Fluids at Rest, Archimedes' Principle and Viscosity Practice Questions — JAMB UTME

Question 1Question →

A U-tube open at both ends contains mercury of density $13\text{}600\text{ kg/m}^3$ . Water of density $1000\text{ kg/m}^3$ is poured into one arm until the water column reaches a height of $27.2\text{ cm}$ . What is the difference in height, in $\text{cm}$ , between the mercury surfaces in the two arms?

Show answer & explanation

Answer: 2

Answer

The difference in height between the mercury surfaces in the two arms is

2.0\text{ cm}

.

At the boundary level where water meets mercury, the pressure produced by the

27.2\text{ cm}

water column must equal the pressure of the mercury column above that same horizontal level. Using

h_w \rho_w = h_m \rho_m

, we solve for the mercury height difference:

h_m = \frac{27.2 \times 1000}{13600} = 2.0\text{ cm}

.

Step-by-Step Solution

1

Equate the hydrostatic pressure exerted by the water column to the hydrostatic pressure exerted by the balancing mercury column at the interface level.

h_w \rho_w g = h_m \rho_m g

At the same horizontal level within a continuous fluid at rest, the pressures must be equal.

2

Cancel the acceleration due to gravity (

g

) from both sides of the equation.

h_w \rho_w = h_m \rho_m

Gravity acts equally on both liquid columns.

3

Substitute the known values (

h_w = 27.2\text{ cm}

,

\rho_w = 1000\text{ kg/m}^3

,

\rho_m = 13600\text{ kg/m}^3

) into the pressure relation.

27.2 \times 1000 = h_m \times 13600

Inserting the physical quantities isolates the unknown mercury column height

h_m

.

4

Solve for the height difference

h_m

of the mercury levels.

h_m = \frac{27200}{13600} = 2.0\text{ cm}

Dividing the water pressure head product by the density of mercury yields the height of the mercury column.

Key Concept

Hydrostatic pressure equilibrium in immiscible fluids (U-tube manometer)

Question 2Question →

A solid block of mass $0.5\text{ kg}$ is completely immersed in water and displaces $0.2\text{ kg}$ of water. What is the magnitude of the upthrust exerted by the water on the block? (Take $g = 10\text{ m/s}^2$ )

2\text{ N}

3\text{ N}

5\text{ N}

7\text{ N}

Show answer & explanation

Answer: $2\text{ N}$

Answer

The magnitude of the upthrust exerted on the block is

2\text{ N}

.

By Archimedes' principle, the buoyant force (upthrust) acting on a submerged object equals the weight of the liquid displaced by the object. Since the mass of displaced water is

0.2\text{ kg}

and

g = 10\text{ m/s}^2

, the weight of the displaced water is

0.2\text{ kg} \times 10\text{ m/s}^2 = 2\text{ N}

.

Step-by-Step Solution

1

Identify Archimedes' Principle

Upthrust (

U

) is equal to the weight of the fluid displaced by the immersed body.

Archimedes' principle states that the buoyant force on a submerged body equals the weight of the fluid it displaces.

2

Calculate the weight of the displaced water

W_{\text{displaced}} = m_{\text{water}} \times g = 0.2\text{ kg} \times 10\text{ m/s}^2 = 2\text{ N}

Weight is calculated as mass multiplied by acceleration due to gravity.

3

State the upthrust

U = 2\text{ N}

The upthrust is directly equal to the weight of the displaced water.

Key Concept

Archimedes' Principle and Upthrust

Estimated Time:45s

Question 3Question →

A solid object of volume $0.002\text{ m}^3$ is completely immersed in water of density $1000\text{ kg/m}^3$ . What is the magnitude of the upthrust exerted on the object by the water? [Take $g = 10\text{ m/s}^2$ ]

Show answer & explanation

Answer: 20

Answer

The magnitude of the upthrust exerted on the object is

20\text{ N}

.

According to Archimedes' principle, any body completely or partially submerged in a fluid experiences an upward force (upthrust) equal to the weight of the fluid displaced. The weight of the displaced fluid is calculated using

U = V \cdot \rho \cdot g

. Substituting

V = 0.002\text{ m}^3

,

\rho = 1000\text{ kg/m}^3

, and

g = 10\text{ m/s}^2

yields

U = 0.002 \times 1000 \times 10 = 20\text{ N}

.

Step-by-Step Solution

1

Identify the given quantities from the problem statement.

V = 0.002\text{ m}^3

,

\rho = 1000\text{ kg/m}^3

, and

g = 10\text{ m/s}^2

.

These are the essential inputs required to calculate the weight of the displaced liquid.

2

Apply Archimedes' principle to find upthrust force.

U = V \rho g = 0.002 \times 1000 \times 10 = 20\text{ N}

.

Archimedes' principle states that the upthrust force equals the weight of the fluid displaced by the object.

Key Concept

Archimedes' Principle and Upthrust

Question 4Question →

A metallic sphere weighs $5.0\text{ N}$ in air. When completely immersed in water, its apparent weight is $3.0\text{ N}$ . When completely immersed in an unknown liquid $X$ , its apparent weight is $3.4\text{ N}$ . What is the density of liquid $X$ in $\text{kg/m}^3$ ? (Take the density of water as $1000\text{ kg/m}^3$ and acceleration due to gravity $g = 10\text{ m/s}^2$ ).

