂ d₂. Since F₂ > F₁, it must be that d₂ < d₁. The small input force is applied over a large distance, while the large output force acts over a small distance.

Q9. Fluid flows from one location to another primarily because:

(A) The densities at the two locations are equal

(B) A difference in pressure exists between the locations

(C) Gravity always pulls the fluid downhill

(D) The viscosity is high

Answer: (B) For a fluid to flow between two locations, there must be a pressure difference. Fluids accelerate from high pressure toward lower pressure.

Q10. For an incompressible fluid flowing in a pipe, if the cross-sectional area increases, what happens to the flow speed?

(A) It increases proportionally

(B) It decreases proportionally

(C) It remains the same

(D) It becomes zero

Answer: (B) According to the continuity equation (A v = constant for incompressible flow), volumetric flow rate must remain the same along the pipe. This means that when cross-sectional area increases, the velocity must decrease proportionally to keep the product A × v constant. Therefore, if the area increases the speed decreases

Q11. Which statement correctly relates absolute pressure (Pₐ), surface (atmospheric) pressure (P₀) and gauge pressure (Pᵍ) at some depth in a fluid?

(A) Pₐ = P₀ − Pᵍ

(B) Pₐ = P₀ + Pᵍ

(C) P₀ = Pₐ + Pᵍ

(D) Pᵍ = P₀ + Pₐ

Answer: (B) Absolute pressure at a point in a fluid equals the pressure at the surface (atmospheric pressure) plus the gauge pressure due to the weight of fluid above. Gauge pressure measures how much additional pressure the fluid column contributes; adding it to the surface pressure gives the total (absolute) pressure.

Q12. Why does gauge pressure in a fluid increase with depth?

(A) Because the fluid is compressible

(B) Because the weight of the vertical column of fluid above adds to the pressure

(C) Because cross-sectional area increases with depth

(D) Because absolute pressure decreases with depth

Answer: (B) Gauge pressure increases with depth because the weight of the overlying column of fluid adds force per unit area. Hydrostatic pressure depends on fluid density, gravitational acceleration and depth and is independent of container shape.

Q13. Which of the following is independent of the shape of the container: the pressure at a given depth, the total weight of the fluid or the volume of fluid displaced?

(A) Only the pressure at a given depth is independent of shape

(B) Only the weight of the fluid is independent of shape

(C) Both pressure at depth and weight are independent of shape

(D) Pressure at a given depth depends on the container’s shape

Answer: (A) Hydrostatic pressure at a given depth depends only on depth, fluid density and g. The total weight of the fluid and the volume displaced depend on the container’s geometry, but pressure at a particular depth is the same regardless of shape.

Q14. According to hydrostatic principles, if you descend twice as deep below the surface of a liquid of constant density, what happens to the gauge pressure?

(A) It halves

(B) It doubles

(C) It quadruples

(D) It remains constant

Answer: (B) For a fluid of constant density, gauge pressure is proportional to depth (P = ρ g h). Doubling the depth doubles the pressure.

Q15. A swimming pool is 5 m deep. Taking the density of water as 1 000 kg/m³ and g = 9.8 m/s², estimate the gauge pressure at the bottom of the pool.

(A) 4.9 kPa

(B) 49 kPa

(C) 490 kPa

(D) 9.8 kPa

Answer: (B) Gauge pressure at 5 m depth in water is ρ g h = (1 000 kg/m³)(9.8 m/s²)(5 m) ≈ 49 000 Pa = 49 kPa.

Q16. When a dam holds back water to a depth h, the average pressure on the dam face equals ρ g h̄, where h̄ is the average depth. If a reservoir is 80 m deep at a dam, what average pressure does the water exert?

(A) 39.2 kPa

(B) 392 kPa

(C) 3.92 MPa

(D) 3.92 kPa

Answer: (B) The average hydrostatic pressure on a dam of depth h is calculated at the midpoint depth (h̄ = h/2). For h = 80 m, h̄ = 40 m, so P = ρ g h̄ = 1 000 × 9.8 × 40 ≈ 3.92 × 10⁵ Pa = 392 kPa.

Q17. A hot-air balloon of volume 1200 m³ is filled with helium (ρ = 0.18 kg/m³). Take weight of helium as 2.1 kN. Outside air density is 1.30 kg/m³. The envelope plus basket has a total mass of 80 kg. What is the maximum payload?

