PHYS 101 - ELEMENTS OF PHYSICS Buoyant forces and Archimedes' principle Question Bank Set 3

PHYS 101 - ELEMENTS OF PHYSICS

- Buoyant forces and Archimedes’

principle

Question Bank - Set 3

Liberty University

Question 1

Question

A cylindrical object with a radius of 5 cm and a height of 10 cm is submerged

in water. The density of the object is 2000 kg/m3and the density of water is

1000 kg/m3. Calculate the buoyant force acting on the object.

Solution

Step 1: First, calculate the volume of the cylindrical object: The volume of

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

Substituting r= 5 cm and h= 10 cm (or 0.1 m), we get:

V=π×(0.05 m)2×0.1 m

V=π×0.0025 m2×0.1 m

V= 0.000785 m3

Step 2: Calculate the weight of the object: The weight of the object is

equal to the volume times the density times the acceleration due to gravity

(g= 9.81 m/s2). Substituting density = 2000 kg/m3:

W= 2000 kg/m3×0.000785 m3×9.81 m/s2

W= 2000 ×0.000785 ×9.81 kg

W= 15.5073 N

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

displaced is given by the volume of water displaced times the density of water

times the acceleration due to gravity. The volume of water displaced is the same

as the volume of the cylinder, since it is fully submerged.

Wwater = 1000 kg/m3×0.000785 m3×9.81 m/s2

Wwater = 1000 ×0.000785 ×9.81 N

Wwater = 7.737 N

Step 4: Calculate the buoyant force: The buoyant force is equal to the

weight of the water displaced. Therefore, the buoyant force acting on the object

is 7.737 N .

Question 2

Question

A cube of wood with a density of 0.7 g/cm3and side length of 4 cm is floating

in water. What is the mass of the cube and what fraction of the cube’s volume

is submerged in the water?

Solution

Step 1: Determine the mass of the cube.

The density of the cube is given as 0.7 g/cm3and the side length is 4 cm. The

volume of the cube can be calculated as V=l3, where lis the side length.

Therefore,

V= 43= 64 cm3.

To find the mass, we can use the formula m=ρV , where ρis the density.

Substituting in our values,

m= 0.7×64 = 44.8 g.

Step 2: Determine the fraction of the cube’s volume submerged.

Let the volume submerged be Vs. Since the cube is floating, the buoyant force

Fbacting on the cube is equal to the weight of the cube Fg. Using Archimedes’

principle, Fb=ρwaterVsg, where ρwater is the density of water and gis the

acceleration due to gravity. The weight of the cube is mcubeg. Equating these,

we get

ρwaterVsg=mcubeg.

Substitute the values we found earlier:

1×Vs×9.81 = 44.8×9.81.

Solving for Vs:

Vs=44.8×9.81

9.81 = 44.8 cm3.

Thus, the fraction of the cube’s volume submerged in water is Vs/V =

44.8/64 = 0.7, or 70

Question 3

Question

A cylindrical metal object with a density of 8000 kg/m3and a volume of 0.04

m3is submerged in a container of water. If the metal object is suspended by a

rope and completely submerged in water, calculate the tension in the rope.

Solution

Step 1: First, we need to determine the weight of the metal object. The weight

of an object can be calculated using the formula W=mg, where mis the mass

of the object and gis the acceleration due to gravity (approximately 9.81 m/s2).

Given that the density of the metal object is 8000 kg/m3and its volume is 0.04

m3, we can calculate the mass of the object as follows:

m= Density ×Volume = 8000 kg/m3×0.04 m3= 320 kg

Therefore, the weight of the metal object is:

W=mg = 320 kg ×9.81 m/s2= 3139.2 N

Step 2: Now, we can determine the buoyant force acting on the metal object.

The buoyant force is equal to the weight of the fluid displaced by the object,

which can be calculated using the formula Fbuoyant =ρ·V·g, where ρis the

density of the fluid (1000 kg/m3for water), Vis the volume of the object, and

gis the acceleration due to gravity:

Fbuoyant = 1000 kg/m3×0.04 m3×9.81 m/s2= 392.4 N

Step 3: Since the metal object is suspended by a rope and completely sub-

merged in water, the tension in the rope is equal to the difference between the

weight of the object and the buoyant force:

Tension = W−Fbuoyant = 3139.2 N −392.4 N = 2746.8 N

Therefore, the tension in the rope when the metal object is completely sub-

merged in water is 2746.8 N.