Show answer & explanation

Answer: 800

Answer

The density of liquid

X

is

800\text{ kg/m}^3

.

The upthrust in water (

2.0\text{ N}

) gives the volume of the sphere as

2.0 \times 10^{-4}\text{ m}^3

. Using the upthrust in liquid X (

1.6\text{ N}

), the density of liquid X is calculated as

\rho_X = \frac{1.6}{(2.0 \times 10^{-4})(10)} = 800\text{ kg/m}^3

.

Step-by-Step Solution

1

Calculate upthrust in water

U_w = 5.0\text{ N} - 3.0\text{ N} = 2.0\text{ N}

Upthrust equals the loss in weight of the submerged body in water.

2

Determine the volume of the metallic sphere

V = \frac{U_w}{\rho_w g} = \frac{2.0}{1000 \times 10} = 2.0 \times 10^{-4}\text{ m}^3

According to Archimedes' principle, upthrust in water equals the weight of displaced water.

3

Calculate upthrust in liquid X

U_X = 5.0\text{ N} - 3.4\text{ N} = 1.6\text{ N}

Loss of weight in liquid X gives the upthrust exerted by liquid X.

4

Calculate the density of liquid X

\rho_X = \frac{U_X}{V g} = \frac{1.6}{(2.0 \times 10^{-4}) \times 10} = 800\text{ kg/m}^3

Rearranging

U_X = \rho_X V g

allows solving for the unknown fluid density.

Key Concept

Archimedes' Principle and Apparent Weight

Estimated Time:2m 0s

Question 5Question →

A spherical particle of radius $3.0 \times 10^{-3}\text{ m}$ and density $8.0 \times 10^3\text{ kg/m}^3$ is released from rest and falls vertically through a tall column of a viscous liquid of density $2.0 \times 10^3\text{ kg/m}^3$ . If the coefficient of viscosity of the fluid is $0.40\text{ Pa}\cdot\text{s}$ and the acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the magnitude of its terminal velocity in $\text{m/s}$ .

Show answer & explanation

Answer: 0.3

Answer

The magnitude of the terminal velocity of the falling sphere is

0.3\text{ m/s}

.

When a body falls at terminal velocity through a viscous medium, its weight is balanced by the sum of buoyancy upthrust and Stokes' viscous drag force. Applying

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

with sphere radius

r = 0.003\text{ m}

, sphere density

\rho_s = 8000\text{ kg/m}^3

, fluid density

\rho_f = 2000\text{ kg/m}^3

, viscosity

\eta = 0.40\text{ Pa}\cdot\text{s}

, and

g = 10\text{ m/s}^2

yields

v_t = 0.3\text{ m/s}

.

Step-by-Step Solution

1

Formulate the dynamic equilibrium condition at terminal velocity.

At terminal velocity, the net acceleration is zero, leading to the force balance equation

W = U + F_v

, where

W

is the gravitational weight of the sphere,

U

is the buoyant upthrust, and

F_v

is the retarding viscous force.

Terminal velocity occurs when the downward force of gravity is precisely balanced by the sum of upward resistive and buoyancy forces.

2

Substitute algebraic expressions for weight, upthrust, and Stokes' viscous drag.

W = \frac{4}{3}\pi r^3 \rho_s g

,

U = \frac{4}{3}\pi r^3 \rho_f g

, and

F_v = 6\pi \eta r v_t

.

Archimedes' principle defines the upthrust force equal to the weight of displaced liquid, while Stokes' law governs viscous resistance on spherical bodies.

3

Solve the equilibrium equation for terminal velocity

v_t

.

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

.

Equating

\frac{4}{3}\pi r^3 (\rho_s - \rho_f) g = 6\pi \eta r v_t

and simplifying cancels common factors of

\pi

and

r

.

4

Substitute the specified numerical parameters into the derived expression.

v_t = \frac{2 \times (3.0 \times 10^{-3})^2 \times (8000 - 2000) \times 10}{9 \times 0.40} = \frac{2 \times 9.0 \times 10^{-6} \times 6000 \times 10}{3.6} = \frac{1.08}{3.6} = 0.3\text{ m/s}

.

Direct calculation yields the exact value of terminal velocity.

Key Concept

Terminal Velocity, Stokes' Law, and Archimedes' Principle

Question 6Question →

A solid cylinder of length $10\text{ cm}$ floats vertically at the interface of two immiscible liquids: oil of density $800\text{ kg/m}^3$ and water of density $1000\text{ kg/m}^3$ . If $4\text{ cm}$ of its length is submerged in water while the remaining $6\text{ cm}$ is submerged in oil, what is the density of the cylinder?

880\text{ kg/m}^3

920\text{ kg/m}^3

900\text{ kg/m}^3

400\text{ kg/m}^3

Show answer & explanation

Answer: $880\text{ kg/m}^3$

Answer

The density of the cylinder is

880\text{ kg/m}^3

.