A) 0.8 kN

B) 8.0 kN

C) 12.4 kN

D) 13.1 kN

Answer: C) Buoyant force (weight of displaced air):

F_b = ρ_air × V × g = (1.30)(1200)(9.8) ≈ 15.3 kN

F_net = F_b − W_He = 13.1 kN

Weight of envelope + basket:

W_env = 80 × 9.8 ≈ 0.8 kN

Maximum payload:

F_payload = F_net − W_env ≈ 13.1 − 0.8 = 12.3 kN ≈ 12.4 kN

Q18. A Venturi tube has a narrow section where water flows at 8 m/s and the pressure is 180 kPa. In the wide section, the speed is 4 m/s. (ρ = 1000 kg/m³). Find the pressure in the wide section.

A) 156 kPa

B) 180 kPa

C) 204 kPa

D) 232 kPa

Answer: C) Apply Bernoulli between narrow (1) and wide (2) sections with negligible height difference:

P₁ + ½ρ v₁² = P₂ + ½ρ v₂²

Rearrange:

P₂ − P₁ = ½ρ (v₁² − v₂²)

Substitute:

P₂ = 180 kPa + 0.5 × 1000 × (64 − 16)

P₂ = 180 kPa + 24 kPa = 204 kPa

Q19. A fluid flows up a tube to a point 5 m higher while speed drops from 5 m/s to 1 m/s. Which statement is correct about the pressure difference?

A) Pressure 49 kPa greater at higher point

B) Pressure 37 kPa lower at higher point

C) No pressure difference

D) Depends only on velocity

Answer: B) Apply Bernoulli between lower point (1) and higher point (2):

P₁ + ½ρ v₁² + ρ g z₁ = P₂ + ½ρ v₂² + ρ g z₂

Rearrange:

P₂ − P₁ = ½ρ (v₁² − v₂²) − ρ g (z₂ − z₁)

Insert: ρ = 1000, v₁ = 5, v₂ = 1, z₂ − z₁ = 5.

Kinetic term = 12,000 Pa (raises P)

Hydrostatic term = 49,000 Pa (lowers P)

Net: P₂ − P₁ = 12 kPa − 49 kPa = −37 kPa

Therefore, the pressure at the higher point is about 37 kPa lower.

Q20. A tank has water level 20 m above a small hole at its base. What is the exit speed of the jet?

A) 9.9 m/s

B) 14 m/s

C) 19.8 m/s

D) 28 m/s

Answer: C) Torricelli’s theorem: v = √(2 g h) = √(2 × 9.8 × 20) ≈ 19.8 m/s.

Q21. A tank has a hole 4 m above the ground. Water level is 8 m above the hole. Where does the water jet land from the base?

A) 2 m

B) 5 m

C) 8 m

D) 11 m

Answer: D) Exit speed from the hole: v = √(2 g h) = √(2 × 9.8 × 8) ≈ 12.5 m/s.

Vertical drop from hole to ground: y = 4 m ⇒ time to fall t = √(2y/g) ≈ 0.90 s.

Horizontal range: R = v × t ≈ 12.5 × 0.90 ≈ 11.3 m ≈ 11 m

Q22. A wooden block of density 600 kg/m³ and volume 1.0 m³ is tethered completely submerged in water. What is the tension in the rope? (g = 9.8 m/s², upward positive)

A) −3.9 kN

B) 3.9 kN

C) 5.9 kN

D) 9.8 kN

Answer: A) Buoyant force: F_b = ρ_water × V × g = 1000 × 1 × 9.8 = 9.8 kN (upward).

Weight of the block: W = ρ_block × V × g = 600 × 1 × 9.8 = 5.9 kN (downward).

Static equilibrium: F_b − W − T = 0

T = F_B − W = 9.8 − 5.9 = 3.9 kN

Magnitude of T = 3.9 kN

With sign T = −3.9 kN (magnitude 3.9 kN downward).

Q23. The same block is submerged in oil (ρ = 800 kg/m³). What happens to the tension?

A) Tension smaller than in water

B) Tension larger than in water

C) Tension zero

D) Tension same

Answer: A) Buoyant force in oil = ρ_oil × V × g = 800 × 9.8 = 7.8 kN.

Weight = 5.9 kN.

Net upward = 1.9 kN → rope tension (downward) = 1.9 kN, smaller than in water.

Q24. A student measures density of an unknown liquid by immersing a cube (m = 0.50 kg, V = 5.0 × 10⁻⁵ m³). Apparent weight in liquid = 3.9 N, weight in air = 4.9 N. What is the buoyant force?

