PHYS 101 - ELEMENTS OF PHYSICS Buoyant forces and Archimedes' principle Question Bank Set 10

PHYS 101 - ELEMENTS OF PHYSICS

- Buoyant forces and Archimedes’

principle

Question Bank - Set 10

Liberty University

Question 1

Question

A cube of wood with a density of 800 kg/m3and a side length of 0.1 m is

floating in a pool of water. Calculate the fraction of the cube’s volume that is

submerged in water.

Solution

Step 1: Determine the density of water. The density of water is approximately

1000 kg/m3.

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

calculated using the formula m=ρV , where ρis the density of the cube and

Vis the volume of the cube. Given that ρ= 800 kg/m3and a= 0.1 m, the

volume of the cube is V=a3= (0.1)3= 0.001 m3. Therefore, the mass of the

cube is m= 800 ×0.001 = 0.8 kg.

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

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

can be calculated using the formula Fb=ρwaterVsubmergedg, where ρwater is the

density of water, Vsubmerged is the volume of the cube submerged in water, and g

is the acceleration due to gravity. The weight of the cube can also be calculated

as weight =mg = 0.8×9.8=7.84 N. For the cube to be in equilibrium, the

buoyant force equals the weight of the cube, so Fb= 7.84 N.

Step 4: Calculate the volume of the cube submerged in water. Using the

known densities and the formula for the buoyant force, we have: ρwaterVsubmergedg=

7.84 N. Substituting the values ρwater = 1000 kg/m3and g= 9.8 m/s2, we can

solve for Vsubmerged. We find Vsubmerged =7.84

1000×9.8= 0.0008 m3.

Step 5: Calculate the fraction of the cube’s volume submerged in water.

The fraction of the cube’s volume submerged in water can be calculated as the

submerged volume divided by the total volume of the cube. Therefore, the

fraction submerged is Vsubmerged

V=0.0008

0.001 =4

5= 0.8 or 80

Thus, 80

Question 2

Question

A cylindrical object with a density of 900 kg/m3and a radius of 0.5 m is

floating in water. Calculate the depth to which the object is submerged in

water. (Density of water = 1000 kg/m3, acceleration due to gravity = 9.81

m/s2)

Solution

Step 1: Find the volume of the object using the formula for the volume of a

cylinder:

V=πr2h

where ris the radius of the cylinder and his the height (depth of submersion

in this case).

Step 2: Substitute the given values into the formula:

V=π(0.5 m)2×h

V=π×0.25 m2×h= 0.25πh m3

Step 3: Calculate the weight of the object using the formula:

Wobject =ρobject ×V×g

where ρobject is the density of the object, Vis the volume of the object, and g

is the acceleration due to gravity.

Step 4: Substitute the given values into the formula:

Wobject = 900 kg/m3×0.25πh m3×9.81 m/s2

Step 5: Calculate the buoyant force acting on the object:

Fbuoyant =ρwater ×Vsubmerged ×g

where ρwater is the density of water, Vsubmerged is the volume of the object

submerged in water, and gis the acceleration due to gravity.

Step 6: Since the object is floating, the weight of the object is equal to the

buoyant force:

Wobject =Fbuoyant

Step 7: Solve for the depth of submersion, h, by equating the weight of the

object to the buoyant force:

900 kg/m3×0.25πh m3×9.81 m/s2= 1000 kg/m3×π×0.5 m ×h×9.81 m/s2

Step 8: Solve for hto find the depth to which the object is submerged in

water.

Question 3

Question

A cube of side length 0.3 m and density 900 kg/m3is placed in a container filled

with water of density 1000 kg/m3. Calculate the buoyant force acting on the

cube when it is fully submerged in the water.

Solution

Step 1: First, calculate the volume of the cube. The volume of a cube is given

by V=s3, where s is the side length of the cube. Given that the side length of

the cube is 0.3 m, we have:

V= (0.3)3= 0.027 m3

Step 2: Next, calculate the mass of the cube. The mass of an object is given

by m=ρV , where ρis the density of the object and V is the volume. Given

that the density of the cube is 900 kg/m3, we have:

m= 900 ×0.027 = 24.3kg

Step 3: Calculate the weight of the cube. The weight of an object is given

by W=mg, where m is the mass of the object and g is the acceleration due to

gravity (approximately 9.81 m/s2).

W= 24.3×9.81 = 238.683 N

Step 4: Calculate the buoyant force. According to Archimedes’ principle,

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

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

to the volume of the cube. Given that the density of water is 1000 kg/m3, the

weight of the water displaced is:

Wwater =ρwaterV= 1000 ×0.027 = 27 kg ×9.81 = 264.33 N

Therefore, the buoyant force acting on the cube is 264.33 N, upwards.

Question 4

Question

A block of wood with a density of 600 kg/m3is floating in a container of water.

The block has a volume of 0.02 m3and a mass of 8 kg. Calculate the buoyant

force acting on the block and determine whether it will sink or float.

Solution

Step 1: Calculate the weight of the block The weight of the block can be calcu-

lated using the formula:

weight = mass ×gravitational acceleration

weight = 8 kg ×9.81 m/s2= 78.48 N

Step 2: Calculate the weight of the displaced water The weight of the dis-

placed water is equal to the weight of the block. This can be calculated using

the formula:

weight of water = weight of block = 78.48 N

Step 3: Calculate the volume of water displaced The volume of water dis-

placed is equal to the volume of the block. This can be calculated using the

formula:

volume of water = volume of block = 0.02 m3

Step 4: Calculate the buoyant force The buoyant force acting on the block

can be calculated using the formula:

buoyant force = density of water×gravitational acceleration×volume of water displaced

buoyant force = 1000 kg/m3×9.81 m/s2×0.02 m3= 196.2 N

Step 5: Compare the buoyant force and the weight of the block Since the

buoyant force (196.2 N) is greater than the weight of the block (78.48 N), the

block will float in the container of water.

Question 5

Question

A cube of wood with a density of 0.80 g/cm3and a side length of 10 cm is sub-

merged in water. Calculate the buoyant force acting on the cube and determine

whether it will sink or float.

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. Given that s= 10 cm,

we have:

V= (10 cm)3= 1000 cm3

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

using the formula m=ρV , where ρis the density of the cube. Given that

ρ= 0.80 g/cm3and V= 1000 cm3, we have:

m= 0.80 g/cm3×1000 cm3= 800 g

Step 3: Determine the buoyant force. The buoyant force acting on an ob-

ject submerged in a fluid is equal to the weight of the fluid displaced by the

object. The buoyant force can be calculated using the formula Fbuoyant =

ρfluidVdisplacedg, where ρfluid is the density of the fluid, Vdisplaced is the vol-

ume of fluid displaced, and gis the acceleration due to gravity. For water,

ρfluid = 1 g/cm3. Since the cube is completely submerged, the volume of water

displaced is equal to the volume of the cube, which is 1000 cm3. Therefore, the

buoyant force is:

Fbuoyant = 1 g/cm3×1000 cm3×9.81 m/s2= 9810 N

Step 4: Determine if the cube will sink or float. Since the buoyant force

of 9810 N is greater than the weight of the cube which is 800 N, the net force

acting on the cube is upward and it will float in water.

Question 6

Question

A wooden cube with a side length of 10 cm and a density of 0.8 g/cm3is placed

in a container filled with water. What is the minimum additional load that

must be placed on top of the cube to make it sink?

Solution

Step 1: Calculate the weight of the cube to determine the buoyant force it

experiences. Given that the density of water is 1 g/cm3, the weight of the

wooden cube is:

Weight = Volume ×Density ×Acceleration due to gravity

Volume = (Side length)3= (10 cm)3= 1000 cm3

Weight = 1000 cm3×0.8 g/cm3×9.8 m/s2= 7840 N

Step 2: Determine the buoyant force acting on the cube. The buoyant force

is equal to the weight of the water displaced by the cube:

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

The volume of the cube under water is equal to the volume of the cube that is

submerged:

Volume submerged = Volume of cube ×Density of object

Density of water

Volume submerged = 1000 cm3× 0.8 g/cm3

1 g/cm3!= 800 cm3

Buoyant force = 800 cm3×1 g/cm3×9.8 m/s2= 7840 N

Step 3: The cube will float until the weight of the additional load exceeds the

buoyant force. Let the additional load be denoted by Fload. Since the buoyant

force on the cube is already equal to its weight, the additional load must exceed

the weight of the cube for it to sink:

Fload >7840 N

Therefore, the minimum additional load required to make the cube sink is

7841 N .

Question 7

Question

A cube of side length aand density ρcube is placed in a liquid of density ρliquid.