According to the Law of Flotation, a floating body displaces its own weight of the fluids in which it floats. For a body submerged across two immiscible liquids, the total buoyant force is the sum of the upthrusts from both fluids:

U = (\rho_w V_w + \rho_o V_o)g

. Equating this to the total weight

W = \rho_c V_{total} g

yields

\rho_c (0.10\text{ m}) = 1000(0.04\text{ m}) + 800(0.06\text{ m}) = 88

, which gives

\rho_c = 880\text{ kg/m}^3

.

Step-by-Step Solution

1

Set up the condition for flotation.

Weight of the cylinder = Total upthrust exerted by both liquids.

For a body floating in equilibrium, its total weight is balanced by the sum of buoyant forces from all surrounding fluids.

2

Express the weight and upthrusts in terms of density, cross-sectional area

A

, length

L

, and acceleration due to gravity

g

.

\rho_c A L g = (\rho_w A h_w + \rho_o A h_o) g

The weight of the cylinder is

\rho_c V_{total} g

and upthrust from each liquid is

\rho_{liquid} V_{submerged} g

.

3

Cancel common factors

A

and

g

from both sides.

\rho_c L = \rho_w h_w + \rho_o h_o

Since the cylinder has a uniform cross-sectional area, volume ratio simplifies to length ratio.

4

Substitute the given values (

L = 0.10\text{ m}

,

h_w = 0.04\text{ m}

,

h_o = 0.06\text{ m}

,

\rho_w = 1000\text{ kg/m}^3

,

\rho_o = 800\text{ kg/m}^3

).

\rho_c (0.10) = 1000(0.04) + 800(0.06) = 40 + 48 = 88\text{ kg/m}^2

Evaluating the weighted contribution of buoyancy from each fluid.

5

Solve for the density of the cylinder

\rho_c

.

\rho_c = \frac{88}{0.10} = 880\text{ kg/m}^3

Dividing both sides by the total length of

0.10\text{ m}

gives the density.

Key Concept

Archimedes' Principle for Floating Bodies in Immiscible Liquids

Question 7Question →

A uniform wooden cube of edge length $0.20\text{ m}$ floats in water of density $1000\text{ kg/m}^3$ with $0.15\text{ m}$ of its vertical height submerged. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what minimum mass, in kilograms, must be placed on the top surface of the cube so that its upper face becomes just flush with the water surface?

Show answer & explanation

Answer: 2

Answer

The minimum mass required to submerge the cube completely flush with the water surface is

2.0\text{ kg}

.

By the Law of Flotation, a floating body displaces its own weight of fluid. Initially, the cube displaces a volume of

0.20\text{ m} \times 0.20\text{ m} \times 0.15\text{ m} = 0.006\text{ m}^3

of water, corresponding to an upthrust of

60\text{ N}

(or mass of

6.0\text{ kg}

). When completely submerged, the total volume displaced is

0.20^3 = 0.008\text{ m}^3

, providing a total upthrust of

80\text{ N}

(or mass equivalent of

8.0\text{ kg}

). The additional mass required on top is therefore the difference:

8.0\text{ kg} - 6.0\text{ kg} = 2.0\text{ kg}

.

Step-by-Step Solution

1

Calculate the cross-sectional area of the cube

A = (0.20\text{ m})^2 = 0.04\text{ m}^2

The base area is needed to find the volume of the block submerged and unsubmerged.

2

Calculate the volume of the cube above the water surface

V_{\text{above}} = 0.04\text{ m}^2 \times (0.20\text{ m} - 0.15\text{ m}) = 0.002\text{ m}^3

To push the cube level with the surface, the additional weight added on top must balance the extra upthrust created by submerging this remaining volume.

3

Calculate the additional mass required

m = \rho_{\text{water}} \times V_{\text{above}} = 1000\text{ kg/m}^3 \times 0.002\text{ m}^3 = 2.0\text{ kg}

By Archimedes' principle, the additional downward mass must equal the mass of the extra water displaced when fully submerged.

Key Concept

Archimedes' Principle and Law of Flotation

Question 8Question →

A solid block of mass $0.60\text{ kg}$ and density $600\text{ kg/m}^3$ is held fully submerged in water of density $1000\text{ kg/m}^3$ by a light vertical string attached to the bottom of a container. What is the tension in the string? (Take $g = 10\text{ m/s}^2$ )

10.0\text{ N}

4.0\text{ N}

6.0\text{ N}

16.0\text{ N}

Show answer & explanation

Answer: $4.0\text{ N}$

Answer

The tension in the string is

4.0\text{ N}

.

First, the volume of the block is computed as

V = \frac{m}{\rho} = \frac{0.60}{600} = 1.0 \times 10^{-3}\text{ m}^3

. According to Archimedes' principle, the upthrust exerted by the displaced water is

U = \rho_{\text{water}} V g = 1000 \times 1.0 \times 10^{-3} \times 10 = 10.0\text{ N}

. The weight of the block is

W = mg = 0.60 \times 10 = 6.0\text{ N}

. For the block to remain completely submerged in equilibrium, the upward upthrust must balance the downward forces (the weight of the block and the tension

T

pulling downward). Thus,

T = U - W = 10.0\text{ N} - 6.0\text{ N} = 4.0\text{ N}

.