A) 1.0 N

B) 3.9 N

C) 4.9 N

D) 8.8 N

Answer: A) Buoyant force = W_air − W_liquid = 4.9 − 3.9 = 1.0 N.

Q25. A metal block of mass 5 kg and volume 5 × 10⁻⁴ m³ is placed in mercury (ρ = 13,600 kg/m³). What will happen to the block?

A) It sinks completely

B) It floats

C) It remains neutrally buoyant

D) No buoyant force acts

Answer: B) Density of block = 5 / (5 × 10⁻⁴) = 10,000 kg/m³.

Since ρ_block < ρ_Hg, the block floats, displacing less volume than its total.

Q26. A pipe narrows from radius 10 cm to radius 5 cm. If the water speed in the wide section is 2 m/s, what is the speed in the narrow section?

A) 2 m/s

B) 4 m/s

C) 8 m/s

D) 16 m/s

Answer: C) Continuity equation gives: A₁v₁ = A₂v₂.

Area ratio = (10²)/(5²) = 4.

v₂ = 4 × 2 = 8 m/s.

Q27. In a pipe, if the diameter is halved, how does the flow speed change (for incompressible steady flow)?

A) Halved

B) Doubled

C) Quadrupled

D) Remains unchanged

Answer: C) From continuity equation : A₁v₁ = A₂v₂.

Area reduces by (½)² = ¼.

To conserve flow, velocity ×4.

Q28. A pipe splits into two branches with area ratio 1:2. If velocity in the wider branch is 2 m/s, what is the velocity in the narrow branch? Assume flow rate is the same in both the branches

A) 1 m/s

B) 2 m/s

C) 3 m/s

D) 4 m/s

Answer: D) Q = A * v

Let's denote the narrow branch with subscript 1 and the wider branch with subscript 2.

Q1 = Q2

A1 * v1 = A2 * v2

Here's the information we're given:

The area ratio is 1:2, so we can say A1 = A and A2 = 2A.

The velocity in the wider branch (v2) is 2 m/s.

Now, we can substitute these values into our equation:

A * v1 = (2A) * 2 m/s

We can cancel out the area (A) from both sides of the equation:

v1 = 2 * 2 m/s

v1 = 4 m/s

Therefore, the velocity in the narrow branch is 4 m/s. This makes intuitive sense: for the same amount of fluid to pass through a smaller area, it must move faster.

Q29. A pump raises water to a height of 20 m. What is the minimum gauge pressure the pump must provide? (ρ = 1000 kg/m³, g = 9.8 m/s²)

A) 19.6 kPa

B) 196 kPa

C) 980 kPa

D) 9.8 kPa

Answer: B) ΔP = ρ g h = 1000 × 9.8 × 20 = 196,000 Pa = 196 kPa.

Q30. A water tower is 50 m tall. What is the gauge pressure at its base?

A) 50 kPa

B) 100 kPa

C) 490 kPa

D) 4.9 MPa

Answer: C) P = ρ g h = 1000 × 9.8 × 50 = 4.9 × 10⁵ Pa = 490 kPa.

Q31. A building has floors 3 m apart. By how much does the hydrostatic pressure drop per floor?

A) 0.29 kPa

B) 2.9 kPa

C) 29 kPa

D) 290 kPa

Answer: C) ΔP = ρ g h = 1000 × 9.8 × 3 = 29,400 Pa ≈ 29 kPa.

Q32. A rock has mass 5 kg in air. When submerged in water, its apparent mass is 3 kg. What is the buoyant force?

A) 2 N

B) 20 N

C) 30 N

D) 50 N

Answer: B) The buoyant force is equal to the weight of the "lost" mass.

Weight in air = 5 × 9.8 = 49 N.

Apparent weight = 3 × 9.8 = 29 N.

Buoyant force = 49 − 29 = 20 N.

It is crucial to understand that mass is never truly lost. A scale measures force, not mass directly. The 'apparent mass' is lower because the upward buoyant force from the water supports part of the rock's weight, reducing the net downward force the scale measures.