The cube floats with a fraction xof its volume submerged in the liquid. De-

termine the relationship between the densities ρcube,ρliquid, and the fraction

Solution

Step 1: Determine the forces acting on the cube. The forces acting on the cube

are its weight Wdownwards and the buoyant force Fbuoy upwards. The weight

is given by:

W=ρcube ·g·Vcube

where Vcube =a3is the volume of the cube.

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

cube submerged is given by:

Vsub =x·Vcube =x·a3

Step 3: Determine the buoyant force. The buoyant force is equal to the

weight of the liquid displaced by the cube:

Fbuoy =ρliquid ·g·Vsub

Step 4: Apply Archimedes’ principle. According to Archimedes’ principle,

the buoyant force Fbuoy acting on the cube is equal to the weight Wof the cube:

ρliquid ·g·Vsub =ρcube ·g·Vcube

Step 5: Substituting the expressions for Vsub and Vcube:

ρliquid ·g·x·a3=ρcube ·g·a3

Step 6: Simplify the equation. Solving for x, we find:

x=ρcube

ρliquid

Therefore, the relationship between the densities ρcube,ρliquid, and the frac-

tion xis x=ρcube

ρliquid .

Question 8

Question

A cube of wood with sides of length 10 cm and density 600 kg/m3is floating in

water. Calculate the height of the cube that is above the water’s surface. The

density of water is 1000 kg/m3.

Solution

Let’s denote the height of the cube above the water’s surface as h.

Step 1: First, let’s calculate the volume of the cube. The volume of the

cube is given by Vcube = (10 cm)3= 0.001 m3.

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

is given by mcube = density ×volume.

Substituting the values, we have: mcube = 600 kg/m3×0.001 m3= 0.6 kg.

Step 3: Now, let’s consider the forces acting on the cube when it is floating

in water. The weight of the cube Wcube =mcube ×g, where g= 9.8 m/s2is the

acceleration due to gravity.

The buoyant force Fbuoyant acting on the cube is equal to the weight of the

water displaced by the cube. Fbuoyant = density of water×volume of cube submerged×

Step 4: In equilibrium, the weight of the cube is balanced by the buoyant

force. Thus, we have Wcube =Fbuoyant.

Substitute the expressions and solve for the volume of the cube submerged:

mcube ×g= density of water ×volume of cube submerged ×g

0.6 kg ×9.8 m/s2= 1000 kg/m3×areacross-section ×h×9.8 m/s2

Solve for hto find the height of the cube above the water’s surface.

Question 9

Question

A cylindrical container with a radius of 0.5 m and a height of 1 m is completely

filled with water. A solid metal sphere with a radius of 0.3 m and a density

of 8000 kg/m3is then lowered into the water until it is completely submerged.

Calculate the buoyant force acting on the sphere when it is in the water.

Solution

Step 1: Calculate the volume of the sphere The volume of a sphere is given by

the formula:

V=4

3πr3

Substitute the radius of the sphere:

V=4

3π(0.3)3

V≈0.1131 m3

Step 2: Calculate the weight of the sphere The weight of the sphere can be

calculated using the formula:

Weight = Mass ×Acceleration due to gravity

The mass of the sphere can be calculated using the formula:

Mass = Density ×Volume

Substitute the density and volume of the sphere:

Mass = 8000 kg/m3×0.1131 m3

Mass ≈904.8 kg

Now, calculate the weight of the sphere:

Weight = 904.8 kg ×9.81 m/s2

Weight ≈8886.3 N

Step 3: Calculate the weight of the water displaced The weight of the water

displaced is equal to the weight of the water that would fill the volume of the

sphere. The volume of the sphere is equal to the volume of water displaced.

Thus, the weight of the water displaced is equal to the weight of the sphere:

Weight of water displaced = 8886.3 N

Step 4: Calculate the buoyant force According to Archimedes’ principle, the

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

the fluid that the object displaces. Therefore, the buoyant force acting on the

sphere is equal to the weight of the water displaced:

Buoyant force = 8886.3 N

Therefore, the buoyant force acting on the sphere when it is in the water is

approximately 8886.3 N.

Question 10

Question

A cube of wood with a side length of 10 cm and a density of 0.8 g/cm3is

placed in a liquid of density 1.2 g/cm3. Calculate the volume of the cube that

is submerged in the liquid.

Solution

Step 1: Calculate the weight of the cube Given density of wood, ρwood = 0.8

g/cm3and side length l= 10 cm. The mass of the cube can be calculated as:

mass = ρwood ×volume = ρwood ×l3

mass = 0.8 g/cm3×(10 cm)3= 800 g

The weight of the cube can be found by multiplying the mass by the acceleration

due to gravity g= 9.81 m/s2:

weight = mass ×g= 800 g ×9.81 m/s2= 7848 N

Step 2: Calculate the buoyant force acting on the cube The buoyant force

is equal to the weight of the liquid displaced by the cube. The volume of liquid

displaced is equal to the volume of the cube submerged, denoted as Vsubmerged.

The buoyant force can be calculated as:

buoyant force = ρliquid ×g×Vsubmerged

Now, we need to find Vsubmerged.

Step 3: Calculate the volume submerged The volume submerged can be

found using Archimedes’ principle which states that the buoyant force is equal

to the weight of the displaced fluid:

buoyant force = weight of fluid displaced

⇒ρliquid ×g×Vsubmerged =ρwood ×g×l3

⇒1.2 g/cm3×Vsubmerged = 0.8 g/cm3×(10 cm)3

⇒Vsubmerged =0.8×1000

1.2= 666.67 cm3

Therefore, the volume of the cube that is submerged in the liquid is 666.67

cm3.

Question 11

Question

A cube of wood with a density of 700 kg/m3and a side length of 0.05 m is

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

water.

Solution

Let’s denote: - Vcube as the volume of the cube, - das the depth to which the

cube is submerged, - ρwater as the density of water, - Fbuoyant as the buoyant

force acting on the cube, - Fgravity as the weight of the cube.

Step 1: Calculate the volume of the cube. Since the cube has a side length

of 0.05 m, the volume can be calculated as:

Vcube = (0.05 m)3= 0.000125 m3

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

calculated using the formula:

Fgravity =m·g

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

mass of the cube can be calculated as:

m=ρcube ·Vcube

m= 700 kg/m3·0.000125 m3= 0.0875 kg

The weight of the cube is:

Fgravity = 0.0875 kg ·9.8 m/s2= 0.8575 N

Step 3: Calculate the buoyant force. According to Archimedes’ principle,

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

by the cube. The buoyant force can be calculated as:

Fbuoyant =ρwater ·Vsubmerged ·g

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

is floating, the buoyant force is equal to the weight of the cube:

Fbuoyant =Fgravity

ρwater ·Vsubmerged ·g=ρcube ·Vcube ·g

ρwater ·Vsubmerged =ρcube ·Vcube

Vsubmerged =ρcube ·Vcube

ρwater

Vsubmerged =700 kg/m3·0.000125 m3

1000 kg/m3= 0.0875 m3

Step 4: Calculate the depth to which the cube is submerged. The depth to

which the cube is submerged can be calculated using the formula:

Vsubmerged =A·d

where Ais the area of the base of the cube and dis the depth. Since the base

of the cube is a square with side length 0.05 m, the area can be calculated as:

A= (0.05 m) ×(0.05 m) = 0.0025 m2

0.0875 m3= 0.0025 m2×d

d=0.0875 m3

0.0025 m2= 0.035 m

Therefore, the depth to which the cube is submerged in the water is 0.035

Question 12

Question

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

in water. What is the buoyant force acting on the cube?

Solution

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

V=L3, where Lis the length of a side. In this case, the side length L= 0.1

m, so the volume of the cube is:

V= (0.1 m)3= 0.001 m3

Step 2: Determine the weight of the cube. The weight of the cube is given by

the formula W=mg, where mis the mass of the cube and gis the acceleration

due to gravity (9.81 m/s2). The mass of the cube can be calculated using the

formula m=ρV , where ρis the density of the material. Substituting in the

given values:

m= 600 kg/m3×0.001 m3= 0.6 kg

Thus, the weight of the cube is:

W= 0.6 kg ×9.81 m/s2= 5.886 N

Step 3: Calculate the buoyant force. According to Archimedes’ principle,

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

fluid displaced by the object. Since the cube is floating, the weight of the cube

is equal to the buoyant force. Therefore, the buoyant force acting on the cube

is:

Buoyant force = 5.886 N

Question 13

Question

A cylindrical container has a height of 1.5 meters and a radius of 0.5 meters.

