PHYS 101 - ELEMENTS OF PHYSICS Buoyant forces and Archimedes' principle Question Bank Set 9

PHYS 101 - ELEMENTS OF PHYSICS

- Buoyant forces and Archimedes’

principle

Question Bank - Set 9

Liberty University

Question 1

Question

A cylindrical container with a radius of 1.5 m is filled with water. A solid iron

cube with a side length of 0.5 m is placed at the bottom of the container. What

is the buoyant force acting on the iron cube?

Given: Density of water = 1000 kg/m3Density of iron = 7800 kg/m3Ac-

celeration due to gravity = 9.81 m/s2

Solution

Step 1: Calculate the volume of the iron cube. The volume of a cube is given

by Vcube = side length3. Given that the side length of the iron cube is 0.5 m,

the volume is:

Vcube = (0.5 m)3= 0.125 m3

Step 2: Calculate the weight of the iron cube. The weight of the iron cube is

given by Wcube = mass ×gravity. The mass of the iron cube can be calculated

using its volume and density: miron = density ×volume. Therefore, the weight

of the iron cube is:

Wcube = 7800 kg/m3×0.125 m3×9.81 m/s2= 9596.25 N

Step 3: Calculate the weight of the water displaced by the iron cube. Ac-

cording to Archimedes’ principle, the buoyant force is equal to the weight of the

water displaced by the object. The volume of water displaced by the iron cube

is the same as the volume of the cube (0.125 m3). Therefore, the weight of the

water displaced is:

Wwater = density of water×Vcube×gravity = 1000 kg/m3×0.125 m3×9.81 m/s2= 1226.25 N

Step 4: Calculate the buoyant force acting on the iron cube. The buoyant

force is equal to the weight of the water displaced, so the buoyant force acting

on the iron cube is:

Fbuoyant = 1226.25 N

Question 2

Question

A cube of steel with a side length of 10 cm and a density of 7,800 kg/m3is

placed in a container filled with water. What is the buoyant force acting on the

cube?

Solution

Step 1: Calculate the volume of the cube. Given the side length of the cube,

the volume can be calculated as:

Volume of cube = (side length)3

Volume of cube = (0.1 m)3= 0.001 m3

Step 2: Determine the volume of water displaced by the cube. The volume

of water displaced by the cube is equal to the volume of the cube, since the cube

is fully submerged.

Volume of water displaced = 0.001 m3

Step 3: Calculate the buoyant force. The buoyant force is equal to the weight

of the water displaced by the cube.

Buoyant force = Density of water×Volume of water displaced×Acceleration due to gravity

Buoyant force = 1000 kg/m3×0.001 m3×9.8 m/s2

Buoyant force = 9.8 N

Therefore, the buoyant force acting on the cube is 9.8 N.

Question 3

Question

A metal block of mass 500 kg and volume 0.1 m3is immersed in a fluid. De-

termine the buoyant force acting on the block if the density of the fluid is 800

kg/m3.

Solution

Step 1: Calculate the density of the metal block using the formula ρ=m

V, where

ρis the density, mis the mass, and Vis the volume.

ρ=500 kg

0.1 m3= 5000 kg/m3

Step 2: Determine the apparent weight of the block in the fluid using the

formula Wapparent =m·g−Fbuoyant, where Wapparent is the apparent weight,

mis the mass, gis the acceleration due to gravity, and Fbuoyant is the buoyant

force.

Wapparent = 500 kg ·9.8 m/s2−Fbuoyant

Step 3: Apply Archimedes’ principle, which states that the buoyant force

Fbuoyant acting on a submerged object is equal to the weight of the fluid displaced

by the object.

Fbuoyant = density of fluid ·g·Vdisplaced

Step 4: Calculate the volume of fluid displaced by the block using the formula

Vdisplaced =m

ρfluid , where ρfluid is the density of the fluid.

Vdisplaced =500 kg

800 kg/m3= 0.625 m3

Step 5: Substitute the values into the expression for Fbuoyant.

Fbuoyant = 800 kg/m3·9.8 m/s2·0.625 m3

Step 6: Calculate the buoyant force.

Fbuoyant = 4900 N

Therefore, the buoyant force acting on the block is 4900 N.

Question 4

Question

A cylindrical wooden log with a diameter of 0.5 m and a length of 2 m floats in

water. The density of water is 1000 kg/m3. Determine the height by which the

log protrudes from the water surface.

Solution

Step 1: Find the volume of the wooden log.

To find the volume of the log, we use the formula for the volume of a cylinder:

Vlog =πr2h

where r= 0.25 m (radius) and h= 2 m (height).

Vlog =π(0.25)2×2

Vlog =π

4×2

Vlog =π

2m3

Step 2: Find the weight of the wooden log.

The weight of the log can be calculated using the formula:

Wlog =mlog ×g

where mlog is the mass of the log and g= 9.81 m/s2. Since density ρ=

mass/volume, we have mlog =ρwood ×Vlog, with ρwood being the density of

wood. Given that the density of wood is approximately 700 kg/m3, we have:

mlog = 700 ×π

mlog = 350πkg

Therefore:

Wlog = 350π×9.81

Wlog = 3433.5πN

Step 3: Find the buoyant force on the log.

According to Archimedes’ principle, the buoyant force Fbuoy on an object in a

fluid is equal to the weight of the fluid displaced by the object. We can calculate

the buoyant force using the formula:

Fbuoy =ρwater ×Vsubmerged ×g

where ρwater is the density of water. Since the log is floating, the volume of the

log submerged in water is equal to the volume of water it displaces.

Vsubmerged =Vlog

Therefore:

Fbuoy = 1000 ×π

2×9.81

Fbuoy = 4905πN

Step 4: Determine the height of the log above water level.

When the log is floating, the buoyant force Fbuoy acting on it is equal to the

weight Wlog of the log.

Fbuoy =Wlog

4905π= 3433.5π

1471.5π= 1000πr2h

1.4715 = r2h

Given r= 0.25 m, we can solve for h(the height by which the log protrudes

from the water surface).

1.4715 = 0.252h

1.4715 = 0.0625h

h=1.4715

0.0625

h≈23.544 m

Therefore, the log protrudes approximately 23.544 meters above the water sur-

face.

Question 5

Question

A cube of wood with side length 10 cm and density 0.6 g/cm3is floating in a

container of water. What is the minimum side length of a cube of gold (density

= 19.3 g/cm3) that will cause the wooden cube to sink when both cubes are

submerged in water?

Solution

Step 1: First, we need to determine the weight of the wooden cube. The weight

of the wooden cube is equal to the weight of the water displaced by the cube,

which is equal to the buoyant force acting on the cube. The volume of the

wooden cube can be calculated as Vwood = (10 cm)3= 1000 cm3. The weight of

the wooden cube is then:

Wwood =Vwood ·densitywood ·g= 1000 cm3×0.6 g/cm3×9.8 m/s2

Step 2: Next, we calculate the minimum volume of gold cube required to

sink the wooden cube. Let the side length of the gold cube be scm. The volume

of the gold cube is Vgold =s3. The buoyant force acting on the gold cube is

equal to the weight of the water displaced by the gold cube. This weight is

equal to the sum of the weight of the water displaced by the wooden cube and

the weight of the wooden cube, since both are submerged.

Step 3: Thus, to sink the wooden cube, the buoyant force acting on the gold

cube (Vgold ×densitygold ×g) must be greater than the weight of the wooden

cube. This can be expressed as:

Vgold ×densitygold ×g > Wwood

Step 4: Substitute in the expressions for Vgold,Wwood, and the given densities

to solve for the minimum side length sof the gold cube.

Question 6

Question

A cube of iron with a side length of 10 cm and a density of 7.87 g/cm3is

submerged in a container of mercury. If the cube floats with 30

Solution

Step 1: Determine the volume of the cube. The volume of a cube is given by

the formula V=s3, where sis the side length. Substituting s= 10 cm, we find:

V= 103cm3= 1000 cm3

Step 2: Calculate the mass of the iron cube. The mass of an object is given

by the product of its volume and density. The mass mof the cube can be

calculated as follows:

m= Volume ×Density = 1000 ×7.87 = 7870 g

Step 3: Determine the volume of the cube submerged in mercury. Since 30

Vsubmerged = 0.3×1000 cm3= 300 cm3

Step 4: Calculate the buoyant force acting on the cube. The buoyant force

Fbon an object submerged in a fluid is equal to the weight of the fluid displaced.