Question 4

Question

A metal block of mass 500 kg and volume 0.5 m3is submerged in a container of

water. Calculate the buoyant force acting on the block and determine if it will

float or sink.

Solution

Step 1: Calculate the weight of the block. Given that the mass of the block is

500 kg and the acceleration due to gravity is 9.81 m/s2, the weight Wcan be

calculated using the formula W=m·g, where mis the mass of the block and

gis the acceleration due to gravity.

W= 500 kg ×9.81 m/s2

W= 4905 N

Step 2: Calculate the buoyant force. The buoyant force Fbacting on an

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

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

the block. Since the density of water is 1000 kg/m3, the buoyant force can be

calculated as:

Fb= density of water ×volume of water displaced ×g

Fb= 1000 kg/m3×0.5 m3×9.81 m/s2

Fb= 4905 N

Step 3: Determine if the block will float or sink. The block will float if the

buoyant force is greater than or equal to the weight of the block. In this case,

since 4905 N = 4905 N, the buoyant force is equal to the weight of the block.

Therefore, the block will remain submerged in the water.

Question 5

Question

A cube of wood with a side length of 0.1 m and a density of 800 kg/m3is floating

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

that the density of water is 1000 kg/m3.

Solution

Step 1: Let’s first determine the volume of the cube. The volume of a cube is

given by the formula V=s3, where sis the side length. So, for the cube in this

problem:

V= (0.1 m)3= 0.001 m3

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

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

the values:

m= 800 kg/m3×0.001 m3= 0.8 kg

Step 3: Using Archimedes’ principle, 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=ρwaterVsubmergedg, where ρwater is the density of water, Vsubmerged

is the volume submerged by the cube, and gis the acceleration due to gravity.

Since the cube is in equilibrium, the weight of the cube is balanced by the

buoyant force:

mcubeg=ρwaterVsubmergedg

Step 4: We can now solve for the volume submerged by the cube.

Vsubmerged =mcube

ρwater

=0.8 kg

1000 kg/m3= 0.0008 m3

Step 5: Finally, we can calculate the depth to which the cube is submerged

in water. Since the cube is floating, the volume submerged is equal to the cross-

sectional area of the cube multiplied by the depth of submersion. Therefore:

Area ×depth = 0.0008 m3

0.1 m ×depth = 0.0008 m3

depth = 0.0008 m3

0.1 m = 0.008 m = 8 mm

Thus, the cube is submerged to a depth of 8 mm in water.

Question 6

Question

A cube of wood with a density of 0.8 g/cm3and side length of 10 cm is floating

in water. If 60

Solution

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

volume of water displaced to find the density of water.

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

of the cube is given by Vcube = side length3= (10 cm)3= 1000 cm3. Since 60

Step 2: Use the volume of water displaced to find the density of water. Let

Vwater be the volume of water displaced. According to Archimedes’ principle,

the buoyant force (Fbuoyant) is equal to the weight of the water displaced. The

density of the cube is given as 0.8 g/cm3. Let mcube be the mass of the cube. We

know that mcube = density ×Vcube. The weight of the cube is Wcube =mcubeg,

where gis the acceleration due to gravity. Since the cube is floating, the buoyant

force Fbuoyant is equal to the weight of the cube. Therefore, Fbuoyant =mcubeg.

The density of water (ρwater) can be found using Fbuoyant =ρwater×g×Vwater.

Substitute mcubegfor Fbuoyant:mcubeg=ρwater ×g×Vwater. Thus, density of

water, ρwater =mcube

Vwater =density of cube×Vcube

Vwater . Substitute the values: ρwater =

0.8 g/cm3×1000 cm3

600 cm3=800

600 = 1.33 g/cm3.

Therefore, the density of water is 1.33 g/cm3.