The container is filled with water to a height of 1.2 meters. Determine the

buoyant force acting on a solid iron sphere of radius 0.2 meters submerged in

the water. The density of iron is 7800 kg/m3and the density of water is 1000

kg/m3.

Solution

Step 1: Find the volume of water displaced by the iron sphere. The volume of

the iron sphere can be calculated using the formula for the volume of a sphere:

Vsphere =4

3πr3

where ris the radius of the sphere. Substitute r= 0.2 m into the formula to

find the volume of the iron sphere:

Vsphere =4

3π(0.2)3=4

3π(0.008) = 0.03351 m3

Step 2: Calculate the volume of water displaced by the iron sphere. The

volume of water displaced is equal to the volume of the iron sphere that is

submerged in the water.

Vwater =Vsphere = 0.03351 m3

Step 3: Determine the weight of water displaced. The weight of water

displaced can be calculated using the formula:

Weight = Density ×Volume ×Acceleration due to gravity

Given that the density of water is 1000 kg/m3and the acceleration due to

gravity is 9.8 m/s2, the weight of water displaced is:

Weight = 1000 ×0.03351 ×9.8 = 328.93 N

Step 4: Determine the weight of the iron sphere. The weight of the iron

sphere can be calculated using the formula:

Weight = Density ×Volume ×Acceleration due to gravity

Given that the density of iron is 7800 kg/m3and the volume of the iron sphere

is the same as the volume of water displaced, the weight of the iron sphere is:

Weight = 7800 ×0.03351 ×9.8 = 2530.51 N

Step 5: Determine the buoyant force acting on the iron sphere. The buoyant

force is equal to the weight of the water displaced, which we calculated to be

328.93 N. Thus, the buoyant force acting on the iron sphere is 328.93 N.

Question 14

Question

A cube of side length 25 cm and density 950 kg/m3is floating in water. What

is the depth of the cube submerged in the water? (Density of water = 1000

kg/m3, acceleration due to gravity = 9.81 m/s2)

Solution

Step 1: Begin by finding 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. Given that the

side length of the cube is 25 cm (0.25 m), we have:

V= (0.25 m)3= 0.015625 m3

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

by the formula W=mg, where mis the mass of the object and gis the

acceleration due to gravity. The mass of the cube can be calculated using the

formula m= density ×V. Substituting in the values:

m= 950 kg/m3×0.015625 m3= 14.84375 kg

Now, we can find the weight of the cube:

W= 14.84375 kg ×9.81 m/s2= 145.5984375 N

Step 3: Determine the buoyant force acting on the cube. According to

Archimedes’ principle, 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 can be calculated using the formula Fb= density of water×Vsubmerged ×g,

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

the depth of the cube submerged in water. The volume submerged is A×h,

where Ais the area of the cube’s face. A=s2= (0.25 m)2= 0.0625 m2Thus,

Vsubmerged = 0.0625 m2×h= 0.0625hm3Substitute the values and calculate

the buoyant force:

Fb= 1000 kg/m3×0.0625hm3×9.81 m/s2= 614.0625hN

Step 4: Set up the equilibrium condition. For the cube to be floating in

equilibrium, the weight of the cube must be equal to the buoyant force acting

on it. Thus, we have:

W=Fb

145.5984375 N = 614.0625hN

Step 5: Solve for the depth of the cube submerged.

614.0625h= 145.5984375

h=145.5984375

614.0625 = 0.237 m

Therefore, the depth of the cube submerged in water is 23.7 cm.

Question 15

Question

A cube of wood with a density of 0.75 g/cm3and side length of 5 cm is floating

in water. Calculate the depth to which the cube floats in water.

Solution

Step 1: Determine the density of water Given that the density of water is

ρwater = 1.0 g/cm3.

Step 2: Calculate the fraction of the cube submerged in water We can use

Archimedes’ principle to find the fraction of the cube submerged in water:

Vsubmerged

Vtotal

=ρwater

ρwood

Plugging in the values, we get

Vsubmerged

53=1.0

0.75

Solving for Vsubmerged gives

Vsubmerged =1.0

0.75 ×53

Step 3: Calculate the depth to which the cube floats in water The depth

to which the cube floats in water is half the height of the submerged portion.

Therefore, the depth is given by

Depth = Vsubmerged

Step 4: Substitute the value of Vsubmerged into the formula and calculate the

depth

Depth = 1.0

0.75 ×5

Calculating this will give us the depth to which the cube floats in water.

Question 16

Question

A solid cube with a side length of 10 cm and a density of 800 kg/m3is floating

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. Substituting s= 0.1 m into the

formula, we have

V= (0.1 m)3= 0.001 m3

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

using the formula m=ρV , where ρis the density of the cube. Substituting

ρ= 800 kg/m3and V= 0.001 m3into the formula,

m= 800 kg/m3×0.001 m3= 0.8 kg

Step 3: Determine the weight of the cube: The weight of the cube can be

calculated using the formula Fgravity =m·g, where gis the acceleration due to

gravity (9.81 m/s2). Substituting m= 0.8 kg,

Fgravity = 0.8 kg ×9.81 m/s2= 7.848 N

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

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

displaced by the object. Since the cube is floating, the weight of the water

displaced is equal to the weight of the cube. Therefore, the buoyant force is

Fbuoyant = 7.848 N

Therefore, the buoyant force acting on the cube is 7.848 N.

Question 17

Question

A solid cube of side length aand density ρsis placed in a liquid of density

ρl, with a < 2h, where his the depth of the liquid. The cube floats partially

submerged in the liquid, with a height xof the cube above the liquid surface.

Determine the expression for xin terms of a,h,ρs, and ρl.

Solution

Step 1: Determine the buoyant force on the cube when it is partially submerged.

The buoyant force Fbacting on the cube is equal to the weight of the liquid

displaced by the cube. Since the cube is partially submerged, the volume of the

liquid displaced is given by Asubmerged ·x, where Asubmerged =a2is the area of

the cube submerged in the liquid.

Therefore, the buoyant force Fbis given by:

Fb=ρl·g·Asubmerged ·x

Step 2: Determine the weight of the cube. The weight of the cube is equal

to the weight of the cube’s volume of liquid, which is a2·a·ρl·g=ρs·a3·g.

Step 3: Set up the equilibrium condition. For the cube to float partially

submerged, the buoyant force Fbmust equal the weight of the cube. Hence, we

have:

ρl·g·a2·x=ρs·a3·g

Step 4: Solve for x. From the equilibrium condition, we can solve for x:

x=ρs·a

ρl·a=ρs

ρl·a

Therefore, the expression for xin terms of a,ρs, and ρlis x=ρs

ρl·a.

Question 18

Question

A spherical balloon with a radius of 5 m is filled with helium. The volume of

the balloon is 524.47 m3and the density of helium is 0.179 kg/m3. Calculate

the buoyant force acting on the balloon when it is released from rest at the

bottom of a lake. The density of water is 1000 kg/m3and the acceleration due

to gravity is 9.81 m/s2.

Solution

Step 1: Calculate the weight of the helium in the balloon.

The weight can be calculated using the formula: W=m·g, where mis the

mass of the helium and gis the acceleration due to gravity.

The mass of the helium can be calculated using the formula: m= density ×

volume.

Substitute the given values to find the mass of the helium:

m= 0.179 kg/m3×524.47 m3

m≈93.81 kg

Step 2: Calculate the buoyant force acting on the balloon.

The buoyant force is equal to the weight of the water displaced by the balloon

(Archimedes’ principle). The volume of water displaced is equal to the volume

of the balloon.

The weight of the water displaced can be calculated using the formula: W=

density ×volume ×g.

Substitute the given values to find the weight of the water displaced:

W= 1000 kg/m3×524.47 m3×9.81 m/s2

W≈5141649.3 N

Step 3: Determine the net force acting on the balloon.