The weight of the mercury displaced by the submerged portion of the cube can

be calculated using its density (ρHg) and the volume submerged (Vsubmerged):

Fb=ρHg ×Vsubmerged ×g

where gis the acceleration due to gravity.

Step 5: Find the density of mercury. Equating the weight of the iron cube

to the buoyant force on it gives:

m×g=ρHg ×Vsubmerged ×g

Solving for ρHg:

ρHg =m

Vsubmerged

=7870 g

300 cm3= 26.23 g/cm3

Therefore, the density of mercury is 26.23 g/cm3.

Question 7

Question

A solid cube of aluminum with sides of length 10 cm is placed in a container

of water. The cube floats with 3 cm of its height above the water surface.

Calculate the density of aluminum. (Density of water = 1000 kg/m3)

Solution

Step 1: Find the volume of the cube submerged in water. The submerged

volume is given by the area of the base of the cube multiplied by the height

submerged. The area of the base is (10 cm)2= 100 cm2. The height submerged

is 10 cm −3 cm = 7 cm. Therefore, the volume submerged is 100 cm2×7 cm =

700 cm3= 0.0007 m3.

Step 2: Find the weight of water displaced by the cube. The buoyant force

acting on the cube is equal to the weight of water displaced by the cube. The

weight of water displaced is equal to the volume submerged multiplied by the

density of water. So, the weight of water displaced is 0.0007 m3×1000 kg/m3=

0.7 kg.

Step 3: Find the weight of the aluminum cube. The weight of the aluminum

cube is equal to its mass multiplied by the acceleration due to gravity. Let mbe

the mass of the cube and gbe the acceleration due to gravity (approximately

9.81 m/s2). Then, the weight of the aluminum cube is m×9.81 N.

Step 4: Use Archimedes’ principle to find the density of aluminum. Since the

cube floats, the weight of the aluminum cube is balanced by the buoyant force.

Therefore, m×9.81 = 0.7 kg ×9.81 m/s2. Solving for m, we have m=0.7

9.81 kg.

The density of aluminum is given by density = mass

volume =m

0.001 m3. Substitute

the values to find the density of aluminum.

Step 5: Evaluate the density of aluminum. The density of aluminum is

0.7/9.81

0.001 kg/m3. Simplify this expression to find the density in kg/m3.

Question 8

Question

A cube of wood with a density of 600 kg/m3and a side length of 0.1 m is floating

in water. Determine the percentage of the cube’s volume that is submerged in

the water.

Solution

Step 1: Calculate the density of water. Given that the density of water is 1000

kg/m3, this is the density that the cube is displacing.

Step 2: Apply Archimedes’ principle to find the volume of water displaced.

The buoyant force on the cube is equal to the weight of the water displaced,

which can be calculated as:

Fbuoyant = weight of water displaced = density of water×volume of water displaced×g

Step 3: Use the density of the cube to find its total mass. The mass of the

cube can be calculated as:

mass of cube = density of cube ×volume of cube

Step 4: Use the total mass of the cube and acceleration due to gravity to

find the weight of the cube.

weight of cube = mass of cube ×g

Step 5: Equate the weight of the cube to the buoyant force to find the volume

of water displaced.

weight of cube = weight of water displaced

Step 6: Use the volume of water displaced to find the percentage of the cube’s

volume submerged in water. The percentage submerged can be calculated as:

% submerged = volume of water displaced

volume of cube ×100%

Question 9

Question

A cube of solid iron with side length 10 cm is submerged in a container of water.

Find the buoyant force acting on the cube and determine whether the cube will

sink or float.

(Note: The density of iron is 7.87 g/cm3and the density of water is 1.00

g/cm3.)

Solution

Step 1: Calculate the volume of the iron cube. The volume of a cube is given by

V=s3, where sis the side length. In this case, s= 10 cm. Thus, the volume

of the iron cube is:

V= 103cm3= 1000 cm3

Step 2: Calculate the mass of the iron cube. The mass of the iron cube

can be found using the density formula: m=ρV , where ρis the density of the

material. Given that the density of iron is 7.87 g/cm3, the mass of the iron cube

is:

m= 7.87 ×1000 = 7870 g

Step 3: Calculate the weight of the iron cube. The weight of the iron cube is

simply its mass multiplied by the acceleration due to gravity (g= 9.81 m/s2):

W=mg = 7870 ×9.81 = 77252.7 N

Step 4: Calculate the buoyant force acting on the iron cube. According to

Archimedes’ principle, the buoyant force is equal to the weight of fluid displaced

by the object. The volume of water displaced by the iron cube is the same as

the volume of the iron cube, which is 1000 cm3. The mass of water displaced is

given by m=ρV , with ρbeing the density of water. Thus, the buoyant force

is:

Fbuoyant =ρwaterV g = 1 ×1000 ×10−6×9.81 = 0.00981 N

Step 5: Analyze whether the cube will sink or float. If the buoyant force is

greater than the weight of the cube, the cube will float. Otherwise, it will sink.

In this case, the buoyant force is 0.00981 N and the weight of the iron cube is

77252.7 N, so the iron cube will sink.

Therefore, the buoyant force acting on the iron cube is 0.00981 N and the

cube will sink in water.

Question 10

Question

A cube with sides of length 5 cm and density 800 kg/m3is floating in water.

Calculate the depth of immersion of the cube.

Solution

Step 1: Determine the buoyant force acting on the cube. The buoyant force

on an object submerged in a fluid is equal to the weight of the fluid displaced

by the object. The weight of the fluid displaced can be calculated using the

formula:

Fbuoyant =ρfluid ·Vsubmerged ·g

where: ρfluid = 1000 kg/m3(density of water), Vsubmerged is the volume of the

cube submerged, and g= 9.81 m/s2is the acceleration due to gravity.

Step 2: Calculate the volume of the cube submerged. The volume of the

cube submerged can be calculated as the product of the cross-sectional area

submerged and the depth of immersion. Since the cube is floating, the buoyant

force is equal to the weight of the cube. Therefore, the volume of the cube

submerged is equal to the volume of the cube.

Vsubmerged =Vcube = (side length)3= (0.05 m)3

Step 3: Substitute the values and calculate the buoyant force.

Fbuoyant = (1000 kg/m3)·(0.05 m)3·9.81 m/s2

Step 4: Calculate the density of the cube. The density of the cube is given

as 800 kg/m3.

Step 5: Set up an equation using Archimedes’ principle. According to

Archimedes’ principle, the buoyant force acting on the cube is equal to the

weight of the cube. Therefore, we can set up the equation:

Fbuoyant =Fgravity

ρfluid ·Vsubmerged ·g=ρcube ·Vcube ·g

Step 6: Solve for the depth of immersion. Since the volume of the cube

submerged is equal to the total volume of the cube, we can solve for the depth

of immersion using the formula:

depth of immersion = Vsubmerged

Across-sectional

Step 7: Substitute the values and calculate the depth of immersion.

Question 11

Question

A cube of side length aand density ρcis floating in a liquid of density ρl. The

cube is initially completely submerged.

If the cube is slowly brought to the surface of the liquid, determine the

minimum force required to keep the cube from sinking.

Solution

Step 1: Let’s first find the weight of the cube when it is fully submerged. The

weight of the cube is equal to the weight of the liquid that the cube displaces.

Given the density of the cube as ρcand the density of the liquid as ρl, the

volume of the cube displacing the liquid is a3. This means the weight of the

liquid displaced is ρl·a3·g, where gis the acceleration due to gravity.

Therefore, the weight of the cube when fully submerged is ρl·a3·g.

Step 2: When the cube is brought to the surface of the liquid, it displaces

a volume of liquid equal to its own volume. This volume is a3. Therefore, the

buoyant force acting on the cube when it is at the surface is ρl·a3·g.