Question 7

Question

A cylindrical object of height h, radius r, and density ρois floating vertically in

a container filled with a liquid of density ρl. The top of the cylinder is xunits

above the liquid surface. Determine the expression for the buoyant force acting

on the object when it is partially submerged in the liquid.

Solution

Step 1: First, calculate the volume of the cylinder that is submerged in the

liquid. This volume can be expressed as Vsubmerged =πr2x.

Step 2: The weight of the object is given by Wobject =ρo·Vtotal ·g, where

Vtotal is the total volume of the cylinder.

Step 3: The buoyant force Fbuoyant acting on the object is equal to the weight

of the liquid displaced, which is given by Fbuoyant =ρl·Vsubmerged ·g.

Step 4: Equate the weight of the object and the buoyant force to find the

expression for the buoyant force:

ρo·πr2h·g=ρl·πr2x·g

Step 5: Solve for xto obtain the expression for the buoyant force:

x=ρo·h

ρl

Therefore, the buoyant force acting on the object when it is partially sub-

merged in the liquid is ρo·πr2·ρo·h

ρl

·g.

Question 8

Question

A cube of aluminum with a side length of 10 cm and a density of 2.7 g/cm3

is floating in a container of water. Determine the depth to which the cube is

submerged in the water.

Solution

Step 1: The buoyant force acting on the aluminum cube is equal to the weight

of the water displaced by the cube. We can find the volume of the submerged

cube using Archimedes’ principle.

Step 2: The weight of the cube is equal to its mass multiplied by the accel-

eration due to gravity (w=m·g).

Step 3: The volume of the cube can be calculated using the formula V=l3,

where lis the side length of the cube.

Step 4: Set up an equation for the buoyant force (Fb) using the volume of

the cube, the density of water (ρw= 1 g/cm3), and the acceleration due to

gravity (g= 9.81 m/s2).

Step 5: Use the equation for density (ρ=m

V) to find the mass of the cube.

Step 6: The weight of the water displaced by the cube is equal to the weight

of the cube.

Step 7: The depth to which the cube is submerged can be calculated by

dividing the volume of the submerged cube by the surface area of the cube in

contact with the water.

Step 8: Calculate the depth to which the cube is submerged in the water.

Question 9

Question

A cylindrical object of radius Rand height his floating in a container of water.

The density of the object is ρo, and the density of water is ρw. Determine the

fraction of the object’s volume that is submerged in the water.

Solution

Let’s denote the fraction of the object’s volume that is submerged in the water

as f. To find f, we need to consider the forces acting on the object.

Step 1: Determine the volume of the object submerged in water. Assume the

object is fully submerged, then the volume of water displaced would be equal

to the volume of the cylinder:

Vdisplaced =πR2h

Step 2: Write the equation for the buoyant force. The buoyant force (Fb) is

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

the weight of the fluid displaced:

Fb=ρw·g·Vdisplaced

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

by:

Wobject =ρo·g·Vtotal

where Vtotal =πR2his the total volume of the object.

Step 4: Set up the equilibrium condition. For the object to be floating, the

buoyant force must be equal to the weight of the object:

Fb=Wobject

Step 5: Substitute the equations and solve for f.

ρw·g·πR2h=ρo·g·πR2h·f

f=ρw

ρo

Therefore, the fraction of the object’s volume submerged in water is ρw

ρo.

Question 10

Question

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

water. What is the minimum distance from the bottom of the cube to the water

surface?

Solution

Step 1: We first find the mass of the wood cube. The mass of the cube is given

by the density formula:

mass = density ×volume

mass = 0.8 g/cm3×(10 cm)3

mass = 0.8×1000 g

mass = 800 g

Step 2: Next, we find the volume of water displaced by the cube. The volume

of water displaced is equal to the volume of the cube submerged in water. Since

the cube is floating, the weight of the wood cube equals the buoyant force. The

volume of water displaced is:

V= side2×distance from bottom to water surface

V= (10 cm)2×h

V= 100 cm2×h

Step 3: The buoyant force is given by the formula:

Buoyant force = density of water ×acceleration due to gravity ×V

Buoyant force = 1 g/cm3×9.8 m/s2×100 cm2×h

Step 4: The weight of the wood cube is equal to the buoyant force:

Weight of cube = Buoyant force

mass of cube×acceleration due to gravity = density of water×acceleration due to gravity×V

800 g ×9.8 m/s2= 1 g/cm3×9.8 m/s2×100 cm2×h

Step 5: Solve for h, the distance from the bottom of the cube to the water

surface:

h=800 ×9.8

1×100 cm

h= 78 cm

Therefore, the minimum distance from the bottom of the cube to the water

surface is 78 cm.