The net force is the difference between the buoyant force and the weight of the

helium:

Net force = Weight of water displaced −Weight of helium

Net force = 5141649.3 N −93.81 kg ×9.81 m/s2

Net force = 5131703.49 N

Therefore, the buoyant force acting on the balloon when it is released from

rest at the bottom of the lake is approximately 5131703.49 N.

Question 19

Question

A cube of side length 0.5 m and density 800 kg/m3is floating in water. A 2 kg

weight is placed on top of the cube and the equilibrium water level changes by

6.25 cm. What is the density of the weight?

Solution

Step 1: Calculate the original volume of the cube.

The volume of the cube is given by V= (0.5 m)3= 0.125 m3.

Step 2: Calculate the weight of the cube.

The weight of the cube can be found using the formula W=ρV g, where ρis

the density of the cube, Vis the volume of the cube, and gis the acceleration

due to gravity.

Substituting the given values, W= (800 kg/m3)(0.125 m3)(9.8 m/s2) = 980 N.

Step 3: Calculate the new volume of the cube and weight.

Since the weight is floating, the weight of the cube plus the weight placed on

top (2 kg) must equal the weight of the displaced water. Therefore, the new

weight of the system is 980 N + 2 kg ×9.8 m/s2= 1000.6 N.

The volume of water displaced is given by Vwater =Wnew

ρwaterg, where ρwater is the

density of water.

Substituting the given values, Vwater =1000.6 N

(1000 kg/m3)(9.8 m/s2)= 0.102 m3.

Step 4: Calculate the side length of the new cube.

The new volume of the cube is Vnew = 0.125 m3−0.102 m3= 0.023 m3. The

side length of the new cube can be found by taking the cube root of this volume:

snew =3

√0.023 m ≈0.27 m.

Step 5: Calculate the density of the weight.

The volume of the weight is the difference between the original cube volume

and the new cube volume: Vweight = 0.125 m3−0.027 m3= 0.098 m3.

The density of the weight is then ρweight =Wweight

Vweight g. Substituting the values,

ρweight =2 kg×9.8 m/s2

0.098 m3×9.8 m/s2≈20.41 kg/m3.

Question 20

Question

A cube of wood with a density of 800 kg/m3and a side length of 0.1 m is floating

in a pool of water. Determine the depth to which the cube is submerged in the

water.

Solution

Step 1: Begin by finding the density of water. We know that the density of

water is approximately 1000 kg/m3.

Step 2: Next, use Archimedes’ principle to find the depth submerged. The

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

by the cube. The buoyant force is given by FB=ρwater ·g·Vsubmerged, where

ρwater is the density of water, gis the acceleration due to gravity, and Vsubmerged

is the volume submerged.

Step 3: Calculate the volume of the cube submerged. The volume of the

cube submerged is equal to Vsubmerged = side length ×side length ×depth.

Step 4: Equate the buoyant force to the weight of the cube to solve for the

depth submerged. The weight of the cube is given by Wcube =ρwood ·g·Vcube,

where ρwood is the density of the wood and Vcube is the volume of the cube.

Step 5: Substituting the values and solving for the depth submerged. From

Step 2, FB=ρwater ·g·side length2·depth From Step 4, Wcube =ρwood ·

g·side length3Setting these two equal gives: ρwater ·side length2·depth =

ρwood ·side length3

Step 6: Finally, solve for the depth submerged. Solving the equation from

Step 5 for depth gives: depth = ρwood

ρwater ·side length

Step 7: Plug in the given values and calculate the depth submerged. depth =

800 kg/m3

1000 kg/m3·0.1 m = 0.08 m

Therefore, the cube is submerged to a depth of 0.08 meters in the water.

Question 21

Question

A cylindrical metal object with a density of 5000 kg/m3and a height of 0.5 m is

floating in water. If the diameter of the cylinder is 0.2 m, calculate the buoyant

force acting on the object and determine if the object will sink or float when

placed in a container of oil with a density of 800 kg/m3.

Solution

Step 1: Calculate the volume of the metal cylinder. The volume of a cylinder

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

Given that the diameter is 0.2 m, the radius ris half of the diameter, so r= 0.1

m. Plugging the values into the formula, we get:

V=π(0.1)2×0.5

V= 0.0157 m3

Step 2: Calculate the weight of the metal cylinder. The weight of an object

is given by the formula W=mg, where mis the mass and gis the acceleration

due to gravity (approximately 9.81m/s2). Given that the density ρis 5000

kg/m3, the mass mof the cylinder is:

m=ρV = 5000 ×0.0157 = 78.5kg

Now, calculate the weight Wof the cylinder:

W= 78.5×9.81 = 770.2N

Step 3: Calculate the buoyant force in water. The buoyant force Fbin water

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

displaced is equal to the volume of the cylinder, which is 0.0157 m3. The density

of water is 1000 kg/m3. Calculating the buoyant force:

Fb=ρwaterV g = 1000 ×0.0157 ×9.81 = 153.717 N

Step 4: Determine if the object will float or sink in oil. For an object to

float, the buoyant force acting on it must be greater than or equal to its weight.

Given that the buoyant force in oil is less than the weight of the object, the

object will sink when placed in a container of oil.

Question 22

Question

A metal cube with a side length of 3 cm and a mass of 100 g is submerged in a

container of water. What is the buoyant force acting on the cube? (Density of

water = 1000 kg/m3, acceleration due to gravity = 9.81 m/s2)

Solution

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

by V= (side length)3. Substituting side length = 3 cm = 0.03 m, we have:

V= (0.03 m)3= 0.000027 m3

Step 2: Calculate the density of the metal. Density is given by the for-

mula density =mass

volume . Substituting mass = 100 g= 0.1kg and volume =

0.000027 m3, we have:

density =0.1kg

0.000027 m3= 3703.7kg/m3

Step 3: Calculate the buoyant force. According to Archimedes’ principle,

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

displaced by the object. The weight of the fluid displaced is equal to the volume

of the fluid displaced multiplied by the density of the fluid and acceleration

due to gravity. The weight of the water displaced by the cube is given by

weight = density of water×V×g. Substituting density of water = 1000 kg/m3,

V= 0.000027 m3, and g= 9.81 m/s2, we have:

weight = 1000 ×0.000027 ×9.81 = 0.2646 N

Step 4: Compare the density of the cube to the density of water. Since

the density of the cube is greater than the density of water, the cube will sink.

Therefore, the buoyant force will be less than the weight of the cube. The

buoyant force is given by the weight of the water displaced, which is 0.2646 N.

Question 23

Question

A cube of wood with sides of length 5 cm and a density of 0.8 g/cm3is submerged

in water. Calculate the buoyant force acting on the cube and determine if it

will float or sink.

Solution

Step 1: Calculate the mass of the wooden cube.

Given that the density of the wood is 0.8 g/cm3and the volume of the cube is

(5 cm)3= 125 cm3, we can find the mass of the cube using the formula m=ρV ,

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

Density = 0.8 g/cm3

Volume = 125 cm3

m= 0.8×125 = 100 g

Step 2: Calculate 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 weight of the fluid displaced is given

by the formula Fb=ρfluidVsubmergedg, where ρfluid is the density of the fluid,

Vsubmerged is the volume of the object submerged, and gis the acceleration due

to gravity.

Density of water = 1 g/cm3

Volume submerged = 125 cm3

Fb= 1 ×125 ×9.81 = 1226.25 N

Step 3: Determine if the cube will float or sink.

An object will float if the buoyant force is greater than or equal to the weight of

the object. Since the buoyant force is 1226.25 N and the weight of the cube is

m×g= 100 ×9.81 = 981 N, we can see that the cube will float as the buoyant

force is greater than the weight of the cube.

Question 24

Question

A cube of side length Land density ρ1is placed in a container of water, with a

fraction of its volume submerged. Another cube of side length 2Land density ρ2

is placed in the same container of water, with a different fraction of its volume

submerged. If the buoyant force acting on the larger cube is 400 N greater than

the buoyant force acting on the smaller cube, determine the densities ρ1and ρ2

in terms of the density of water ρw.

Solution

Step 1: Recall the expression for the buoyant force Fbacting on an object

submerged in a fluid:

Fb=ρw·g·Vsub

where ρwis the density of water, gis the acceleration due to gravity, and Vsub

is the volume of the object submerged in the fluid.

Step 2: The volume of a cube is given by V=L3. Let V1be the volume of

the smaller cube submerged, and V2be the volume of the larger cube submerged.