Step 3: For the cube to stay at the surface of the liquid and not sink, the

force exerted on the cube from above (the force keeping it from sinking) must

be equal to the weight of the cube when fully submerged plus the buoyant force

acting on it at the surface.

Therefore, the minimum force required to keep the cube from sinking is

ρl·a3·g+ρl·a3·g= 2ρl·a3·g.

Question 12

Question

A brass cylinder with a density of 850 kg/m3and a height of 0.2 m is completely

submerged in water. The cylinder has a radius of 0.05 m. Calculate the buoyant

force acting on the cylinder and the depth at which it is submerged in the water.

Solution

Step 1: Calculate the volume of the brass cylinder. The volume of a cylinder is

given by V=πr2h, where ris the radius and his the height. Substituting the

given values, we have:

V=π(0.05 m)2×0.2 m

V=π×0.0025 m2×0.2 m

V= 0.0005πm3

Step 2: Calculate the mass of the brass cylinder. The mass of an object

is given by m= density ×volume. Substituting the given density of the brass

cylinder and the volume calculated above, we have:

m= 850 kg/m3×0.0005πm3

m= 0.425πkg

Step 3: Calculate the weight of the brass cylinder. The weight of an object

is given by W=m×g, where gis the acceleration due to gravity (9.81 m/s2).

W= 0.425π×9.81 N/kg

W≈4.17πN

Step 4: Calculate the buoyant force acting on the cylinder. The buoyant

force acting on an object submerged in a fluid is equal to the weight of the fluid

displaced by the object. The volume of water displaced is equal to the volume

of the cylinder. The density of water is 1000 kg/m3.

Fbuoyant = density of water ×volume ×g

Fbuoyant = 1000 kg/m3×0.0005πm3×9.81 N/kg

Fbuoyant ≈4.905πN

Step 5: Calculate the depth at which the cylinder is submerged. The buoyant

force is equal to the weight of the water displaced by the cylinder when it is

submerged. Let the depth of submersion be d. The cross-sectional area of the

cylinder is πr2. The buoyant force is also equal to the weight of the water in

the volume of the cylinder submerged at depth d.

Fbuoyant = density of water ×volume submerged ×g

4.905π≈1000 ×π×0.052×d×9.81

Solving for d:

d≈4.905

1000 ×0.052×9.81

d≈0.996 m

Therefore, the buoyant force acting on the cylinder is approximately 4.905πN

and the depth at which it is submerged in the water is approximately 0.996 m.

Question 13

Question

A cube of side length 0.5 m and density 800 kg/m3is submerged in water.

Calculate the buoyant force acting on the cube.

Solution

Step 1: Determine the volume of the cube. The volume of a cube is given by

the formula V=s3, where sis the side length of the cube. Substituting s= 0.5

m, we find:

V= (0.5 m)3= 0.125 m3

Step 2: Calculate the mass of the cube. The mass of the cube can be

calculated using the formula m=ρ×V, where ρis the density of the cube.

Substituting ρ= 800 kg/m3and V= 0.125 m3, we get:

m= 800 kg/m3×0.125 m3= 100 kg

Step 3: Determine the buoyant force acting on the cube. The buoyant

force acting on an object submerged in a fluid is equal to the weight of the

fluid displaced by the object. The buoyant force is given by the formula Fb=

ρwater ×Vdisplaced ×g, where ρwater is the density of water, Vdisplaced is the volume

of water displaced by the cube, and gis the acceleration due to gravity.

Step 4: Calculate the volume of water displaced. Since the cube is fully

submerged, the volume of water displaced is equal to the volume of the cube,

which is 0.125 m3.

Step 5: Substitute values and solve for the buoyant force. Substitute ρwater =

1000 kg/m3,Vdisplaced = 0.125 m3, and g= 9.81 m/s2into the formula:

Fb= 1000 kg/m3×0.125 m3×9.81 m/s2= 1226.25 N

Therefore, the buoyant force acting on the cube is 1226.25 N.

Question 14

Question

A cube of wood with side length 10 cm and density 800 kg/m3is floating in

a container of water. The cube is completely submerged with one face parallel

to the surface of the water. What is the depth of the cube below the water

surface?

Solution

Step 1: Determine the buoyant force acting on the cube. The buoyant force

(Fb) acting on an object immersed in a fluid is equal to the weight of the fluid

displaced by the object. This can be calculated using Archimedes’ principle:

Fb=ρwater ·g·Vsubmerged

where: - ρwater = 1000 kg/m3is the density of water, - g= 9.81 m/s2is the

acceleration due to gravity, - Vsubmerged is the volume of the cube submerged in

water.

Step 2: Calculate the volume of the cube submerged in water. Since one

face of the cube is parallel to the surface of the water, the depth of submersion

(h) is equal to the side length of the cube. Therefore, Vsubmerged = 0.12·hm3.

Step 3: Determine the weight of the cube. The weight of the cube is equal

to its mass multiplied by the acceleration due to gravity:

Weight of cube = Density of wood ·V·g

Step 4: Apply the condition for floating. For an object to float, the buoyant

force acting on it must be equal to its weight. Therefore, Fb= Weight of cube.

Step 5: Solve for the depth of submersion. Equating the buoyant force to

the weight of the cube:

ρwater ·g·0.12·h= 800 ·0.13·g

Solving for hgives:

h=800

1000 = 0.8 m

Therefore, the depth of the cube below the water surface is 0.8 meters.

Question 15

Question

A solid sphere of radius Rand density ρsis placed in a liquid of density ρlwith

half of the sphere submerged in the liquid. If the sphere is in equilibrium, find

the ratio of the density of the sphere to the density of the liquid, ρs

ρl.

Solution

1. We start by analyzing the forces acting on the sphere in equilibrium. The

forces acting on the sphere are: - The weight of the sphere, acting downward,

given by W=4

3πR3ρsg- The buoyant force, acting upward, given by Fb=Vlρlg,

where Vl=1

2·4

3πR3is the volume of the liquid displaced by the submerged part

of the sphere

2. Since the sphere is in equilibrium, the net force on the sphere is zero.

This means that the weight of the sphere is equal to the buoyant force. Mathe-

matically, this can be expressed as:

3πR3ρsg=1

2·4

3πR3ρlg

3. Simplifying the equation from step 2, we get:

ρs=1

2ρl

4. Therefore, the ratio of the density of the sphere to the density of the

liquid is:

ρs

ρl

2ρl

ρl

5. Hence, the ratio of the density of the sphere to the density of the liquid

is ρs

ρl=1

Question 16

Question

A cylindrical container of height hand radius ris completely filled with a liquid

of density ρ1. A solid sphere of radius Rand density ρ2is submerged in the

liquid and floats at a depth d. Determine the density ρ2of the sphere in terms

of ρ1,h,r,R, and d.

Solution

Step 1: The buoyant force on the sphere is equal to the weight of the liquid

displaced by the sphere. Step 2: The buoyant force is given by Fb=Vsphereρ1g,

where Vsphere is the volume of the sphere submerged in the liquid. Step 3: The

weight of the sphere is given by Fg=Vsphereρ2g. Step 4: The difference in

depth between the top of the sphere and the liquid surface is d. Step 5: Using

similar triangles, we can relate dto Rand rusing the equation R

r=h−d

h. Step

6: The volume of the sphere is given by Vsphere =4

3πR3, and the volume of

liquid displaced is equal to Vsphere. Step 7: Equating the buoyant force and the

weight of the sphere, we have Vsphereρ1g=Vsphereρ2g. Step 8: Substituting the

expressions for Vsphere and rearranging, we find ρ2=ρ1R3

r3h

h−d, which is the

density of the sphere in terms of the given parameters ρ1,h,r,R, and d.

Question 17

Question

A cylindrical object with a radius of 5 cm and a height of 20 cm is floating

in water with its axis perpendicular to the surface of the water. If the density

of the object is 0.8 g/cm3, determine the buoyant force acting on the object.