Question 11

Question

A rectangular block of wood with dimensions 10 cm ×5 cm ×3 cm is floating in

water. Calculate the density of the wood if the block is immersed in water up

to a depth of 2 cm. The density of water is 1000 kg/m3.

Solution

Step 1: First, we need to calculate the volume of the wooden block. The volume

of the wooden block is given by:

V=l×w×h

where l,w, and hare the length, width, and height of the block, respectively.

Plugging in the values, we get:

V= 10 cm ×5 cm ×3 cm = 150 cm3

Step 2: Next, we need to calculate the volume of water displaced by the

block. When the block is submerged in water, it displaces its own weight in

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

submerged. The volume of water displaced is given by:

Vdisplaced =l×w×d

where dis the depth to which the block is immersed in water. Plugging in the

values, we get:

Vdisplaced = 10 cm ×5 cm ×2 cm = 100 cm3

Step 3: We know that the density of water is 1000 kg/m3. The mass of the

volume of water displaced by the block is given by:

mwater =ρwater ×Vdisplaced

Plugging in the values, we get:

mwater = 1000 kg/m3×100 cm3

1000000 cm3/m3= 0.1 kg

Step 4: The buoyant force acting on the block is due to the weight of the

water displaced. The buoyant force is given by:

Fbuoyant =mwater ×g

where gis the acceleration due to gravity (9.81 m/s2). Plugging in the values,

we get:

Fbuoyant = 0.1 kg ×9.81 m/s2= 0.981 N

Step 5: The weight of the block is balanced by the buoyant force and the

weight of the block. The weight of the block is calculated using the equation:

Fblock =mblock ×g

The weight of the block is equal to the weight of water displaced since the block

is floating. Plugging in the values, we get:

Fblock = 0.1 kg ×10 = 0.981 N

Step 6: From Step 5, we see that the weight of the block is equal to the

buoyant force. This means that the density of the block is equal to the density

of water, which is 1000 kg/m3.

Question 12

Question

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

container of water. Determine the depth to which the cube sinks in the water.

The density of water is 1 g/cm3.

Solution

Step 1: First, we need to find the weight of the cube. Given that the density of

wood is 0.8 g/cm3and the volume of the cube is (10 cm)3= 1000 cm3, we can

find the mass of the cube as follows:

Mass of cube = Density ×Volume = 0.8 g/cm3×1000 cm3= 800 g

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

due to gravity, g= 9.8 m/s2:

Weight of cube = 800 g ×9.8 m/s2= 7840 N

Step 2: Next, we need to determine the buoyant force acting on the cube.

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

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

volume of the cube. Therefore, the buoyant force can be calculated as follows:

Buoyant force = Density of water×Volume of cube×g= 1 g/cm3×1000 cm3×9.8 m/s2= 9800 N

Step 3: The depth to which the cube sinks can be determined by considering

the equilibrium of forces. When the cube is at rest in the water, the weight of

the cube is balanced by the buoyant force:

Weight of cube = Buoyant force

7840 N = 9800 N

Since the weight of the cube is less than the buoyant force, the cube will sink

until the upward buoyant force equals the weight of the cube. Therefore, the

cube will sink to a depth such that the weight of the water displaced by the

volume submerged is equal to the weight of the cube. Let dbe the depth to

which the cube sinks. The volume of water displaced is 1000 cm3×d.

Weight of water displaced = Density of water×Volume of water displaced×g= 1 g/cm3×1000 cm3×d×9.8 m/s2= 9800dN

Setting the weight of water displaced equal to the weight of the cube:

9800d= 7840

d=7840

9800 = 0.8 m

Therefore, the cube will sink to a depth of 0.8 meters in the water.