Step 3: We have that the buoyant force acting on the smaller cube is Fb1=

ρw·g·V1and the buoyant force acting on the larger cube is Fb2=ρw·g·V2.

We are given that Fb2−Fb1= 400 N.

Step 4: Substituting in the expressions for the buoyant forces:

ρw·g·V2−ρw·g·V1= 400

ρw·g·8L3−ρw·g·L3= 400

Step 5: Simplifying, we get:

7ρw·g·L3= 400

Step 6: Now, we know that the mass of the cubes is related to their densities

and volumes:

Masssmall =ρ1·Vsmall =ρ1·L3

Masslarge =ρ2·Vlarge =ρ2·(2L)3

Step 7: Since the cubes are in water and in equilibrium, the weight of the

cubes must be balanced by the buoyant force:

Weightsmall =Fb1=ρw·g·V1=ρw·g·L3

Weightlarge =Fb2=ρw·g·V2=ρw·g·8L3

Step 8: Equating the weights:

ρ1·L3·g=ρw·g·L3

ρ2·8L3·g=ρw·g·8L3

Step 9: Since we are given that the buoyant force acting on the larger cube

is greater by 400 N, we can solve for ρ1and ρ2in terms of ρwby setting up and

solving a system of equations:

(7ρw= 400

ρ1=ρw

Step 10: Solving the system of equations, we find:

ρw=400

ρ2=8

ρw

400/7=56

400

Therefore, the densities of the cubes are ρ1=400

7and ρ2=56

400 in terms of

the density of water ρw.

Question 25

Question

A sphere of radius Rand density ρ1is partially submerged in a liquid of density

ρ2. If the sphere is in equilibrium, determine the ratio ρ1

ρ2in terms of the depth

hto which the sphere is submerged.

Solution

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

liquid displaced. Let’s denote the volume of the sphere submerged in the liquid

as Vs, and the volume of the sphere above the liquid surface as Va. The volume

of the liquid displaced is then Vs.

Step 2: The weight of the liquid displaced is equal to the buoyant force.

The weight of the liquid displaced is Vsρ2g, where gis the acceleration due to

gravity. This must equal the weight of the sphere submerged in the liquid.

Step 3: The weight of the sphere submerged in the liquid can be expressed

as:

W=Vsρ1g=4

3πR3ρ1g

Step 4: Setting the weight of the liquid displaced equal to the weight of the

sphere submerged, we have:

Vsρ2g=4

3πR3ρ1g

Step 5: Given that the sphere is partially submerged, we can express Vsin

terms of Rand h. We know that Vs=1

24/3πR3−1

2πR2h. Simplifying this

expression, we get Vs=2

3πR3−πR2h.

Step 6: Substituting this expression for Vsinto the equation from Step 4,

we get:

2

3πR3−πR2hρ2g=4

3πR3ρ1g

Step 7: Simplifying this equation, we find:

2

3−2h

Rρ2= 4ρ1

Step 8: Finally, isolating the ratio ρ1

ρ2, we get:

ρ1

ρ2

42

3−2h

R−1

Question 26

Question

A solid sphere of radius Ris floating in a liquid with half of its volume sub-

merged. The density of the liquid is ρland the density of the sphere is ρs.

Determine an expression for the density of the sphere in terms of the radius R,

the density of the liquid ρl, and the acceleration due to gravity g.

Solution

Step 1: Let’s first determine the volume of the sphere that is submerged in

the liquid. Since half of the sphere is submerged, the volume of the sphere

submerged is 1

2of the total volume:

Vsubmerged =1

2·4

3πR3=2

3πR3

Step 2: The buoyant force Fbacting on the sphere is equal to the weight of

the liquid displaced. Using Archimedes’ principle, we have:

Fb=ρl·Vsubmerged ·g

Step 3: The weight of the sphere Wsis equal to the weight of the displaced

liquid, so:

Ws=ρs·4

3πR3·g

Step 4: Since the sphere is floating, the buoyant force Fbis equal to the

weight of the sphere Ws. Equating the two expressions for Fband Ws, we have:

ρl·2

3πR3·g=ρs·4

3πR3·g

Step 5: Simplifying the equation above, we get:

ρs= 2ρl

Therefore, the density of the sphere is given by ρs= 2ρl.

Question 27

Question

A spherical balloon with a radius of 0.5 meters is filled with helium. The density

of helium is 0.178 kg/m3and the density of air is 1.225 kg/m3. Calculate the

maximum mass that the balloon can lift off the ground.

Solution

Step 1: Find the volume of the balloon. The volume of a sphere is given by the

formula:

Vsphere =4

3πr3

Plugging in the radius r= 0.5 meters:

Vsphere =4

3π(0.5)3=4

3π×0.125 = 1

6πm3

Step 2: Find the mass of the displaced air by the balloon. The buoyant force

acting on the balloon is equal to the weight of the air displaced by the balloon.

The mass of the air displaced is given by:

mair = Densityair ×Vsphere = 1.225 ×1

6π=5

6πkg

Step 3: Calculate the weight of the air displaced. The weight of the air

displaced can be found using the formula Fbuoyant =mg, where mis the mass

of the air displaced.

Fbuoyant = 9.81 ×5

6π≈8.14πN

Step 4: Calculate the maximum mass the balloon can lift. The weight of the

helium inside the balloon must be less than the buoyant force in order for the

balloon to lift. Let mhelium be the maximum mass the balloon can lift, then:

mhelium ≤Fbuoyant

mhelium ≤8.14π

Therefore, the maximum mass that the balloon can lift off the ground is

approximately 8.14πkg.

Question 28

Question

A solid aluminum sphere with radius 10 cm is floating in water with 20

Solution

Step 1: First, we need to determine the density of water, which is ρwater =

1000 kg/m3.

Step 2: We know that the buoyant force (Fbuoyant) acting on the sphere

is equal to the weight of the water displaced by the submerged portion of the

sphere. The volume of water displaced is given by the volume of the submerged

portion of the sphere:

Vsubmerged =4

3πr3×20

100 =4

3π(0.10 m)3×0.20 = 8

3000πm3.

Step 3: The weight of the water displaced is equal to the buoyant force:

Fbuoyant =ρwaterVsubmergedg,

where gis the acceleration due to gravity.

Step 4: The weight of the aluminum sphere is given by:

Fweight =ρaluminumVsphereg,

where Vsphere =4

3π(0.10 m)3is the total volume of the sphere.

Step 5: Since the sphere is floating, the buoyant force must equal the weight

of the sphere:

ρwaterVsubmergedg=ρaluminumVsphereg.

Step 6: Now, we can solve for the density of the aluminum sphere:

ρaluminum =ρwaterVsubmerged

Vsphere

=1000 ×8

3000 π

3π(0.10)3=2666.67

0.004 ≈666,670 kg/m3.

Step 7: Therefore, the density of the aluminum sphere is approximately

666,670 kg/m3.

Question 29

Question

A cube of side length 2.5 m and density 800 kg/m3is floating in a liquid with

density 1000 kg/m3. Calculate the height of the cube that is submerged in the

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

Solution

Step 1: Calculate the weight of the cube. The weight of the cube is equal

to the gravitational force acting on it. This can be calculated by multiplying

the volume of the cube by the density of the cube and the acceleration due to

gravity.

Volume of cube = (side length)3= (2.5 m)3= 15.625 m3

Weight of cube = Volume ×Density ×g= 15.625 m3×800 kg/m3×9.81 m/s2

Step 2: Calculate the buoyant force acting on the cube. The buoyant force

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

This can be calculated by multiplying the volume of the cube submerged in the

liquid by the density of the liquid and the acceleration due to gravity. Let hbe

the height of the cube submerged in the liquid.

Volume of submerged cube = Area of base ×height = (2.5 m)2×h= 6.25hm3

Buoyant force = Volume×Density of liquid×g= 6.25hm3×1000 kg/m3×9.81 m/s2

Step 3: Apply Archimedes’ principle. According to Archimedes’ principle,

the weight of the cube is equal to the buoyant force acting on it when it is in

equilibrium.

Weight of cube = Buoyant force

15.625 m3×800 kg/m3×9.81 m/s2= 6.25hm3×1000 kg/m3×9.81 m/s2

Step 4: Solve for h.

15.625 ×800 = 6.25h×1000

h=15.625 ×800

6.25 ×1000

h= 2 m

Therefore, the cube is submerged to a height of 2 meters in the liquid.