(Assume the density of water is 1 g/cm3)

Solution

Step 1: Calculate the volume of the cylinder: The volume of a cylinder is given

by the formula V=πr2h, where ris the radius and his the height. Substitute

r= 5 cm and h= 20 cm into the formula:

V=π×(5 cm)2×20 cm

V= 500πcm3

Step 2: Calculate the mass of the object: The mass of the object can be

found by multiplying the volume by the density. Substitute ρ= 0.8 g/cm3into

the formula:

Mass = ρ×V

Mass = 0.8 g/cm3×500πcm3

Mass = 400πg

Step 3: Calculate the weight of the object: The weight of the object can be

found using the formula W=mg, where mis the mass and gis the acceleration

due to gravity (approximately 9.81 m/s2). Convert mass from grams to kg:

Mass = 400πg=0.4πkg

Substitute m= 0.4πkg into the formula:

W= 0.4πkg ×9.81 m/s2

W= 3.924πN

Step 4: Calculate the buoyant force: The buoyant force exerted on an object

in a fluid is equal to the weight of the fluid displaced by the object. It can be

calculated using the formula Fb=ρfluid ×Vdisplaced ×g, where ρfluid is the density

of the fluid, Vdisplaced is the volume of fluid displaced, and gis the acceleration

due to gravity. In this case, since the object is floating, the buoyant force is

equal to the weight of the object. Therefore, the buoyant force is equal to the

weight of the object calculated in Step 3:

Fb= 3.924πN

Question 18

Question

A hollow sphere made of a certain material has a radius of 0.5 m and a mass of

20 kg. When the sphere is placed in a tank of water, it floats with 60

Solution

Step 1: Let’s first calculate the volume of the sphere that is submerged in water.

The volume of the sphere is given by the formula V=4

3πr3, where ris the radius

of the sphere. Given that the radius r= 0.5 m, the total volume of the sphere

is:

Vtotal =4

3π(0.5)3=4

3π×0.125 = 1

6πm3

Since 60

Vsubmerged = 0.6×Vtotal = 0.6×1

6π=1

10πm3

Step 2: Next, let’s calculate the buoyant force acting on the sphere. The

buoyant force is equal to the weight of the water displaced by the sphere. Since

the sphere is floating, the buoyant force Fbis equal to the weight of the sphere

Wsphere. The weight of the sphere Wsphere can be calculated as:

Wsphere =mg

Wsphere = 20 kg ×9.8 m/s2= 196 N

Therefore, the buoyant force Fb= 196 N.

Step 3: We can now calculate the weight of water displaced by the submerged

volume of the sphere. The weight of water displaced is given by the formula

Wwater =Vsubmerged ·ρwater ·g, where ρwater is the density of water (1000 kg/m3),

and gis the acceleration due to gravity (9.8 m/s2).

Wwater =1

10π·1000 ·9.8 = 30πN

Step 4: Finally, we can calculate the density of the material of the sphere.

According to Archimedes’ principle, the buoyant force Fbis equal to the weight

of the fluid displaced, which is Wwater in this case. Therefore, the density of the

material of the sphere ρsphere can be calculated as:

ρsphere =Wsphere

Vsubmerged ·g

ρsphere =196

10 π·9.8=196

3π≈20 kg/m3

Therefore, the density of the material of the sphere is approximately 20

kg/m3.

Question 19

Question

A cylinder with a density of 800 kg/m3and a height of 10 cm floats on a liquid

with a density of 1200 kg/m3. Calculate the diameter of the cylinder if the top

surface of the cylinder is at the liquid surface. Assume that the pressure at the

liquid surface is 1 atm.

Solution

Step 1: Determine the volume of the submerged portion of the cylinder. Let

Vsub be the volume of the submerged portion of the cylinder. Since the cylinder

floats at the liquid surface, the weight of the liquid displaced by the submerged

portion equals the weight of the cylinder.

Weight of cylinder = Weight of liquid displaced

Volume of cylinder ×Density of cylinder ×g=Vsub ×Density of liquid ×g

π(d

2)2×0.1×800 = Vsub ×1200

πd2

4×800 = Vsub ×1200

Vsub =πd2

Step 2: Calculate the weight of the cylinder and the buoyant force. The

weight of the cylinder is equal to the weight of the liquid displaced by the

submerged portion of the cylinder.

Weight of cylinder = Volume of cylinder ×Density of cylinder ×g

π(d

2)2×0.1×800 = πd2

4×0.1×800 ×g

πd2

4×800 = πd2

4×0.1×800 ×g

Weight of cylinder = πd2

4×800 ×9.81

Step 3: Solve for the diameter of the cylinder. Since the top of the cylinder

is at the liquid surface, the net force acting on the cylinder is zero.

Weight of cylinder = Buoyant force

πd2

4×800 ×9.81 = Vsub ×1200 ×g

πd2

4×800 ×9.81 = πd2

3×1200 ×9.81

d=r4

d≈1.15 meters

Therefore, the diameter of the cylinder is approximately 1.15 meters.

Question 20

Question

A cube of wood with a side length of 10 cm and a density of 800 kg/m3floats in

water. Determine the depth to which the cube is submerged in water. Assume

the density of water is 1000 kg/m3.

Solution

Step 1: First, let’s determine the weight of the cube. The weight of the cube is

equal to the force of gravity acting on it:

Weight of cube = Mass ×Acceleration due to gravity

Given that density ρcube = 800 kg/m3, side length L= 10 cm = 0.1m, and

density of water ρwater = 1000 kg/m3, the mass of the cube is:

Mass of cube = ρcube ×Volume of cube

Step 2: Calculate the volume of the cube. The volume of the cube is given

by Vcube =L3.

Substituting L= 0.1minto the formula gives:

Vcube = (0.1)3= 0.001 m3

Step 3: Now, compute the mass of the cube.

Mass of cube = 800 ×0.001 = 0.8kg

Step 4: Calculate the weight of the cube. The acceleration due to gravity is

approximately 9.81 m/s2. Hence,

Weight of cube = 0.8×9.81 = 7.848 N

Step 5: Next, we need to find the buoyant force acting on the cube. Ac-

cording to Archimedes’ principle, the buoyant force is equal to the weight of the

water displaced by the cube. The buoyant force is given by:

Buoyant force = Weight of water displaced = ρwater ×g×Vsubmerged

where Vsubmerged is the volume of the cube submerged in water.

Step 6: Determine the volume of the cube submerged in water. Let dbe

the depth to which the cube is submerged. Then, the volume submerged can

be expressed as Vsubmerged = 0.12×d.

Step 7: We can now derive an expression for the buoyant force.

1000 ×9.81 ×0.01d= 7.848

Solving for dgives:

d=7.848

981 ≈0.008 m

Therefore, the cube is submerged to a depth of 0.008 meters in water.

Question 21

Question

A cube of wood with sides of length 10 cm and density 0.8 g/cm3floats in water.

Calculate the depth to which the cube is submerged in the water.

Solution

Step 1: The buoyant force Fbacting on the cube is equal to the weight of

the water displaced by the cube. Step 2: The weight of the water displaced

by the cube is equal to the weight of the cube itself. Step 3: The weight of

the cube is given by W=mg, where mis the mass of the cube and gis the

acceleration due to gravity. Step 4: The mass of the cube is given by m=ρV ,

where ρis the density of the cube and Vis the volume of the cube. Step

5: The volume of the cube is V= (10 cm)3. Step 6: Substituting the values

into the equations, we find m= 0.8 g/cm3×(10 cm)3and W= 0.8 g/cm3×

(10 cm)3×9.8 m/s2. Step 7: The buoyant force Fbis equal to the weight of the

cube, so Fb= 0.8 g/cm3×(10 cm)3×9.8 m/s2. Step 8: The buoyant force Fb

is also equal to the weight of the water displaced by the cube, which is given

by ρwater ×g×Vsubmerged. Step 9: Setting these two equations equal to each

other, we have ρwater ×g×Vsubmerged = 0.8 g/cm3×(10 cm)3×9.8 m/s2. Step

10: Solving for Vsubmerged, we find Vsubmerged =0.8 g/cm3×(10 cm)3×9.8 m/s2

ρwater ×g. Step

11: The depth to which the cube is submerged can be calculated using the

formula Vsubmerged =A×d, where Ais the area of one face of the cube and d

is the depth to which the cube is submerged. Step 12: Substituting the values,

we find 10 cm2×d=0.8 g/cm3×(10 cm)3×9.8 m/s2

ρwater ×g. Step 13: Solving for d, we get

d=0.8 g/cm3×(10 cm)3×9.8 m/s2

ρwater ×g×10 cm2. Step 14: Finally, we can calculate the depth dby

plugging in the values for ρwater = 1 g/cm3and g= 9.8 m/s2.