Question 13

Question

A cube of wood with sides measuring 6 cm is floating in a tub of water. The

density of water is 1000 kg/m3and the density of wood is 700 kg/m3. Determine

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

Solution

Step 1: Start by determining the volume of the cube. The volume of the cube is

given by the formula: Vcube =s3, where sis the length of each side of the cube.

Given that s= 6 cm, convert this to meters before calculating the volume.

Vcube = (0.06 m)3= 0.000216 m3

Step 2: Calculate the weight of the wooden cube. The weight of the wooden

cube is given by the formula: Wcube =mcube ·g, where mcube is the mass of

the cube and gis the acceleration due to gravity. The mass of the cube can be

calculated using the formula mcube =ρwood ·Vcube.

mcube = 700 kg/m3×0.000216 m3= 0.1512 kg

Wcube = 0.1512 kg ×9.81 m/s2= 1.48 N

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

is given by the formula: Fbuoyant =ρwater ·Vsubmerged ·g, where ρwater is the

density of water, Vsubmerged is the volume of water displaced by the submerged

part of the cube, and gis the acceleration due to gravity. Because the cube is

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

Fbuoyant = 1.48 N

Step 4: Find the volume of water displaced (submerged volume). Using

Archimedes’ principle, the volume of water displaced is equal to the volume of

the cube that is submerged. Let the depth to which the cube is submerged be

d, then Vsubmerged =s2·d.

Vsubmerged = 0.062m×d= 0.0036 dm3

Step 5: Equate the weight of the cube to the buoyant force and solve for d.

1.48 N = 1000 kg/m3×0.0036 dm3×9.81 m/s2

d=1.48

1000 ×0.0036 ×9.81 = 0.043 m = 4.3 cm

Therefore, the cube is submerged to a depth of 4.3 cm in the water.

Question 14

Question

A uniform solid sphere of radius 10 cm and mass 3 kg is placed in a container

filled with water. The sphere is released and floats at a certain depth. Calculate

the depth at which the sphere floats in the water, given that the density of water

is 1000 kg/m3and the acceleration due to gravity is 9.8 m/s2.

Solution

Step 1: First, let’s calculate the buoyant force acting on the sphere. The buoyant

force Fbuoyant is given by the formula

Fbuoyant =ρ·Vdisplaced ·g,

where ρis the density of the fluid, Vdisplaced is the volume of the fluid displaced

by the object, and gis the acceleration due to gravity.

Step 2: The volume of the water displaced by the sphere is equal to the

volume of the sphere submerged in the water. Since the sphere floats, the

buoyant force will exactly balance the weight of the sphere. The weight of the

sphere Wis given by

W=m·g,

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

Step 3: The volume of the sphere Vsphere is given by

Vsphere =4

3πr3,

where ris the radius of the sphere.

Step 4: The mass of the water displaced mdisplaced can be calculated using

the density of water and the volume of the sphere submerged in water.

mdisplaced =ρ·Vsphere.

Step 5: Since the sphere floats, the weight of the displaced water must be

equal to the weight of the sphere. This gives us the equation

ρ·Vsphere ·g=m·g.

Step 6: Substituting the expressions for Vsphere and mdisplaced into the equa-

tion and solving for the depth dat which the sphere floats, we get

3πr3·ρ·g=m·g.

3π(0.1)3·ρ·g·d=m·g.

Step 7: Solving for d, we find

d=m

3πr3·ρ.

Step 8: Substituting the given values m= 3 kg, r= 0.1 m, and ρ=

1000 kg/m3into the formula, we can calculate the depth at which the sphere

floats in the water.

d=3

3π(0.1)3·1000.

Step 9: Calculating the depth d, we get

d=3

3π·0.001 ·1000 =3

4π≈0.238 m.

Therefore, the sphere floats at a depth of approximately 0.238 meters in the

water.

Question 15

Question

A hot air balloon has a volume of 3000 m3. The air inside the balloon is heated

to a temperature of 150

C. If the air outside the balloon is at 20

C, calculate

the buoyant force acting on the balloon. Assume the density of air at 20

C is

1.2 kg/m3and the density of air at 150

C is 0.8 kg/m3.