Question 30

Question

A cube of iron with sides of length 10 cm is completely submerged in water. The

density of iron is 7.87 g/cm3, and the density of water is 1.00 g/cm3. Calculate

the buoyant force acting on the iron cube.

Solution

Step 1: First, we need to calculate the volume of the iron cube. The volume of

a cube is given by the formula V=a3, where ais the length of a side. Given

that the side length of the cube is 10 cm, the volume of the iron cube is:

Vcube = (10 cm)3= 1000 cm3

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

can be found using the formula m=ρ×V, where ρis the density of iron.

Substitute the values:

miron = 7.87 g/cm3×1000 cm3= 7870 g

Step 3: Now, we calculate the weight of the iron cube. The weight Wof

an object is given by the formula W=m×g, where gis the acceleration due

to gravity (approximately 9.81 m/s2). Convert the mass of the iron cube to

kilograms:

miron = 7870 g = 7.87 kg

Calculate the weight:

Wiron = 7.87 kg ×9.81 m/s2= 77.3 N

Step 4: Finally, we 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 cube is also

1000 cm3. The weight of this volume of water is given by:

Wwater =ρwater ×Vcube ×g= 1.00 g/cm3×1000 cm3×9.81 m/s2= 9810 N

Step 5: Since the buoyant force is equal to the weight of the water displaced,

the buoyant force acting on the iron cube is 9810 N.

Question 31

Question

A cube of wood with a density of 700 kg/m3and a side length of 0.1 m is floating

in a tub of water. What is the depth of immersion of the cube?

Solution

Step 1: The depth of immersion of an object floating in a fluid is determined

by the buoyant force acting on the object. The buoyant force can be calculated

using Archimedes’ principle, which states that the buoyant force is equal to the

weight of the fluid displaced by the object. Step 2: The weight of the fluid

displaced by the cube is equal to the weight of the cube itself. We can calculate

the weight of the cube using the formula W=mg, where mis the mass of the

cube and gis the acceleration due to gravity. Step 3: Since the cube is floating,

the buoyant force FBacting on the cube is equal in magnitude to the weight W

of the cube. This can be expressed as FB=W. Step 4: The buoyant force can

also be calculated using the formula FB=ρVimmersedg, where ρis the density of

the fluid (water), Vimmersed is the volume of the cube submerged in water, and g

is the acceleration due to gravity. Step 5: The volume of the cube submerged in

water can be calculated as Vimmersed = (side length)3×depth of immersion. Step

6: Equating the two expressions for the buoyant force, we have ρVimmersedg=

mg. Step 7: Substitute the given values into the equation and solve for the

depth of immersion, h. Remember that ρwater = 1000 kg/m3. Step 8: After

substituting the values into the equation and solving for h, we find the depth of

immersion of the cube to be approximately 0.071 m or 7.1 cm.

Question 32

Question

A cube of side length 0.1 m is submerged in water. The upper surface of the

cube is 4 cm below the water surface. Find the buoyant force acting on the

cube. (Density of water = 1000 kg/m3)

Solution

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

the cube submerged is the same as the volume of water displaced, which is given

by the formula:

V=Ah

where Ais the area of one face of the cube and his the height submerged.

Given that the side length of the cube is 0.1 m and the cube is submerged 4 cm

(0.04 m) under the water surface, the area Aand height hcan be calculated as

follows:

A= (0.1)2

h= 0.04

Step 2: Calculate the volume of water displaced. Substitute the values into

the formula:

V= (0.1)2×0.04

Step 3: Calculate the mass of water displaced. The mass of water displaced

can be found using the formula:

m=V×ρ

where ρis the density of water (1000 kg/m3).

Step 4: Calculate the weight of the water displaced. The weight of the water

displaced is equal to the buoyant force acting on the cube, which is given by:

Fb=m×g

where gis the acceleration due to gravity (9.81 m/s2).

Step 5: Substitute the values and calculate the buoyant force. Substitute

the values of mand ginto the formula:

Fb=V×ρ×g

Question 33

Question

A wooden cube of side length 10 cm floats in water with 4 cm of its height above

the water surface. What is the density of the wood?

Solution

Step 1: Calculate the volume of the cube submerged in the water. Step 2:

Calculate the weight of the water displaced by the submerged cube. Step 3: Set

up an equation using Archimedes’ principle to solve for the density of the wood.

Step 1: The volume of the cube submerged in water is given by Vsubmerged =

A×hsubmerged, where Ais the area of one face of the cube and hsubmerged is the

height submerged in water. Given that the side length of the cube is 10 cm and

4 cm of its height is submerged in water, we have: A= (10 cm)2= 100 cm2,

hsubmerged = 4 cm. Therefore, Vsubmerged = 100 cm2×4 cm = 400 cm3.

Step 2: The weight of the water displaced by the cube can be calculated

using the density of water, ρwater = 1000 kg/m3. The mass of the water displaced

is then mwater =ρwater ×Vsubmerged. Converting the volume to cubic meters, we

have Vsubmerged = 0.0004 m3. Therefore, mwater = 1000 kg/m3×0.0004 m3=

0.4 kg. The weight of the water displaced is Wwater =mwater ×g, where g=

9.8 m/s2is the acceleration due to gravity. Hence, Wwater = 0.4 kg ×9.8 m/s2=

3.92 N.

Step 3: According to Archimedes’ principle, the buoyant force acting on the

cube is equal to the weight of the water displaced by the cube. Therefore, the

buoyant force, Fbuoyant =Wwater = 3.92 N. The buoyant force can also be ex-

pressed as Fbuoyant =ρwater ×g×Vsubmerged. Setting these two expressions equal

to each other, we have: 3.92 N = 1000 kg/m3×9.8 m/s2×0.0004 m3. Solving

for the density of the wood, ρwood, we get: ρwood =3.92 N

0.00392 m3= 1000 kg/m3.

Therefore, the density of the wood is 1000 kg/m3.

Question 34

Question

A cube of side length aand density ρcube is floating in water with 1/3 of its

volume submerged. Find the ratio of the densities of the cube and water, given

that the cube is in equilibrium.

Solution

Step 1: Recall that for an object floating in a fluid, the buoyant force on the

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

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

by the submerged volume of the cube. Step 3: The weight of the cube, acting

downwards, is equal to the weight of the water displaced by the submerged

volume of the cube, acting upwards. Step 4: The weight of the cube is given

by Wcube =ρcube ·g·Vsub, where ρcube is the density of the cube, gis the

acceleration due to gravity, and Vsub is the submerged volume of the cube. Step

5: The weight of the displaced water is given by Wwater =ρwater ·g·Vsub, where

ρwater is the density of water. Step 6: Since the cube is in equilibrium, the

weight of the cube and the weight of the displaced water are equal. Therefore,

ρcube ·Vsub =ρwater ·Vsub. Step 7: Simplifying the equation gives ρcube =ρwater.

Step 8: Therefore, the ratio of the densities of the cube and water is 1 : 1 .

Question 35

Question

A cube of wood with a density of 700 kg/m3and a side length of 0.20 m floats

in a container of water. What fraction of the cube’s volume is above water?

Solution

Step 1: First, we need to find the density of water. The density of water is

approximately 1000 kg/m3.

Step 2: Next, we can find the weight of the cube using the formula w=

mg, where mis the mass of the cube and gis the acceleration due to gravity

(9.81 m/s2).

Step 3: The volume of the cube can be calculated using V= side length3.

Step 4: The weight of the water displaced by the cube is equal to the weight

of the cube. Using the formula for weight of water displaced (wdisplaced =

ρwater ·Vsubmerged ·g) where ρwater is the density of water, Vsubmerged is the

volume of the cube submerged in water, and gis the acceleration due to gravity.

Step 5: The volume of the cube submerged in water is equal to the fraction

of the cube submerged multiplied by the total volume of the cube.

Step 5: Calculate the fraction of the cube’s volume submerged in water.

The fraction of the cube’s volume submerged in water can be calculated as the

submerged volume divided by the total volume of the cube. Therefore, the

fraction submerged is Vsubmerged

V=0.0008

0.001 =4

5= 0.8 or 80

Thus, 80

Question 2

Question

A cylindrical object with a density of 900 kg/m3and a radius of 0.5 m is

floating in water. Calculate the depth to which the object is submerged in

water. (Density of water = 1000 kg/m3, acceleration due to gravity = 9.81

m/s2)

Solution

Step 1: Find the volume of the object using the formula for the volume of a

cylinder:

V=πr2h

where ris the radius of the cylinder and his the height (depth of submersion

in this case).