Question 22

Question

A cube of side length aand density ρ1is floating in a liquid of density ρ2. Given

that the cube is immersed in the liquid to a depth h, determine the portion of

the cube that is submerged in the liquid.

Solution

Step 1: First, let’s consider the forces acting on the cube. The forces are the

gravitational force pulling the cube downwards and the buoyant force push-

ing the cube upwards. Step 2: The weight of the cube can be calculated as

W=ρ1a3g, where gis the acceleration due to gravity. Step 3: The buoyant

force acting on the cube can be calculated as Fbuoyant =ρ2Vsubmergedg, where

Vsubmerged is the volume of the cube that is submerged. Step 4: According to

Archimedes’ principle, the buoyant force is equal to the weight of the liquid

displaced. So, Fbuoyant =ρ2Vsubmergedg=ρ2a2hg. Step 5: Setting the weight of

the cube equal to the buoyant force, we have ρ1a3g=ρ2a2hg. Step 6: Solving

for h, we find h=ρ1

ρ2a. Step 7: Therefore, the portion of the cube submerged

in the liquid is h

a=ρ1

ρ2.

Question 23

Question

A cube of wood with a density of 650 kg/m3and side length 0.10 m is floating

in a liquid with a density of 800 kg/m3. Calculate the height of the cube that

is submerged in the liquid.

Solution

Step 1: First, we need to identify the forces acting on the cube. The weight of

the cube is acting downwards, and the buoyant force is acting upwards.

Step 2: The weight of the cube can be calculated using the formula:

Weight = mass ×acceleration due to gravity

Step 3: The mass of the cube can be found using the formula:

mass = density ×volume

Step 4: The volume of the cube is given by:

Volume = side length3

Step 5: Substituting the given values into the formulas, we find:

mass = 650 kg/m3×(0.10 m)3

Step 6: Calculate the mass of the cube.

Step 7: Next, we calculate the weight of the cube using the mass we found

and the acceleration due to gravity.

Step 8: The buoyant force acting on the cube is equal to the weight of the

liquid displaced by the cube.

Step 9: The volume of liquid displaced by the cube is equal to the volume

of the submerged part of the cube.

Step 10: The buoyant force can be calculated using the density of the liquid

and the volume of the submerged cube.

Step 11: Now, we set up the equilibrium condition: weight of the cube =

buoyant force.

Step 12: Solve the equation for the height of the cube that is submerged in

the liquid.

Step 13: Calculate the height of the cube that is submerged in the liquid.

Question 24

Question

A 2 m long, 1 m wide, and 1 m high wooden box is floating in water with its

long dimension perpendicular to the surface. What is the mass of the box if it is

floating with only its top 10 cm above the water surface? (Assume the density

of water is 1000 kg/m3and the acceleration due to gravity is 9.81 m/s2.)

Solution

Step 1: Determine the volume of the wooden box submerged in water. The

volume of the wooden box that is submerged in water can be calculated as the

product of its length, width, and the depth it is submerged. Given that the

depth submerged is 1 m - 0.1 m = 0.9 m,

Vsubmerged = 2 m ×1 m ×0.9 m = 1.8 m3

Step 2: Calculate the weight of the water displaced by the submerged part

of the box. The weight of the water displaced is equal to the weight of the box.

The weight of water displaced is given by mg =ρV g, where ρis the density

of water, Vis the volume of water displaced, and gis the acceleration due to

gravity. Substituting the known values, we get:

m×9.81 = 1000 ×1.8×9.81

Step 3: Solve for the mass of the box.

m=1000 ×1.8×9.81

9.81 = 1800 kg

Therefore, the mass of the box is 1800 kg.

Question 25

Question

A cylindrical container with a diameter of 10 cm and a height of 20 cm is

completely filled with water. A block of wood with a density of 0.8 g/cm3and a

volume of 100 cm3is placed into the water-filled container. The block of wood

floats at a certain level in the water. Calculate the depth to which the block of

wood sinks below the surface of the water.

Solution

Step 1: First, let’s find the weight of the block of wood. Given density of wood,

ρw= 0.8 g/cm3

Given volume of the block of wood, Vw= 100 cm3

The weight of the block of wood can be calculated as:

Weight of wood = ρw×Vw×g

Weight of wood = 0.8 g/cm3×100 cm3×9.81 m/s2

Weight of wood = 784.8 g

Step 2: Now, let’s find the buoyant force acting on the block of wood. The

buoyant force can be calculated using Archimedes’ Principle:

Buoyant force = ρwater ×Vdisplaced ×g

Since the block of wood is floating, the buoyant force is equal to the weight

of the wood. We can rearrange the formula to solve for Vdisplaced:

Vdisplaced =Weight of wood

ρwater ×g

Vdisplaced =784.8 g

1 g/cm3×9.81 m/s2

Vdisplaced =784.8 g

9.81 N/m3

Vdisplaced = 80 cm3

Step 3: Now, let’s calculate the height to which the block of wood sinks

below the surface of the water. The volume of the block of wood submerged in

water is the volume of water displaced, which we found to be 80 cm3. Since the

container is cylindrical, we can use the formula for the volume of a cylinder to

find the height submerged:

Vsubmerged =π×r2×h

80 cm3=π×10 cm

22

×h

80 cm3=π×25 cm2×h

h=80 cm3

25 π

h≈1.02 cm

Therefore, the block of wood sinks approximately 1.02 cm below the surface

of the water.

Question 26

Question

A solid cube of density ρand side length Lis placed in a liquid of density 2ρ.

If the cube floats with half of its height submerged in the liquid, what is the

length of the cube that is submerged?

Solution

Step 1: Let’s first determine the buoyant force acting on the cube. The buoyant

force is given by Fb=ρliquidVsubmergedg, where ρliquid is the density of the

liquid, Vsubmerged is the volume of the cube submerged in the liquid, and gis

the acceleration due to gravity. Given that the cube floats with half of its height

submerged, the volume submerged is Vsubmerged =1

2L2·L=1

2L3.

Step 2: Next, we calculate the weight of the cube. The weight of the cube

is given by W=ρV g, where ρis the density of the cube, Vis the volume of

the cube, and gis the acceleration due to gravity. The volume of the cube is

V=L3.

Step 3: Since the cube is in equilibrium, the weight of the cube must be

balanced by the buoyant force. So, we have W=Fb. Substitute the expressions

for Wand Fb:ρL3g= 2ρ·1

2L3g.

Step 4: Simplifying the equation above, we find that: ρL3g=ρL3g. This

equation implies that the weight of the cube is equal to the buoyant force exerted

on it.

Step 5: Therefore, the length of the cube that is submerged in the liquid is

2L.

Question 27

Question

A metal block with a density of 5000 kg/m3and volume of 0.02 m3is floating

in a pool of water. What is the buoyant force acting on the block?

Solution

Step 1: First, we calculate the weight of the block. Given: Density of the block,

ρblock = 5000 kg/m3Volume of the block, Vblock = 0.02 m3Acceleration due to

gravity, g= 9.81 m/s2

Using the formula for weight:

Weight of the block = ρblock ·Vblock ·g

Weight of the block = 5000 kg/m3·0.02 m3·9.81 m/s2

Weight of the block = 981 N

Step 2: Next, we calculate the weight of the water displaced by the block.