Solution

Step 1: Calculate the difference in air densities inside and outside the balloon.

Given: - Density of air at 20

C (ρ0) = 1.2 kg/m3- Density of air at 150

C (ρ)

= 0.8 kg/m3

The difference in densities is:

∆ρ=ρ0−ρ= 1.2 kg/m3−0.8 kg/m3= 0.4 kg/m3

Step 2: Calculate the mass of air displaced by the balloon. The mass of air

displaced is equal to the difference in densities times the volume of the balloon:

m= ∆ρ×V= 0.4 kg/m3×3000 m3= 1200 kg

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

air is:

Wdisplaced =m×g= 1200 kg ×9.81 m/s2= 11772 N

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

force acting on the balloon is equal to the weight of the displaced air:

Fbuoyant = 11772 N

Therefore, the buoyant force acting on the hot air balloon is 11772 N.

Question 16

Question

A cube of side length Land density ρcube is floating in a liquid of density ρliquid.

The cube is initially floating with a fraction fof its volume submerged beneath

the liquid. Find an expression for the density ρcube of the cube in terms of f,

ρliquid, and L.

Solution

Step 1: First, let’s write the equilibrium condition for the cube as it floats in

the liquid. The buoyant force must be equal to the weight of the submerged

portion of the cube. Step 2: The volume of the cube is V=L3and the

fraction submerged is f, so the volume submerged is Vsubmerged =fV . Step

3: The weight of the cube is given by Wcube =ρcubeV g, and the weight of

the submerged portion of the cube is Wsubmerged =ρliquidVsubmergedg. Step

4: Using the equilibrium condition, we have ρcubeV g =ρliquidfV g. Step 5:

Substituting the expressions for Vand Vsubmerged into the equation, we get

ρcubeL3=ρliquidfL3. Step 6: Finally, solving for ρcube, we find that ρcube =

ρliquidf.

Question 17

Question

A cylindrical ship with a radius of 10 meters and a length of 100 meters floats

in seawater (density of 1025 kg/m3). The ship’s average density is 900 kg/m3.

Determine the fraction of the ship’s volume that is beneath the waterline.

Solution

Step 1: Calculate the volume of the ship submerged in water.

The buoyant force Fbacting on the ship is equal to the weight of the water

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

Fb=ρwater ·Vsubmerged ·g

where ρwater = density of water = 1025 kg/m3,Vsubmerged = volume of the ship

submerged in water, g= acceleration due to gravity = 9.81 m/s2.

The weight of the ship Wcan be calculated as:

W=ρavg ·Vtotal ·g

where ρavg = average density of the ship = 900 kg/m3,Vtotal = total volume of

the ship.

Since the ship is floating, the buoyant force Fbis equal to the weight of the

ship W. Therefore,

Fb=W

ρwater ·Vsubmerged ·g=ρavg ·Vtotal ·g

ρwater ·Vsubmerged =ρavg ·Vtotal

Vsubmerged =ρavg

ρwater

·Vtotal

Substitute the given values to find the volume submerged:

Vsubmerged =900 kg/m3

1025 kg/m3·π·(10 m)2·100 m

Vsubmerged =900

1025 ·π·1000

Vsubmerged ≈876.71 m3

Therefore, the volume of the ship submerged in water is approximately

876.71 m3.

Step 2: Calculate the fraction of the ship’s volume beneath the waterline.

The total volume of the ship Vtotal can be calculated as:

Vtotal =π·(10 m)2·100 m

Vtotal =π·1000

Vtotal = 3141.59 m3

The fraction of the ship’s volume beneath the waterline is:

Vsubmerged

Vtotal

=876.71 m3

3141.59 m3

Vsubmerged

Vtotal

≈0.279

Therefore, approximately 27.9

Question 18

Question

A cube of aluminum with sides of length 10 cm and a density of 2700 kg/m3is

submerged in water. Calculate the buoyant force acting on the cube.

Solution

Step 1: Calculate the volume of the aluminum cube.

Volume of cube = side length3= (0.10 m)3= 0.001 m3

Step 2: Calculate the weight of the aluminum cube.