Step 2: Substitute the given values into the formula:

V=π(0.5 m)2×h

V=π×0.25 m2×h= 0.25πh m3

Step 3: Calculate the weight of the object using the formula:

Wobject =ρobject ×V×g

where ρobject is the density of the object, Vis the volume of the object, and g

is the acceleration due to gravity.

Step 4: Substitute the given values into the formula:

Wobject = 900 kg/m3×0.25πh m3×9.81 m/s2

Step 5: Calculate the buoyant force acting on the object:

Fbuoyant =ρwater ×Vsubmerged ×g

where ρwater is the density of water, Vsubmerged is the volume of the object

submerged in water, and gis the acceleration due to gravity.

Step 6: Since the object is floating, the weight of the object is equal to the

buoyant force:

Wobject =Fbuoyant

Step 7: Solve for the depth of submersion, h, by equating the weight of the

object to the buoyant force:

900 kg/m3×0.25πh m3×9.81 m/s2= 1000 kg/m3×π×0.5 m ×h×9.81 m/s2

Step 8: Solve for hto find the depth to which the object is submerged in

water.

Question 3

Question

A cube of side length 0.3 m and density 900 kg/m3is placed in a container filled

with water of density 1000 kg/m3. Calculate the buoyant force acting on the

cube when it is fully submerged in the water.

Solution

Step 1: First, calculate the volume of the cube. The volume of a cube is given

by V=s3, where s is the side length of the cube. Given that the side length of

the cube is 0.3 m, we have:

V= (0.3)3= 0.027 m3

Step 2: Next, calculate the mass of the cube. The mass of an object is given

by m=ρV , where ρis the density of the object and V is the volume. Given

that the density of the cube is 900 kg/m3, we have:

m= 900 ×0.027 = 24.3kg

Step 3: Calculate the weight of the cube. The weight of an object is given

by W=mg, where m is the mass of the object and g is the acceleration due to

gravity (approximately 9.81 m/s2).

W= 24.3×9.81 = 238.683 N

Step 4: Calculate the buoyant force. According to Archimedes’ principle,

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

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

to the volume of the cube. Given that the density of water is 1000 kg/m3, the

weight of the water displaced is:

Wwater =ρwaterV= 1000 ×0.027 = 27 kg ×9.81 = 264.33 N

Therefore, the buoyant force acting on the cube is 264.33 N, upwards.

Question 4

Question

A block of wood with a density of 600 kg/m3is floating in a container of water.

The block has a volume of 0.02 m3and a mass of 8 kg. Calculate the buoyant

force acting on the block and determine whether it will sink or float.

Solution

Step 1: Calculate the weight of the block The weight of the block can be calcu-

lated using the formula:

weight = mass ×gravitational acceleration

weight = 8 kg ×9.81 m/s2= 78.48 N

Step 2: Calculate the weight of the displaced water The weight of the dis-

placed water is equal to the weight of the block. This can be calculated using

the formula:

weight of water = weight of block = 78.48 N

Step 3: Calculate the volume of water displaced The volume of water dis-

placed is equal to the volume of the block. This can be calculated using the

formula:

volume of water = volume of block = 0.02 m3

Step 4: Calculate the buoyant force The buoyant force acting on the block

can be calculated using the formula:

buoyant force = density of water×gravitational acceleration×volume of water displaced

buoyant force = 1000 kg/m3×9.81 m/s2×0.02 m3= 196.2 N

Step 5: Compare the buoyant force and the weight of the block Since the

buoyant force (196.2 N) is greater than the weight of the block (78.48 N), the

block will float in the container of water.

Question 5

Question

A cube of wood with a density of 0.80 g/cm3and a side length of 10 cm is sub-

merged in water. Calculate the buoyant force acting on the cube and determine

whether it will sink or float.

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. Given that s= 10 cm,

we have:

V= (10 cm)3= 1000 cm3

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

using the formula m=ρV , where ρis the density of the cube. Given that

ρ= 0.80 g/cm3and V= 1000 cm3, we have:

m= 0.80 g/cm3×1000 cm3= 800 g

Step 3: Determine the buoyant force. The buoyant force acting on an ob-

ject submerged in a fluid is equal to the weight of the fluid displaced by the

object. The buoyant force can be calculated using the formula Fbuoyant =

ρfluidVdisplacedg, where ρfluid is the density of the fluid, Vdisplaced is the vol-

ume of fluid displaced, and gis the acceleration due to gravity. For water,

ρfluid = 1 g/cm3. Since the cube is completely submerged, the volume of water

displaced is equal to the volume of the cube, which is 1000 cm3. Therefore, the

buoyant force is:

Fbuoyant = 1 g/cm3×1000 cm3×9.81 m/s2= 9810 N

Step 4: Determine if the cube will sink or float. Since the buoyant force

of 9810 N is greater than the weight of the cube which is 800 N, the net force

acting on the cube is upward and it will float in water.

Question 6

Question

A wooden cube with a side length of 10 cm and a density of 0.8 g/cm3is placed

in a container filled with water. What is the minimum additional load that

must be placed on top of the cube to make it sink?

Solution

Step 1: Calculate the weight of the cube to determine the buoyant force it

experiences. Given that the density of water is 1 g/cm3, the weight of the

wooden cube is:

Weight = Volume ×Density ×Acceleration due to gravity

Volume = (Side length)3= (10 cm)3= 1000 cm3

Weight = 1000 cm3×0.8 g/cm3×9.8 m/s2= 7840 N

Step 2: Determine the buoyant force acting on the cube. The buoyant force

is equal to the weight of the water displaced by the cube:

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

The volume of the cube under water is equal to the volume of the cube that is

submerged:

Volume submerged = Volume of cube ×Density of object

Density of water

Volume submerged = 1000 cm3× 0.8 g/cm3

1 g/cm3!= 800 cm3

Buoyant force = 800 cm3×1 g/cm3×9.8 m/s2= 7840 N

Step 3: The cube will float until the weight of the additional load exceeds the

buoyant force. Let the additional load be denoted by Fload. Since the buoyant

force on the cube is already equal to its weight, the additional load must exceed

the weight of the cube for it to sink:

Fload >7840 N

Therefore, the minimum additional load required to make the cube sink is

7841 N .

Question 7

Question

A cube of side length aand density ρcube is placed in a liquid of density ρliquid.

The cube floats with a fraction xof its volume submerged in the liquid. De-

termine the relationship between the densities ρcube,ρliquid, and the fraction

Solution

Step 1: Determine the forces acting on the cube. The forces acting on the cube

are its weight Wdownwards and the buoyant force Fbuoy upwards. The weight

is given by:

W=ρcube ·g·Vcube

where Vcube =a3is the volume of the cube.

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

cube submerged is given by:

Vsub =x·Vcube =x·a3

Step 3: Determine the buoyant force. The buoyant force is equal to the

weight of the liquid displaced by the cube:

Fbuoy =ρliquid ·g·Vsub

Step 4: Apply Archimedes’ principle. According to Archimedes’ principle,

the buoyant force Fbuoy acting on the cube is equal to the weight Wof the cube:

ρliquid ·g·Vsub =ρcube ·g·Vcube

Step 5: Substituting the expressions for Vsub and Vcube:

ρliquid ·g·x·a3=ρcube ·g·a3

Step 6: Simplify the equation. Solving for x, we find:

x=ρcube

ρliquid

Therefore, the relationship between the densities ρcube,ρliquid, and the frac-

tion xis x=ρcube

ρliquid .

Question 8

Question

A cube of wood with sides of length 10 cm and density 600 kg/m3is floating in

water. Calculate the height of the cube that is above the water’s surface. The

density of water is 1000 kg/m3.

Solution

Let’s denote the height of the cube above the water’s surface as h.

Step 1: First, let’s calculate the volume of the cube. The volume of the

cube is given by Vcube = (10 cm)3= 0.001 m3.

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

is given by mcube = density ×volume.

Substituting the values, we have: mcube = 600 kg/m3×0.001 m3= 0.6 kg.

Step 3: Now, let’s consider the forces acting on the cube when it is floating

in water. The weight of the cube Wcube =mcube ×g, where g= 9.8 m/s2is the

acceleration due to gravity.