Step-by-Step Solution

1

Calculate the volume of the block using its mass and density.

V = \frac{m}{\rho_{\text{block}}} = \frac{0.60\text{ kg}}{600\text{ kg/m}^3} = 1.0 \times 10^{-3}\text{ m}^3

The volume of fluid displaced equals the total volume of the fully submerged block.

2

Calculate the upward upthrust force exerted by the water.

U = \rho_{\text{water}} \cdot V \cdot g = 1000\text{ kg/m}^3 \times 1.0 \times 10^{-3}\text{ m}^3 \times 10\text{ m/s}^2 = 10.0\text{ N}

Archimedes' principle states upthrust equals the weight of the displaced fluid.

3

Calculate the downward gravitational weight of the block.

W = m \cdot g = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}

Weight is the force exerted on the mass of the block by gravity.

4

Apply vertical force equilibrium to solve for string tension.

U = W + T \Rightarrow T = U - W = 10.0\text{ N} - 6.0\text{ N} = 4.0\text{ N}

The string is tied to the bottom, so tension acts downward to hold the buoyant block in equilibrium.

Key Concept

Archimedes' Principle and Static Equilibrium of Submerged Bodies

Question 9Question →

A solid sphere of mass $0.40\text{ kg}$ and relative density $2.5$ is held fully submerged in a liquid of density $800\text{ kg/m}^3$ by a light string attached to a fixed support. What is the tension in the string? (Take $g = 10\text{ m/s}^2$ )

2.72\text{ N}

3.60\text{ N}

1.28\text{ N}

2.40\text{ N}

Show answer & explanation

Answer: $2.72\text{ N}$

Answer

The tension in the string is

2.72\text{ N}

.

The sphere has a weight of

4.0\text{ N}

acting downward. When fully submerged, it displaces a volume of liquid equal to its own volume (

1.6 \times 10^{-4}\text{ m}^3

). The displaced liquid of density

800\text{ kg/m}^3

exerts an upward buoyant force (upthrust) of

1.28\text{ N}

. Tension balances the remaining downward force:

T = 4.0\text{ N} - 1.28\text{ N} = 2.72\text{ N}

.

Step-by-Step Solution

1

Calculate the weight of the sphere

W = mg = 0.40\text{ kg} \times 10\text{ m/s}^2 = 4.0\text{ N}

The force of gravity acting downward on the mass.

2

Determine the volume of the sphere using its relative density

Density of sphere

\rho_s = 2.5 \times 1000\text{ kg/m}^3 = 2500\text{ kg/m}^3

. Volume

V = \frac{m}{\rho_s} = \frac{0.40}{2500} = 1.6 \times 10^{-4}\text{ m}^3

.

Relative density is the ratio of the substance's density to the density of water (

1000\text{ kg/m}^3

).

3

Calculate the upthrust exerted by the liquid

U = V \cdot \rho_{\text{liquid}} \cdot g = (1.6 \times 10^{-4}) \times 800 \times 10 = 1.28\text{ N}

By Archimedes' principle, upthrust equals the weight of the fluid displaced by the submerged volume.

4

Calculate tension in the string

T = W - U = 4.0\text{ N} - 1.28\text{ N} = 2.72\text{ N}

For vertical equilibrium of the submerged body, weight acts downward while upthrust and tension act upward.

Key Concept

Archimedes' Principle and Apparent Weight

Estimated Time:1m 30s

Question 10Question →

A uniform cylindrical rod of length $20\text{ cm}$ floats vertically in liquid $X$ of density $800\text{ kg/m}^3$ with $15\text{ cm}$ of its length submerged. When transferred to liquid $Y$ , it floats vertically with $12\text{ cm}$ of its length submerged. What is the density of liquid $Y$ ?

640\text{ kg/m}^3

1000\text{ kg/m}^3

1333\text{ kg/m}^3

500\text{ kg/m}^3

Show answer & explanation

Answer: $1000\text{ kg/m}^3$

Answer

The density of liquid

Y

is

1000\text{ kg/m}^3

.

According to the Law of Flotation, a floating body displaces its own weight of liquid. Therefore,

h_X \rho_X = h_Y \rho_Y

. Substituting

15\text{ cm} \times 800\text{ kg/m}^3 = 12\text{ cm} \times \rho_Y

yields

\rho_Y = 1000\text{ kg/m}^3

.

Step-by-Step Solution

1

Apply the Law of Flotation for a floating body of uniform cross-sectional area

A

.

Weight of rod

W = \text{Upthrust} = A \cdot h_{\text{submerged}} \cdot \rho_{\text{liquid}} \cdot g

.

A floating body displaces a weight of fluid equal to its own total weight.

2

Equate the upthrust in liquid

X

to the upthrust in liquid

Y

.

A \cdot h_X \cdot \rho_X \cdot g = A \cdot h_Y \cdot \rho_Y \cdot g \implies h_X \cdot \rho_X = h_Y \cdot \rho_Y

.

Since the rod is identical and floating freely in both liquids, its weight

W

remains unchanged.