Given: Density of water, ρwater = 1000 kg/m3Volume of the water displaced,

Vdisplaced =Vblock = 0.02 m3

Using the formula for weight:

Weight of the displaced water = ρwater ·Vdisplaced ·g

Weight of the displaced water = 1000 kg/m3·0.02 m3·9.81 m/s2

Weight of the displaced water = 196.2 N

Step 3: Since the block is floating, the buoyant force is equal to the weight

of the water displaced. Therefore, the buoyant force acting on the block is 196.2

Question 28

Question

A cylindrical object with a radius of 0.1 m and a height of 0.3 m is placed in a

container of water. The density of the object is 800 kg/m3, and the density of

water is 1000 kg/m3. Determine the buoyant force acting on the object when it

is fully submerged in water.

Solution

Step 1: Calculate the volume of the cylindrical object using the formula for the

volume of a cylinder, V=πr2h, where ris the radius and his the height.

Volume of cylinder = π·(0.1 m)2·0.3 m = 0.009 m3

Step 2: Calculate the total mass of the object by multiplying its volume by

its density.

Mass of object = Volume of cylinder×Density of object = 0.009 m3×800 kg/m3= 7.2 kg

Step 3: Calculate the volume of water displaced by the submerged object.

Since the object is fully submerged, the volume of water displaced is equal to

the volume of the object.

Volume of water displaced = 0.009 m3

Step 4: Calculate the buoyant force acting on the object using the formula

for buoyant force, Fb=ρ·g·V, where ρis the density of the fluid (water in

this case), gis the acceleration due to gravity, and Vis the volume of water

displaced.

Fb= 1000 kg/m3×9.8 m/s2×0.009 m3= 88.2 N

Therefore, the buoyant force acting on the object when it is fully submerged

in water is 88.2 N.

Question 29

Question

A cube of wood with sides of length 10 cm and density 800 kg/m3is floating

in water as shown in the figure. Determine the distance hthat the cube is

submerged in the water.

cube_in_water.png

Solution

Step 1: The buoyant force Fbacting on an object submerged in a fluid is equal

to the weight of the fluid displaced by the object. This can be calculated using

the formula Fb=ρV g, where ρis the density of the fluid, Vis the volume of

the fluid displaced, and gis the acceleration due to gravity.

Step 2: The weight of the cube is equal to the weight of the water it displaces.

This can be expressed as mg =ρcubeVcubeg, where mis the mass of the cube,

ρcube is the density of the cube, and Vcube is the volume of the cube.

Step 3: Since the cube is floating, the weight of the cube is balanced by the

buoyant force acting on it. Thus, mg =Fb. Substituting the expressions for mg

and Fb, we get ρcubeVcubeg=ρV g.

Step 4: The volume of the cube can be expressed as Vcube =l3, where lis

the side length of the cube.

Step 5: Substituting the expressions for Vcube into the equation from Step

3, we obtain ρcubel3=ρhl2, where his the height submerged in the water.

Step 6: Solving for h, we find h=ρcube

ρl.

Step 7: Substituting the given values, we have h=800 kg/m3

1000 kg/m3·0.1 m.

Step 8: Calculating h, we find h= 0.08 m or 8 cm. Therefore, the cube is

submerged 8 cm into the water.

Question 30

Question

A cube of wood with a density of 600 kg/m3and a side length of 0.1 m is floating

in water. Calculate the depth to which the cube is submerged in water.

Solution

Step 1: Let’s first calculate the weight of the cube, which is given by the formula

W=m·g, where mis the mass of the cube and gis the acceleration due to

gravity (9.81 m/s2). The volume of the cube can be calculated as V= (0.1 m)3=

0.001 m3. The mass of the cube can be calculated as m=ρ·V, where ρis the

density of the cube. Substitute the values to get m= 600 kg/m3×0.001 m3= 0.6

kg.

Therefore, the weight of the cube is W= 0.6 kg ×9.81 m/s2= 5.886 N.

Step 2: According to Archimedes’ principle, the buoyant force acting on the

cube is equal to the weight of the water displaced by the cube. This buoyant

force (Fb) can be calculated using the formula Fb=ρwater ·g·Vsubmerged, where

ρwater is the density of water.

Since the cube is floating, the weight of the cube is balanced by the buoyant

force. Therefore, 5.886 N = ρwater ·9.81 m/s2·0.1 m ·Asubmerged.

Step 3: Solving for Asubmerged, we have 0.1ρwater = 5.886. Thus, Asubmerged =

5.886

0.1×1000×9.81 = 0.006 m.

Therefore, the cube is submerged to a depth of 0.006 m in water.

Question 31

Question

A cube of side length 0.2 m and density 800 kg/m3is floating in water with 1/5

of its volume submerged. Find the density of the water.

Solution

Step 1: Calculate the volume of the cube. Given s= 0.2 m (side length) and

Vsubmerged =1

5Vcube, we can find the total volume of the cube. The volume of

the cube is Vcube =s3= (0.2)3= 0.008 m3. Since 1/5 of the cube’s volume is

submerged, the volume submerged is Vsubmerged =1

5×0.008 = 0.0016 m3.

Step 2: Use Archimedes’ principle to find the density of water. The buoyant

force experienced by the cube is equal to the weight of the water displaced.

The buoyant force can be calculated by Fbuoyant =ρw·Vsubmerged ·g, where ρw

is the density of water (to be found) and gis the acceleration due to gravity.

The weight of the cube is equal to the weight of the water it displaces, so

Fbuoyant =mg, where mis the mass of the cube and gis the acceleration due

to gravity. Since the cube is floating, the weight of the cube equals the buoyant

force: mg =ρcube ·Vsubmerged ·g. Substitute the known values: 800 ×0.0016 ×

9.8 = ρw×0.0016 ×9.8. Solving for ρwgives ρw= 800 kg/m3.

Therefore, the density of water is 800 kg/m3.

Question 32

Question

A cube of side length aand mass mis submerged in a liquid of density ρ. The

cube is initially floating at the surface of the liquid. Calculate the depth to

which the cube sinks when a small weight ∆mis added to the top of the cube.

Given: V=a3,A=a2, where Vis the volume of the cube and Ais the area of

the base of the cube.

Solution

Step 1: The weight of the cube is equal to the weight of the liquid displaced by

the cube. This is based on Archimedes’ principle. Let ρlbe the density of the

liquid.

Weight of cube = m=ρlV g

Step 2: When a small weight ∆mis added, the new total weight of the cube

is m+ ∆m. The cube will sink until the buoyant force equals its total weight.

Buoyant force = (ρl−ρ)V g

Step 3: At equilibrium, the total weight of the cube is balanced by the

buoyant force.

m+ ∆m= (ρl−ρ)V g

⇒∆m= (ρl−ρ)V g −m

Step 4: The depth to which the cube sinks can be calculated using the

concept of pressure. The pressure difference between the top and bottom of the

cube causes a net force that pushes the cube upward.

∆p=ρlgh

Fnet = (ρl−ρ)V g =A∆p

⇒h=(ρl−ρ)V g

Aρlg

⇒h=(ρl−ρ)a3

a2ρl

= (a−ρ

ρl

So, the depth to which the cube sinks when a weight ∆mis added is (a−ρ

ρl)a.

Question 33

Question

A cube of side length 10 m and density 500 kg/m3is floating in a liquid of density

1000 kg/m3. Calculate the depth to which the cube is submerged in the liquid.

(Assume the acceleration due to gravity is 9.81 m/s2.)

Solution

Step 1: The buoyant force on the cube is equal to the weight of the liquid

displaced by the cube. Step 2: The weight of the cube is equal to the weight of

the liquid displaced. Step 3: The weight of the cube can be calculated as mg,

where mis the mass of the cube and gis the acceleration due to gravity. Step

4: The mass of the cube can be calculated as V·ρcube, where Vis the volume

of the cube and ρcube is the density of the cube. Step 5: The volume of the

cube is 103m3. Step 6: Substituting values into the equation for the weight of

the cube gives mg = 103·500 ·9.81. Step 7: The weight of the liquid displaced

is equal to the weight of the cube, so mg =Vsubmerged ·ρliquid ·g. Step 8: The

volume of the cube submerged in the liquid is equal to the submerged depth

times the cross-sectional area of the cube (10 m ×10 m). Step 9: Substituting in

the known values gives 103·500 ·9.81 = 10 ×10 ×depth ×1000 ×9.81. Step 10:

Solving for the depth, we find that the cube is submerged to a depth of 5 m .