Q33. From the above experiment, what is the volume of the rock? (ρ_water = 1000 kg/m³)

A) 2 × 10⁻⁴ m³

B) 3 × 10⁻⁴ m³

C) 2 × 10⁻³ m³

D) 5 × 10⁻³ m³

Answer: C) V = F_b / (ρ g) = 20 / (1000 × 9.8) = 2 × 10⁻³ m³

Q34. In a horizontal flow where elevation differences are negligible, Bernoulli’s principle says that if the fluid speed increases:

(A) The fluid pressure increases

(B) The fluid pressure decreases

(C) The fluid density decreases

(D) There is no relationship between speed and pressure

Answer: (B) Bernoulli’s principle states that in a horizontal streamline (no change in elevation), an increase in flow speed results in a decrease in pressure. This inverse relationship arises because kinetic energy per unit mass increases at the expense of pressure energy.

Q35. For an ideal fluid exiting a small hole, the gauge pressure at the hole is equal to:

(A) Atmospheric pressure

(B) ρ g h (where h is the depth below the surface)

(C) Zero

(D) Twice the atmospheric pressure

Answer: (C) The question asks for the gauge pressure at the hole. For an ideal fluid jet exiting into the atmosphere, the pressure of the fluid streamlines becomes equal to the surrounding atmospheric pressure. Therefore, the gauge pressure of the water just as it exits is zero. The gauge pressure inside the tank at the depth of the hole is ρgh.

Q36. A scuba diver’s compressed air gauge reads 400 kPa. At what approximate depth in seawater (density 1 030 kg/m³) does the external absolute pressure equal this? Assume surface pressure is 101 kPa.

(A) 10 m

(B) 30 m

(C) 40 m

(D) 50 m

Answer: (B) The diver’s gauge reads absolute pressure. Subtracting atmospheric pressure gives a gauge pressure of 299 kPa.

P_g = ρ gh or h = P_g / ρ g

with ρ = 1 030 kg/m³ yields a depth near 30 m.

Q37. A block of wood of density 600 kg/m³ and volume 1.0 m³ is completely submerged in water and tethered by a rope. What is the tension in the rope (upward positive)?

(A) −3.92 kN (the rope must pull downward)

(B) 3.92 kN (the rope must pull downward)

(C) 5.88 kN (the rope must pull upward)

(D) 9.80 kN (the rope must pull downward)

Answer: (A) Weight of wood (downward): W = ρ_block × V × g = 600 × 1.0 × 9.8 = 5.88 kN.

Buoyant force (upward): F_b = ρ_water × V × g = 1000 × 1.0 × 9.8 = 9.80 kN.

Force balance on the block (upward positive): ΣF = F_b − W − T = 0

⇒ T = F_b − W = 9.80 − 5.88 = 3.92 kN downward.

Sign convention: With upward positive defined, the signed tension on the block is T = −3.92 kN (magnitude 3.92 kN downward).

Q38. A 2 kg hollow ball displaces 0.003 m³ of water when completely submerged. Will it float or sink when released?

(A) It will sink because its weight exceeds the buoyant force

(B) It will float because the buoyant force exceeds its weight

(C) It will remain neutrally buoyant

(D) It will oscillate up and down

Answer: (B) The ball displaces 0.003 m³ of water, giving a buoyant force of 29.4 N. Its weight is 19.6 N. Because the buoyant force exceeds the weight, the ball floats.

Q39. A block of wood (ρ = 700 kg/m³) is held completely submerged under water by a string attached to the bottom of a tank. The tank is placed in an elevator that is accelerating upwards at 2.0 m/s². How does the tension in the string change compared to when the elevator is at rest?

A) The tension increases

B) The tension decreases

C) The tension remains the same

D) The tension becomes zero

Answer: A) Force Analysis (Ground/Inertial Frame)

The block experiences three forces:

Buoyant force (Fᴮ), acting upward

Weight of the block (W), acting downward

Tension from the string (T), acting downward to keep the block submerged

Since the block must accelerate upward with the elevator, the net force is:

ΣF = ma

So,

Fᴮ − W − T = m·a

Buoyant Force

In an upward-accelerating elevator, the effective gravity is:

gₑff = g + a

Fᴮ = ρ_water × V × (g + a)

Weight of the Block

The block’s weight does not depend on the elevator’s acceleration. It is:

W = m·g = ρ_block × V × g

Substitute into Net Force Equation

ρ_water × V × (g + a) − ρ_block × V × g − T = (ρ_block × V) × a

Solve for Tension, T

T = ρ_water × V × (g + a) − ρ_block × V × g − ρ_block × V × a

T = (ρ_water − ρ_block) × V × (g + a)

When a = 0,

T_rest = (ρ_water − ρ_block) × V × g

Conclusion

Since the new tension depends on (g + a), and a > 0 when the elevator accelerates upward, the tension increases compared to when the elevator is at rest.