3

Substitute the known values (

h_X = 15\text{ cm}

,

\rho_X = 800\text{ kg/m}^3

,

h_Y = 12\text{ cm}

) and solve for

\rho_Y

.

\rho_Y = \frac{h_X \cdot \rho_X}{h_Y} = \frac{15 \times 800}{12} = 1000\text{ kg/m}^3

.

Rearranging the linear equation yields the density of liquid

Y

.

Key Concept

Law of Flotation and Hydrometer Principle

Estimated Time:1m 15s

Question 11Question →

A beaker containing water of density $1000\text{ kg/m}^3$ rests on a digital weighing scale, giving an initial reading of $1.50\text{ kg}$ . A solid aluminum block of mass $0.80\text{ kg}$ and density $2500\text{ kg/m}^3$ is suspended from a string and completely immersed in the water without touching the bottom or sides of the beaker. What is the new reading on the digital weighing scale, in kilograms? (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 1.82

Answer

The new reading on the digital weighing scale is

1.82\text{ kg}

.

When the aluminum block is fully submerged in the water, it displaces a volume of water equal to its own volume (

V = \frac{0.80}{2500} = 3.2 \times 10^{-4}\text{ m}^3

). The mass of this displaced water is

m_{\text{water}} = 1000 \times 3.2 \times 10^{-4} = 0.32\text{ kg}

. The upthrust exerted by the water upward on the block is equal to the weight of the displaced water (

3.2\text{ N}

). By Newton's Third Law, the block exerts an equal and opposite downward reaction force (

3.2\text{ N}

) on the water. This extra downward force adds an equivalent mass of

0.32\text{ kg}

to the digital scale reading, making the new reading

1.50\text{ kg} + 0.32\text{ kg} = 1.82\text{ kg}

.

Step-by-Step Solution

1

Calculate the volume of the submerged block

Volume

V = 3.2 \times 10^{-4}\text{ m}^3

The volume of fluid displaced by a completely submerged body equals the volume of the body itself.

2

Find the mass of the displaced water

Mass of displaced water

m_{\text{water}} = 0.32\text{ kg}

According to Archimedes' principle, the upthrust equals the weight of the displaced fluid, which corresponds to a displaced mass of

\rho_{\text{water}} V

.

3

Apply Newton's Third Law to determine the change in scale reading

Scale reading increase

\Delta m = 0.32\text{ kg}

The fluid exerts an upward buoyant force on the block, so by Newton's Third Law, the block exerts an equal downward reaction force on the fluid, transferring an effective weight equal to the upthrust onto the scale.

4

Compute the total new scale reading

New scale reading

= 1.82\text{ kg}

Sum the initial mass reading of the beaker system (

1.50\text{ kg}

) and the mass of the displaced water (

0.32\text{ kg}

).

Key Concept

Apparent weight transfer, Archimedes' principle, and Newton's Third Law

Question 12Question →

A diver is swimming at a depth of $3.5\text{ m}$ below the surface of a freshwater lake. If the density of water is $1000\text{ kg/m}^3$ and the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the gauge pressure exerted on the diver in pascals?

Show answer & explanation

Answer: 35000

Answer

35000 Pa

The gauge pressure exerted by a static column of fluid is given by

P = h \rho g

. Using the values

h = 3.5\text{ m}

,

\rho = 1000\text{ kg/m}^3

, and

g = 10\text{ m/s}^2

, the pressure is

P = 3.5 \times 1000 \times 10 = 35000\text{ Pa}

.

Step-by-Step Solution

1

Identify the formula for liquid hydrostatic pressure

P = h \rho g

Gauge pressure at a depth

h

in a static fluid depends on depth, fluid density, and gravitational field strength.

2

Substitute the given numerical values

P = 3.5 \times 1000 \times 10

Substitute depth

h = 3.5\text{ m}

, density

\rho = 1000\text{ kg/m}^3

, and

g = 10\text{ m/s}^2

.

3

Perform the multiplication to determine the pressure

35000\text{ Pa}

Complete the calculation to get the pressure in SI units (Pascals).

Key Concept

Hydrostatic Pressure in Static Fluids

Estimated Time:45s

Question 13Question →

A solid uniform cylinder of height $0.20\text{ m}$ and cross-sectional area $5.0 \times 10^{-3}\text{ m}^2$ floats vertically at the boundary between oil of density $800\text{ kg/m}^3$ and water of density $1000\text{ kg/m}^3$ . If a height of $0.08\text{ m}$ of the cylinder extends into the water layer while the remaining upper portion is completely covered by the oil layer, what is the mass of the cylinder in kilograms?

Show answer & explanation

Answer: 0.88

Answer

The mass of the cylinder is

0.88\text{ kg}

.

According to the Law of Flotation, a floating body displaces its own weight of fluid. When floating at the interface of two immiscible liquids, the total mass of the body equals the sum of the masses of the displaced liquids. Displaced water mass is

\rho_w A h_w = 0.40\text{ kg}

and displaced oil mass is

\rho_o A h_o = 0.48\text{ kg}

, giving a total cylinder mass of

0.88\text{ kg}

.