Question 34

Question

A cube of side length aand density ρ1is partially submerged in a liquid of

density ρ2such that one face of the cube is parallel to the liquid’s surface. If

the cube is floating in equilibrium, determine the depth of submersion of the

cube in terms of a,ρ1, and ρ2.

Solution

Let’s denote the depth of submersion of the cube as d. To find d, we need to

consider the forces acting on the cube.

Step 1: Identify the forces acting on the cube. In this scenario, the

following forces act on the cube: - The weight of the cube acting downward

(Wc). - The buoyant force acting upward (Fb). - The normal force acting on

the cube due to the liquid (N).

Step 2: Set up the forces equation. Since the cube is floating in equi-

librium, the net force acting on it is zero in both the vertical and horizontal

directions. Therefore, we have:

(Wc−Fb= 0 (vertical direction)

N= 0 (horizontal direction)

Step 3: Calculate the weight of the cube (Wc). The weight of the

cube is given by:

Wc=mc·g=ρ1·Vc·g

where mcis the mass of the cube, Vcis its volume, and gis the acceleration due

to gravity.

Step 4: Calculate the buoyant force (Fb). The buoyant force acting on

the cube is given by:

Fb=ρ2·Vc·g

where ρ2is the density of the liquid.

Step 5: Equate the weight and buoyant force. Substitute the expres-

sions for Wcand Fbinto the vertical forces equation:

ρ1·Vc·g−ρ2·Vc·g= 0

Step 6: Find the depth of submersion (d) in terms of the cube’s

side length (a), ρ1, and ρ2.The volume of the submerged part of the cube

can be expressed as Vs=a2d. Thus, we have:

ρ1·a2d·g−ρ2·a2d·g= 0

Solving for dgives:

d=ρ2

ρ1

·a

Therefore, the depth of submersion of the cube in terms of a,ρ1, and ρ2is

ρ2

ρ1

·a.

Question 35

Question

A cube of iron with a density of 7.87 g/cm3is submerged in a container filled

with water. The cube has a side length of 5 cm. Determine the buoyant force

acting on the cube and whether the cube will float or sink in water.

Solution

Step 1: Find the volume of the iron cube. The volume of the cube can be

calculated using the formula: V=s3, where sis the side length of the cube.

Given that the side length of the cube is 5 cm, we have:

V= 53= 125 cm3

Step 2: Calculate the mass of the iron cube. The mass of the iron cube can

be found using the formula: m= density ×V. Substitute the values into the

formula:

m= 7.87 g/cm3×125 cm3= 983.75 g

Step 3: Determine the weight of the iron cube. The weight can be calculated

using the formula: Fgravity =m×g, where gis the acceleration due to gravity

(9.81 m/s2). Convert the mass from grams to kilograms:

m= 983.75 g = 0.98375 kg

Step 4: Calculate the buoyant force acting on the iron cube. The buoyant

force is equal to the weight of the water displaced, so the buoyant force acting

on the iron cube is:

Fbuoyant = 1226.25 N

Question 2

Question

A cube of steel with a side length of 10 cm and a density of 7,800 kg/m3is

placed in a container filled with water. What is the buoyant force acting on the

cube?

Solution

Step 1: Calculate the volume of the cube. Given the side length of the cube,

the volume can be calculated as:

Volume of cube = (side length)3

Volume of cube = (0.1 m)3= 0.001 m3

Step 2: Determine the volume of water displaced by the cube. The volume

of water displaced by the cube is equal to the volume of the cube, since the cube

is fully submerged.

Volume of water displaced = 0.001 m3

Step 3: Calculate the buoyant force. The buoyant force is equal to the weight

of the water displaced by the cube.

Buoyant force = Density of water×Volume of water displaced×Acceleration due to gravity

Buoyant force = 1000 kg/m3×0.001 m3×9.8 m/s2

Buoyant force = 9.8 N

Therefore, the buoyant force acting on the cube is 9.8 N.

Question 3

Question

A metal block of mass 500 kg and volume 0.1 m3is immersed in a fluid. De-

termine the buoyant force acting on the block if the density of the fluid is 800

kg/m3.

Solution

Step 1: Calculate the density of the metal block using the formula ρ=m

V, where

ρis the density, mis the mass, and Vis the volume.

ρ=500 kg

0.1 m3= 5000 kg/m3

Step 2: Determine the apparent weight of the block in the fluid using the

formula Wapparent =m·g−Fbuoyant, where Wapparent is the apparent weight,

mis the mass, gis the acceleration due to gravity, and Fbuoyant is the buoyant

force.

Wapparent = 500 kg ·9.8 m/s2−Fbuoyant

Step 3: Apply Archimedes’ principle, which states that the buoyant force

Fbuoyant acting on a submerged object is equal to the weight of the fluid displaced

by the object.

Fbuoyant = density of fluid ·g·Vdisplaced

Step 4: Calculate the volume of fluid displaced by the block using the formula

Vdisplaced =m

ρfluid , where ρfluid is the density of the fluid.

Vdisplaced =500 kg

800 kg/m3= 0.625 m3

Step 5: Substitute the values into the expression for Fbuoyant.

Fbuoyant = 800 kg/m3·9.8 m/s2·0.625 m3

Step 6: Calculate the buoyant force.

Fbuoyant = 4900 N

Therefore, the buoyant force acting on the block is 4900 N.

Question 4

Question

A cylindrical wooden log with a diameter of 0.5 m and a length of 2 m floats in

water. The density of water is 1000 kg/m3. Determine the height by which the

log protrudes from the water surface.

Solution

Step 1: Find the volume of the wooden log.

To find the volume of the log, we use the formula for the volume of a cylinder:

Vlog =πr2h

where r= 0.25 m (radius) and h= 2 m (height).

Vlog =π(0.25)2×2

Vlog =π

4×2

Vlog =π

2m3

Step 2: Find the weight of the wooden log.

The weight of the log can be calculated using the formula:

Wlog =mlog ×g

where mlog is the mass of the log and g= 9.81 m/s2. Since density ρ=

mass/volume, we have mlog =ρwood ×Vlog, with ρwood being the density of

wood. Given that the density of wood is approximately 700 kg/m3, we have:

mlog = 700 ×π

mlog = 350πkg

Therefore:

Wlog = 350π×9.81

Wlog = 3433.5πN

Step 3: Find the buoyant force on the log.

According to Archimedes’ principle, the buoyant force Fbuoy on an object in a

fluid is equal to the weight of the fluid displaced by the object. We can calculate

the buoyant force using the formula:

Fbuoy =ρwater ×Vsubmerged ×g

where ρwater is the density of water. Since the log is floating, the volume of the

log submerged in water is equal to the volume of water it displaces.

Vsubmerged =Vlog

Therefore:

Fbuoy = 1000 ×π

2×9.81

Fbuoy = 4905πN

Step 4: Determine the height of the log above water level.

When the log is floating, the buoyant force Fbuoy acting on it is equal to the

weight Wlog of the log.

Fbuoy =Wlog

4905π= 3433.5π

1471.5π= 1000πr2h

1.4715 = r2h

Given r= 0.25 m, we can solve for h(the height by which the log protrudes

from the water surface).

1.4715 = 0.252h

1.4715 = 0.0625h

h=1.4715

0.0625

h≈23.544 m

Therefore, the log protrudes approximately 23.544 meters above the water sur-

face.

Question 5

Question

A cube of wood with side length 10 cm and density 0.6 g/cm3is floating in a

container of water. What is the minimum side length of a cube of gold (density

= 19.3 g/cm3) that will cause the wooden cube to sink when both cubes are

submerged in water?

Solution

Step 1: First, we need to determine the weight of the wooden cube. The weight

of the wooden cube is equal to the weight of the water displaced by the cube,

which is equal to the buoyant force acting on the cube. The volume of the

wooden cube can be calculated as Vwood = (10 cm)3= 1000 cm3. The weight of

the wooden cube is then:

Wwood =Vwood ·densitywood ·g= 1000 cm3×0.6 g/cm3×9.8 m/s2

Step 2: Next, we calculate the minimum volume of gold cube required to

sink the wooden cube. Let the side length of the gold cube be scm. The volume

of the gold cube is Vgold =s3. The buoyant force acting on the gold cube is

equal to the weight of the water displaced by the gold cube. This weight is

equal to the sum of the weight of the water displaced by the wooden cube and

the weight of the wooden cube, since both are submerged.