The buoyant force Fbuoyant acting on the cube is equal to the weight of the

water displaced by the cube. Fbuoyant = density of water×volume of cube submerged×

Step 4: In equilibrium, the weight of the cube is balanced by the buoyant

force. Thus, we have Wcube =Fbuoyant.

Substitute the expressions and solve for the volume of the cube submerged:

mcube ×g= density of water ×volume of cube submerged ×g

0.6 kg ×9.8 m/s2= 1000 kg/m3×areacross-section ×h×9.8 m/s2

Solve for hto find the height of the cube above the water’s surface.

Question 9

Question

A cylindrical container with a radius of 0.5 m and a height of 1 m is completely

filled with water. A solid metal sphere with a radius of 0.3 m and a density

of 8000 kg/m3is then lowered into the water until it is completely submerged.

Calculate the buoyant force acting on the sphere when it is in the water.

Solution

Step 1: Calculate the volume of the sphere The volume of a sphere is given by

the formula:

V=4

3πr3

Substitute the radius of the sphere:

V=4

3π(0.3)3

V≈0.1131 m3

Step 2: Calculate the weight of the sphere The weight of the sphere can be

calculated using the formula:

Weight = Mass ×Acceleration due to gravity

The mass of the sphere can be calculated using the formula:

Mass = Density ×Volume

Substitute the density and volume of the sphere:

Mass = 8000 kg/m3×0.1131 m3

Mass ≈904.8 kg

Now, calculate the weight of the sphere:

Weight = 904.8 kg ×9.81 m/s2

Weight ≈8886.3 N

Step 3: Calculate the weight of the water displaced The weight of the water

displaced is equal to the weight of the water that would fill the volume of the

sphere. The volume of the sphere is equal to the volume of water displaced.

Thus, the weight of the water displaced is equal to the weight of the sphere:

Weight of water displaced = 8886.3 N

Step 4: Calculate the buoyant force According to Archimedes’ principle, the

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

the fluid that the object displaces. Therefore, the buoyant force acting on the

sphere is equal to the weight of the water displaced:

Buoyant force = 8886.3 N

Therefore, the buoyant force acting on the sphere when it is in the water is

approximately 8886.3 N.

Question 10

Question

A cube of wood with a side length of 10 cm and a density of 0.8 g/cm3is

placed in a liquid of density 1.2 g/cm3. Calculate the volume of the cube that

is submerged in the liquid.

Solution

Step 1: Calculate the weight of the cube Given density of wood, ρwood = 0.8

g/cm3and side length l= 10 cm. The mass of the cube can be calculated as:

mass = ρwood ×volume = ρwood ×l3

mass = 0.8 g/cm3×(10 cm)3= 800 g

The weight of the cube can be found by multiplying the mass by the acceleration

due to gravity g= 9.81 m/s2:

weight = mass ×g= 800 g ×9.81 m/s2= 7848 N

Step 2: Calculate the buoyant force acting on the cube The buoyant force

is equal to the weight of the liquid displaced by the cube. The volume of liquid

displaced is equal to the volume of the cube submerged, denoted as Vsubmerged.

The buoyant force can be calculated as:

buoyant force = ρliquid ×g×Vsubmerged

Now, we need to find Vsubmerged.

Step 3: Calculate the volume submerged The volume submerged can be

found using Archimedes’ principle which states that the buoyant force is equal

to the weight of the displaced fluid:

buoyant force = weight of fluid displaced

⇒ρliquid ×g×Vsubmerged =ρwood ×g×l3

⇒1.2 g/cm3×Vsubmerged = 0.8 g/cm3×(10 cm)3

⇒Vsubmerged =0.8×1000

1.2= 666.67 cm3

Therefore, the volume of the cube that is submerged in the liquid is 666.67

cm3.

Question 11

Question

A cube of wood with a density of 700 kg/m3and a side length of 0.05 m is

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

water.

Solution

Let’s denote: - Vcube as the volume of the cube, - das the depth to which the

cube is submerged, - ρwater as the density of water, - Fbuoyant as the buoyant

force acting on the cube, - Fgravity as the weight of the cube.

Step 1: Calculate the volume of the cube. Since the cube has a side length

of 0.05 m, the volume can be calculated as:

Vcube = (0.05 m)3= 0.000125 m3

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

calculated using the formula:

Fgravity =m·g

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

mass of the cube can be calculated as:

m=ρcube ·Vcube

m= 700 kg/m3·0.000125 m3= 0.0875 kg

The weight of the cube is:

Fgravity = 0.0875 kg ·9.8 m/s2= 0.8575 N

Step 3: Calculate the buoyant force. According to Archimedes’ principle,

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

by the cube. The buoyant force can be calculated as:

Fbuoyant =ρwater ·Vsubmerged ·g

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

is floating, the buoyant force is equal to the weight of the cube:

Fbuoyant =Fgravity

ρwater ·Vsubmerged ·g=ρcube ·Vcube ·g

ρwater ·Vsubmerged =ρcube ·Vcube

Vsubmerged =ρcube ·Vcube

ρwater

Vsubmerged =700 kg/m3·0.000125 m3

1000 kg/m3= 0.0875 m3

Step 4: Calculate the depth to which the cube is submerged. The depth to

which the cube is submerged can be calculated using the formula:

Vsubmerged =A·d

where Ais the area of the base of the cube and dis the depth. Since the base

of the cube is a square with side length 0.05 m, the area can be calculated as:

A= (0.05 m) ×(0.05 m) = 0.0025 m2

0.0875 m3= 0.0025 m2×d

d=0.0875 m3

0.0025 m2= 0.035 m

Therefore, the depth to which the cube is submerged in the water is 0.035

Question 12

Question

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

in water. What is the buoyant force acting on the cube?

Solution

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

V=L3, where Lis the length of a side. In this case, the side length L= 0.1

m, so the volume of the cube is:

V= (0.1 m)3= 0.001 m3

Step 2: Determine the weight of the cube. The weight of the cube is given by

the formula W=mg, where mis the mass of the cube and gis the acceleration

due to gravity (9.81 m/s2). The mass of the cube can be calculated using the

formula m=ρV , where ρis the density of the material. Substituting in the

given values:

m= 600 kg/m3×0.001 m3= 0.6 kg

Thus, the weight of the cube is:

W= 0.6 kg ×9.81 m/s2= 5.886 N

Step 3: Calculate the buoyant force. According to Archimedes’ principle,

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

fluid displaced by the object. Since the cube is floating, the weight of the cube

is equal to the buoyant force. Therefore, the buoyant force acting on the cube

is:

Buoyant force = 5.886 N

Question 13

Question

A cylindrical container has a height of 1.5 meters and a radius of 0.5 meters.

The container is filled with water to a height of 1.2 meters. Determine the

buoyant force acting on a solid iron sphere of radius 0.2 meters submerged in

the water. The density of iron is 7800 kg/m3and the density of water is 1000

kg/m3.

Solution

Step 1: Find the volume of water displaced by the iron sphere. The volume of

the iron sphere can be calculated using the formula for the volume of a sphere:

Vsphere =4

3πr3

where ris the radius of the sphere. Substitute r= 0.2 m into the formula to

find the volume of the iron sphere:

Vsphere =4

3π(0.2)3=4

3π(0.008) = 0.03351 m3

Step 2: Calculate the volume of water displaced by the iron sphere. The

volume of water displaced is equal to the volume of the iron sphere that is

submerged in the water.

Vwater =Vsphere = 0.03351 m3

Step 3: Determine the weight of water displaced. The weight of water

displaced can be calculated using the formula:

Weight = Density ×Volume ×Acceleration due to gravity

Given that the density of water is 1000 kg/m3and the acceleration due to

gravity is 9.8 m/s2, the weight of water displaced is:

Weight = 1000 ×0.03351 ×9.8 = 328.93 N

Step 4: Determine the weight of the iron sphere. The weight of the iron

sphere can be calculated using the formula:

Weight = Density ×Volume ×Acceleration due to gravity

Given that the density of iron is 7800 kg/m3and the volume of the iron sphere

is the same as the volume of water displaced, the weight of the iron sphere is:

Weight = 7800 ×0.03351 ×9.8 = 2530.51 N

Step 5: Determine the buoyant force acting on the iron sphere. The buoyant

force is equal to the weight of the water displaced, which we calculated to be

328.93 N. Thus, the buoyant force acting on the iron sphere is 328.93 N.