Step-by-Step Solution

1

Find the height of the cylinder submerged in the oil layer.

h_o = 0.20\text{ m} - 0.08\text{ m} = 0.12\text{ m}

The total cylinder height is

0.20\text{ m}

, and

0.08\text{ m}

is submerged in water.

2

Calculate the volumes of water and oil displaced by the cylinder.

V_w = 5.0 \times 10^{-3} \times 0.08 = 4.0 \times 10^{-4}\text{ m}^3

;

V_o = 5.0 \times 10^{-3} \times 0.12 = 6.0 \times 10^{-4}\text{ m}^3

Volume displaced in each fluid equals cross-sectional area multiplied by the submerged height in that fluid.

3

Calculate the mass of the floating cylinder using the Law of Flotation.

m = \rho_w V_w + \rho_o V_o = (1000 \times 4.0 \times 10^{-4}) + (800 \times 6.0 \times 10^{-4}) = 0.40\text{ kg} + 0.48\text{ kg} = 0.88\text{ kg}

For a floating object in static equilibrium, its mass equals the total mass of the fluids displaced by its submerged parts.

Key Concept

Law of Flotation in Layered Liquids

Estimated Time:1m 30s

Question 14Question →

A solid block floats in water of density $1000\text{ kg/m}^3$ with $60\%$ of its total volume submerged. When the same block is placed in an unknown liquid $X$ , $80\%$ of its total volume is submerged. What is the density of liquid $X$ ?

750\text{ kg/m}^3

1333\text{ kg/m}^3

480\text{ kg/m}^3

800\text{ kg/m}^3

Show answer & explanation

Answer: $750\text{ kg/m}^3$

Answer

The density of liquid

X

is

750\text{ kg/m}^3

.

For any floating body, its weight equals the upthrust exerted by the liquid. The upthrust is given by

U = \rho_{\text{fluid}} \cdot V_{\text{submerged}} \cdot g

. Since the weight of the block is unchanged,

\rho_{\text{water}} \cdot V_{\text{sub, water}} = \rho_X \cdot V_{\text{sub, X}}

. Substituting the given values:

1000 \times 0.60V = \rho_X \times 0.80V

, yielding

\rho_X = 750\text{ kg/m}^3

.

Step-by-Step Solution

1

Apply the Law of Flotation to the block in water

Weight of block

W = \rho_w \cdot V_{\text{sub, water}} \cdot g = 1000 \cdot 0.60V \cdot g = 600 V g

A floating object displaces its own weight of fluid.

2

Apply the Law of Flotation to the block in liquid X

Weight of block

W = \rho_X \cdot V_{\text{sub, X}} \cdot g = \rho_X \cdot 0.80V \cdot g

The weight of the block remains constant regardless of the fluid.

3

Equate the two expressions for the weight of the block and solve for

\rho_X

\rho_X \cdot 0.80V \cdot g = 600 V g \implies \rho_X = \frac{600}{0.80} = 750\text{ kg/m}^3

Since both buoyant forces equal the block's weight, set them equal to each other.

Key Concept

Law of Flotation and Archimedes' Principle

Question 15Question →

A solid object has a mass of $0.50\text{ kg}$ and a volume of $2.0 \times 10^{-4}\text{ m}^3$ . If it is completely immersed in a liquid of density $800\text{ kg/m}^3$ , what is the magnitude of the upthrust exerted on the object? [Take $g = 10\text{ m/s}^2$ ]

1.6\text{ N}

5.0\text{ N}

3.4\text{ N}

0.16\text{ N}

Show answer & explanation

Answer: $1.6\text{ N}$

Answer

The magnitude of the upthrust exerted on the object is

1.6\text{ N}

.

By Archimedes' Principle, upthrust is equal to the weight of the liquid displaced:

U = \rho_{\text{liquid}} V g

. Substituting

\rho = 800\text{ kg/m}^3

,

V = 2.0 \times 10^{-4}\text{ m}^3

, and

g = 10\text{ m/s}^2

yields

1.6\text{ N}

.

Step-by-Step Solution

1

Identify the given values and state Archimedes' Principle

Volume of displaced liquid

V = 2.0 \times 10^{-4}\text{ m}^3

, density of liquid

\rho = 800\text{ kg/m}^3

, acceleration due to gravity

g = 10\text{ m/s}^2

. Upthrust formula is

U = \rho V g

.

According to Archimedes' Principle, the upward buoyant force (upthrust) equals the weight of the liquid displaced by the submerged object.

2

Calculate the upthrust

U = 800\text{ kg/m}^3 \times (2.0 \times 10^{-4}\text{ m}^3) \times 10\text{ m/s}^2 = 1.6\text{ N}

.

Multiplying fluid density by submerged volume and gravitational acceleration gives the force in newtons.

Key Concept

Archimedes' Principle and Upthrust

Question 16Question →

A spherical ball bearing of radius $3.0\text{ mm}$ and density $5400\text{ kg/m}^3$ falls vertically through a viscous oil of density $900\text{ kg/m}^3$ and dynamic viscosity coefficient $0.10\text{ Pa}\cdot\text{s}$ . Assuming the motion obeys Stokes' law and taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the terminal velocity of the sphere?