Q40. Water flows from a pipe at ground level with a speed of 5 m/s and a gauge pressure of 3.0 × 10⁵ Pa. The pipe leads to a faucet in a skyscraper 50 m above the ground. Assuming constant pipe radius, what is the gauge pressure at the faucet? (ρ = 1000 kg/m³, g = 9.8 m/s²)

A) −2.0 × 10⁵ Pa

B) −1.9 × 10⁵ Pa

C) 1.0 × 10⁵ Pa

D) 8.0 × 10⁵ Pa

Answer: B) Apply Bernoulli along the same streamline from ground pipe (1) to faucet (2):

P₁ + ½ρ v₁² + ρ g z₁ = P₂ + ½ρ v₂² + ρ g z₂

Constant radius ⇒ v₁ = v₂ = 5 m/s, so the kinetic terms cancel.

Elevation change: z₂ − z₁ = +50 m.

Rearranging (using gauge pressures so P₀ cancels):

P₂ = P₁ − ρ g (z₂ − z₁) = P₁ − ρ g Δh

Substitute values:

P₂ = 3.0 × 10⁵ − (1000 × 9.8 × 50)

P₂ = 3.0 × 10⁵ − 4.9 × 10⁵

P₂ = −1.9 × 10⁵ Pa (gauge).

Interpretation: Negative gauge pressure indicates suction relative to atmosphere at the faucet location. It means the absolute pressure at the faucet is below atmospheric pressure. If there were a leak at the faucet, air would be sucked into the pipe rather than water leaking out

Q41. An iceberg floats in the ocean. Given the density of ice is approximately 920 kg/m³ and the density of seawater is approximately 1025 kg/m³, what percentage of the iceberg's volume is submerged?

(A) About 10%

(B) About 50%

(C) About 90%

(D) 100%

Answer: (C) For a floating object, the buoyant force equals its weight. This leads to the relationship:

% submerged = (ρ_object / ρ_fluid) × 100%

which gives the

Here, (920 ÷ 1025) × 100% ≈ 89.7%

So, about 90% of the iceberg’s volume is underwater.

Useful Input: This is the origin of the expression “the tip of the iceberg.” The vast majority of an iceberg’s mass and volume is hidden beneath the surface of the water, posing a significant hazard to shipping.

Q42. A submarine is neutrally buoyant at a certain depth in the ocean. If it expels some water from its ballast tanks, what will happen? (A) It will begin to sink. (B) It will begin to rise. (C) It will remain at the same depth. (D) It will tilt forward.

Answer: (B) Being neutrally buoyant means the submarine's total weight equals the buoyant force. By expelling water, the submarine reduces its total mass and weight. The buoyant force (which dep

Q43. A crown with a weight of 10 N in air has an apparent weight of 9 N when fully submerged in water. What is the density of the crown? (Use ρ_water = 1000 kg/m³, g ≈ 10 m/s²)

(A) 1000 kg/m³

(B) 5000 kg/m³

(C) 9000 kg/m³

(D) 10000 kg/m³

Answer: (D) The buoyant force is the difference in weights:

F_b = 10 N − 9 N = 1 N

Volume of the crown:

V = F_b ÷ (ρ_water × g)

V = 1 × 10⁻⁴ m³

Mass of the crown:

m = W ÷ g = 10 ÷ 10 = 1 kg

Density of the crown:

ρ = m ÷ V = 1 ÷ (1 × 10⁻⁴) = 10000 kg/m³

So, the density of the crown is 10000 kg/m³.

This is the legendary method supposedly used by Archimedes to determine if the king's crown was made of pure gold without damaging it. By finding its density, he could compare it to the known density of gold.

Q44. An object is floating in a container of pure water. If a large amount of salt is dissolved into the water, what will happen to the object? (A) It will sink to the bottom. (B) It will remain at the same level. (C) It will float higher in the water. (D) It will float lower in the water.

Answer: (C) Dissolving salt in water increases the water's density. According to Archimedes' principle, a floating object displaces a weight of fluid equal to its own weight. Since the fluid is now denser, a smaller volume of it is needed to equal the object's weight. Therefore, the object will float higher.

This is why it is much easier to float in the ocean than in a freshwater swimming pool. The high salt content of seawater, especially in places like the Dead Sea, makes it very dense and highly buoyant.