Step 3: Thus, to sink the wooden cube, the buoyant force acting on the gold

cube (Vgold ×densitygold ×g) must be greater than the weight of the wooden

cube. This can be expressed as:

Vgold ×densitygold ×g > Wwood

Step 4: Substitute in the expressions for Vgold,Wwood, and the given densities

to solve for the minimum side length sof the gold cube.

Question 6

Question

A cube of iron with a side length of 10 cm and a density of 7.87 g/cm3is

submerged in a container of mercury. If the cube floats with 30

Solution

Step 1: Determine the volume of the cube. The volume of a cube is given by

the formula V=s3, where sis the side length. Substituting s= 10 cm, we find:

V= 103cm3= 1000 cm3

Step 2: Calculate the mass of the iron cube. The mass of an object is given

by the product of its volume and density. The mass mof the cube can be

calculated as follows:

m= Volume ×Density = 1000 ×7.87 = 7870 g

Step 3: Determine the volume of the cube submerged in mercury. Since 30

Vsubmerged = 0.3×1000 cm3= 300 cm3

Step 4: Calculate the buoyant force acting on the cube. The buoyant force

Fbon an object submerged in a fluid is equal to the weight of the fluid displaced.

The weight of the mercury displaced by the submerged portion of the cube can

be calculated using its density (ρHg) and the volume submerged (Vsubmerged):

Fb=ρHg ×Vsubmerged ×g

where gis the acceleration due to gravity.

Step 5: Find the density of mercury. Equating the weight of the iron cube

to the buoyant force on it gives:

m×g=ρHg ×Vsubmerged ×g

Solving for ρHg:

ρHg =m

Vsubmerged

=7870 g

300 cm3= 26.23 g/cm3

Therefore, the density of mercury is 26.23 g/cm3.

Question 7

Question

A solid cube of aluminum with sides of length 10 cm is placed in a container

of water. The cube floats with 3 cm of its height above the water surface.

Calculate the density of aluminum. (Density of water = 1000 kg/m3)

Solution

Step 1: Find the volume of the cube submerged in water. The submerged

volume is given by the area of the base of the cube multiplied by the height

submerged. The area of the base is (10 cm)2= 100 cm2. The height submerged

is 10 cm −3 cm = 7 cm. Therefore, the volume submerged is 100 cm2×7 cm =

700 cm3= 0.0007 m3.

Step 2: Find the weight of water displaced by the cube. The buoyant force

acting on the cube is equal to the weight of water displaced by the cube. The

weight of water displaced is equal to the volume submerged multiplied by the

density of water. So, the weight of water displaced is 0.0007 m3×1000 kg/m3=

0.7 kg.

Step 3: Find the weight of the aluminum cube. The weight of the aluminum

cube is equal to its mass multiplied by the acceleration due to gravity. Let mbe

the mass of the cube and gbe the acceleration due to gravity (approximately

9.81 m/s2). Then, the weight of the aluminum cube is m×9.81 N.

Step 4: Use Archimedes’ principle to find the density of aluminum. Since the

cube floats, the weight of the aluminum cube is balanced by the buoyant force.

Therefore, m×9.81 = 0.7 kg ×9.81 m/s2. Solving for m, we have m=0.7

9.81 kg.

The density of aluminum is given by density = mass

volume =m

0.001 m3. Substitute

the values to find the density of aluminum.

Step 5: Evaluate the density of aluminum. The density of aluminum is

0.7/9.81

0.001 kg/m3. Simplify this expression to find the density in kg/m3.

Question 8

Question

A cube of wood with a density of 600 kg/m3and a side length of 0.1 m is floating

in water. Determine the percentage of the cube’s volume that is submerged in

the water.

Solution

Step 1: Calculate the density of water. Given that the density of water is 1000

kg/m3, this is the density that the cube is displacing.

Step 2: Apply Archimedes’ principle to find the volume of water displaced.

The buoyant force on the cube is equal to the weight of the water displaced,

which can be calculated as:

Fbuoyant = weight of water displaced = density of water×volume of water displaced×g

Step 3: Use the density of the cube to find its total mass. The mass of the

cube can be calculated as:

mass of cube = density of cube ×volume of cube

Step 4: Use the total mass of the cube and acceleration due to gravity to

find the weight of the cube.

weight of cube = mass of cube ×g

Step 5: Equate the weight of the cube to the buoyant force to find the volume

of water displaced.

weight of cube = weight of water displaced

Step 6: Use the volume of water displaced to find the percentage of the cube’s

volume submerged in water. The percentage submerged can be calculated as:

% submerged = volume of water displaced

volume of cube ×100%

Question 9

Question

A cube of solid iron with side length 10 cm is submerged in a container of water.

Find the buoyant force acting on the cube and determine whether the cube will

sink or float.

(Note: The density of iron is 7.87 g/cm3and the density of water is 1.00

g/cm3.)

Solution

Step 1: Calculate the volume of the iron cube. The volume of a cube is given by

V=s3, where sis the side length. In this case, s= 10 cm. Thus, the volume

of the iron cube is:

V= 103cm3= 1000 cm3

Step 2: Calculate the mass of the iron cube. The mass of the iron cube

can be found using the density formula: m=ρV , where ρis the density of the

material. Given that the density of iron is 7.87 g/cm3, the mass of the iron cube

is:

m= 7.87 ×1000 = 7870 g

Step 3: Calculate the weight of the iron cube. The weight of the iron cube is

simply its mass multiplied by the acceleration due to gravity (g= 9.81 m/s2):

W=mg = 7870 ×9.81 = 77252.7 N

Step 4: Calculate the buoyant force acting on the iron cube. According to

Archimedes’ principle, the buoyant force is equal to the weight of fluid displaced

by the object. The volume of water displaced by the iron cube is the same as

the volume of the iron cube, which is 1000 cm3. The mass of water displaced is

given by m=ρV , with ρbeing the density of water. Thus, the buoyant force

is:

Fbuoyant =ρwaterV g = 1 ×1000 ×10−6×9.81 = 0.00981 N

Step 5: Analyze whether the cube will sink or float. If the buoyant force is

greater than the weight of the cube, the cube will float. Otherwise, it will sink.

In this case, the buoyant force is 0.00981 N and the weight of the iron cube is

77252.7 N, so the iron cube will sink.

Therefore, the buoyant force acting on the iron cube is 0.00981 N and the

cube will sink in water.

Question 10

Question

A cube with sides of length 5 cm and density 800 kg/m3is floating in water.

Calculate the depth of immersion of the cube.

Solution

Step 1: Determine the buoyant force acting on the cube. The buoyant force

on an object submerged in a fluid is equal to the weight of the fluid displaced

by the object. The weight of the fluid displaced can be calculated using the

formula:

Fbuoyant =ρfluid ·Vsubmerged ·g

where: ρfluid = 1000 kg/m3(density of water), Vsubmerged is the volume of the

cube submerged, and g= 9.81 m/s2is the acceleration due to gravity.

Step 2: Calculate the volume of the cube submerged. The volume of the

cube submerged can be calculated as the product of the cross-sectional area

submerged and the depth of immersion. Since the cube is floating, the buoyant

force is equal to the weight of the cube. Therefore, the volume of the cube

submerged is equal to the volume of the cube.

Vsubmerged =Vcube = (side length)3= (0.05 m)3

Step 3: Substitute the values and calculate the buoyant force.

Fbuoyant = (1000 kg/m3)·(0.05 m)3·9.81 m/s2

Step 4: Calculate the density of the cube. The density of the cube is given

as 800 kg/m3.

Step 5: Set up an equation using Archimedes’ principle. According to

Archimedes’ principle, the buoyant force acting on the cube is equal to the

weight of the cube. Therefore, we can set up the equation:

Fbuoyant =Fgravity