0.90\text{ m/s}

1.08\text{ m/s}

0.68\text{ m/s}

0.18\text{ m/s}

Show answer & explanation

Answer: $0.90\text{ m/s}$

Answer

The terminal velocity of the sphere is

0.90\text{ m/s}

.

At terminal velocity, the downward force of gravity (weight of the sphere) is balanced by the sum of two upward forces: the buoyant force (upthrust) and the viscous drag force given by Stokes' law (

F_v = 6\pi \eta r v_T

). Using the formula

v_T = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

with

r = 3.0 \times 10^{-3}\text{ m}

,

\rho_s - \rho_f = 4500\text{ kg/m}^3

,

\eta = 0.10\text{ Pa}\cdot\text{s}

, and

g = 10\text{ m/s}^2

yields

0.90\text{ m/s}

.

Step-by-Step Solution

1

Convert given parameters to standard SI units

Radius

r = 3.0\text{ mm} = 3.0 \times 10^{-3}\text{ m}

, density of sphere

\rho_s = 5400\text{ kg/m}^3

, density of liquid

\rho_f = 900\text{ kg/m}^3

, viscosity

\eta = 0.10\text{ Pa}\cdot\text{s}

,

g = 10\text{ m/s}^2

.

Ensures dimensional consistency across all terms in the physical equations.

2

Apply the equilibrium condition at terminal velocity

At terminal velocity

v_T

, downward weight equals upward forces:

W = U + F_v

, where

W = \frac{4}{3}\pi r^3 \rho_s g

,

U = \frac{4}{3}\pi r^3 \rho_f g

, and

F_v = 6\pi \eta r v_T

.

Terminal velocity is reached when net acceleration is zero.

3

Rearrange Stokes' law formula for terminal velocity

v_T = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

Isolates the target unknown variable

v_T

.

4

Substitute the physical values and solve

v_T = \frac{2 \times (3.0 \times 10^{-3})^2 \times (5400 - 900) \times 10}{9 \times 0.10} = \frac{2 \times (9.0 \times 10^{-6}) \times 4500 \times 10}{0.90} = \frac{0.81}{0.90} = 0.90\text{ m/s}

.

Calculates the final quantitative answer.

Key Concept

Viscosity and Stokes' Law for terminal velocity of a sphere in a viscous fluid

Estimated Time:2m 0s

Question 17Question →

A solid sphere of mass $0.60\text{ kg}$ and volume $2.0 \times 10^{-4}\text{ m}^3$ is released from rest in a tall vessel filled with a viscous liquid of density $1000\text{ kg/m}^3$ . As the sphere falls, it eventually reaches a constant terminal velocity of $4.0\text{ m/s}$ . Assuming that the viscous drag force is directly proportional to the speed of the sphere, what is the magnitude of the viscous drag force acting on the sphere when its speed is $1.5\text{ m/s}$ ? (Take $g = 10\text{ m/s}^2$ )

0.75\text{ N}

1.50\text{ N}

2.25\text{ N}

2.50\text{ N}

Show answer & explanation

Answer: $1.50\text{ N}$

Answer

1.50\text{ N}

At terminal velocity, the sphere is in translational equilibrium under three forces: downward weight (

6.0\text{ N}

), upward upthrust (

2.0\text{ N}

), and upward viscous drag (

4.0\text{ N}

). Since viscous drag is directly proportional to velocity (

F_v = k v

), the proportionality constant

k

is

1.0\text{ N}\cdot\text{s/m}

. Therefore, at

1.5\text{ m/s}

, the viscous force is

1.0 \times 1.5 = 1.50\text{ N}

.

Step-by-Step Solution

1

Calculate the downward gravitational force (weight) acting on the sphere.

W = m \times g = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}

Weight is the force pulling the sphere downward.

2

Calculate the buoyant force (upthrust) exerted by the liquid using Archimedes' principle.

U = \rho_l \times V \times g = 1000\text{ kg/m}^3 \times (2.0 \times 10^{-4}\text{ m}^3) \times 10\text{ m/s}^2 = 2.0\text{ N}

Upthrust equals the weight of the fluid displaced by the submerged sphere.

3

Determine the viscous drag force at terminal velocity (

v_t = 4.0\text{ m/s}

) using equilibrium of forces.

F_v(v_t) = W - U = 6.0\text{ N} - 2.0\text{ N} = 4.0\text{ N}

At terminal velocity, the net acceleration is zero, so downward weight is balanced by upward upthrust and viscous drag.

4

Find the constant of proportionality

k

for viscous drag (

F_v = k v

).

k = \frac{F_v(v_t)}{v_t} = \frac{4.0\text{ N}}{4.0\text{ m/s}} = 1.0\text{ N}\cdot\text{s/m}

Viscous force is given as directly proportional to speed.

5

Calculate the viscous drag force at a speed of

1.5\text{ m/s}

.

F_v(1.5) = k \times 1.5\text{ m/s} = 1.0\text{ N}\cdot\text{s/m} \times 1.5\text{ m/s} = 1.50\text{ N}

Applying the constant

k

to the specified speed.

Key Concept

Terminal velocity in viscous fluids and Archimedes' Principle

Estimated Time:2m 30s

Question 18Question →

A solid alloy specimen weighs $240\text{ g}$ in air, $180\text{ g}$ when completely immersed in water, and $195\text{ g}$ when completely immersed in an unknown liquid $X$ . What is the relative density of liquid $X$ ?

0.75

0.81

1.33

0.25

Show answer & explanation

Answer: $0.75$

Answer

The relative density of liquid X is

0.75

.

According to Archimedes' principle, relative density of a liquid is the ratio of the upthrust experienced by a solid in that liquid to the upthrust experienced by the same solid in water. Upthrust in water is

240\text{ g} - 180\text{ g} = 60\text{ g}

, and upthrust in liquid X is

240\text{ g} - 195\text{ g} = 45\text{ g}

. Dividing

45

by

60

gives

0.75

.

Step-by-Step Solution

1

Calculate the upthrust in water

240\text{ g} - 180\text{ g} = 60\text{ g}

By Archimedes' principle, upthrust in water equals the mass loss when immersed in water.

2

Calculate the upthrust in liquid X

240\text{ g} - 195\text{ g} = 45\text{ g}

Upthrust in liquid X equals the mass loss when immersed in liquid X.

3

Determine the relative density of liquid X

\text{Relative Density} = \frac{\text{Upthrust in liquid X}}{\text{Upthrust in water}} = \frac{45\text{ g}}{60\text{ g}} = 0.75

Relative density of a liquid is defined as the ratio of the weight of a given volume of the liquid to the weight of an equal volume of water.

Key Concept

Archimedes' Principle and Relative Density of Liquids

Estimated Time:1m 30s

Question 19Question →

A solid block of wood has a mass of $0.60\text{ kg}$ . When placed in a vessel of water, it floats freely on the surface. What is the magnitude of the upthrust exerted by the water on the block? (Take $g = 10\text{ m/s}^2$ )

6.0\text{ N}

0.60\text{ N}

0.0\text{ N}

60\text{ N}

Show answer & explanation

Answer: $6.0\text{ N}$

Answer

The upthrust exerted by the water on the floating block is

6.0\text{ N}

.

According to the Law of Flotation, a body floating freely in a fluid displaces a weight of fluid equal to its own weight. Therefore, the upward force (upthrust) exerted by the fluid is equal to the weight of the block:

U = mg = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}

.

Step-by-Step Solution

1

Determine the weight of the floating block in air.

W = mg = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}

Weight is the force of gravity acting on the block's mass.

2

Apply the Law of Flotation to find the upthrust.

Upthrust

U = W = 6.0\text{ N}

A freely floating body displaces a volume of fluid whose weight is equal to the total weight of the body.

Key Concept

Law of Flotation and Archimedes' Principle

Estimated Time:45s

Question 20Question →

A solid wooden block of density $600\text{ kg/m}^3$ and volume $4.0 \times 10^{-3}\text{ m}^3$ floats in water of density $1000\text{ kg/m}^3$ . A metal block is placed on top of the wooden block so that the wooden block is just completely submerged while the metal block remains entirely above the water surface. What is the mass of the metal block? (Take $g = 10\text{ m/s}^2$ )

1.6\text{ kg}

2.4\text{ kg}

4.0\text{ kg}

6.4\text{ kg}

Show answer & explanation

Answer: $1.6\text{ kg}$

Answer

1.6\text{ kg}

When the wooden block is completely submerged, it displaces

4.0 \times 10^{-3}\text{ m}^3

of water, creating an upthrust of

40\text{ N}

. The weight of the wooden block is

24\text{ N}

. For equilibrium, the total downward weight must equal the upthrust (

24\text{ N} + W_{\text{metal}} = 40\text{ N}

), which gives

W_{\text{metal}} = 16\text{ N}

and a corresponding mass of

1.6\text{ kg}

.

Step-by-Step Solution

1

Calculate the mass and weight of the wooden block.

m_{\text{wood}} = \rho_{\text{wood}} \times V_{\text{wood}} = 600\text{ kg/m}^3 \times 4.0 \times 10^{-3}\text{ m}^3 = 2.4\text{ kg}

, so

W_{\text{wood}} = 2.4 \times 10 = 24\text{ N}

.

The weight of the wood contributes to the total downward force of the floating system.

2

Calculate the total upthrust exerted by the water when the wooden block is completely submerged.

U = \rho_{\text{water}} \times V_{\text{wood}} \times g = 1000\text{ kg/m}^3 \times 4.0 \times 10^{-3}\text{ m}^3 \times 10\text{ m/s}^2 = 40\text{ N}

.

By Archimedes' principle, upthrust equals the weight of the displaced liquid.

3

Apply the law of flotation to solve for the weight and mass of the metal block.

W_{\text{wood}} + W_{\text{metal}} = U \implies 24\text{ N} + W_{\text{metal}} = 40\text{ N} \implies W_{\text{metal}} = 16\text{ N}

. Thus,

m_{\text{metal}} = \frac{16\text{ N}}{10\text{ m/s}^2} = 1.6\text{ kg}

.

For the system to float in equilibrium just submerged, total downward weight must equal total upward buoyant force.

Key Concept

Archimedes' Principle and Law of Flotation

Estimated Time:2m 0s

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