Weight = Mass ×Acceleration due to gravity

Mass = Volume ×Density = 0.001 m3×2700 kg/m3= 2.7 kg

Weight = 2.7 kg ×9.81 m/s2= 26.487 N

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

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

displaced.

Buoyant Force = Weight of water displaced

Volume of water displaced = Volume of cube = 0.001 m3

Density of water = 1000 kg/m3

Weight of water displaced = Volume of water displaced×Density of water×Acceleration due to gravity

Weight of water displaced = 0.001 m3×1000 kg/m3×9.81 m/s2= 9.81 N

Buoyant Force = 9.81 N

Therefore, the buoyant force acting on the aluminum cube is 9.81 N.

Question 19

Question

A metal object of volume 500 cm3and density 2000 kg/m3is immersed in water.

Calculate the buoyant force acting on the object and determine if the object

will float or sink in the water.

Solution

Step 1: Convert the volume of the object to m3: Given that the volume of the

object is 500 cm3, we need to convert this to m3. Since 1 cm3= 1 ×10−6m3,

the volume of the object in m3is:

500 cm3×1×10−6m3/cm3= 0.0005 m3

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

calculated using the formula: mass = density ×volume. Substituting the given

values: mass = 2000 kg/m3×0.0005 m3= 1 kg

Step 3: Calculate the buoyant force: The buoyant force is given by the

formula: Buoyant force = Density of fluid ×Volume submerged ×Acceleration

due to gravity. The density of water is 1000 kg/m3, and the acceleration due to

gravity is 9.81 m/s2.

Since the object is completely immersed in water, the volume submerged is

equal to its total volume, which is 0.0005 m3. Substituting the values: Buoyant

force = 1000 kg/m3×0.0005 m3×9.81 m/s2= 4.905 N

Step 4: Determine if the object will float or sink: The object will float if the

buoyant force is greater than or equal to the weight of the object, and it will

sink if the buoyant force is less than the weight of the object.

Since the mass of the object is 1 kg, its weight is given by: Weight = mass

×gravity = 1 kg ×9.81 m/s2= 9.81 N

Since the buoyant force (4.905 N) is less than the weight of the object (9.81

N), the object will sink in water.

Question 20

Question

A cube of wood with a density of 0.6 g/cm3and side length 10 cm is floating in

a liquid with a density of 1.2 g/cm3. What is the depth of the cube below the

surface of the liquid?

Solution

Step 1: Find the density of water (liquid) using the given density. Step 2: Apply

Archimedes’ principle to find the volume of wood submerged in the liquid. Step

3: Use the volume to calculate the depth below the surface of the liquid.

Step 1: The density of water is 1.2 g/cm3.

Step 2: Let Vbe the volume of wood submerged in the liquid. The buoyant

force Fbacting on the wood is given by:

Fb=ρf·g·V

where ρfis the density of the liquid and gis the acceleration due to gravity.

The weight of the wood is balanced by the buoyant force, so:

Fb=m·g

where mis the mass of the wood. The mass of the wood can be calculated using

its density ρwand volume. Since the wood is submerged, the volume is V. The

mass mis given by:

m=ρw·V

Equating the expressions for Fb, we get:

ρf·g·V=ρw·V

Solving for V:

V=ρw

ρf

·V

Step 3: The depth of the cube below the surface of the liquid can be calcu-

lated by dividing the volume of the cube submerged by the area of the cube’s

face. Let hbe the depth of submersion. The volume of wood submerged is

equivalent to the volume of the submerged part of the cube:

V=A·h

where Ais the area of the cube’s face. Solving for h:

h=V

Substitute the calculated value of Vinto the above equation to find the

depth of the cube below the surface of the liquid.

Question 21

Question

A cylindrical tank has a diameter of 4 meters and a height of 6 meters. If the

tank is filled with water, calculate the buoyant force acting on the tank.

Solution

Step 1: Determine the volume of the cylindrical tank. The volume of a cylinder

is given by the formula V=πr2h, where ris the radius of the cylinder and h

is the height. Given that the diameter of the tank is 4 meters, the radius ris

half of the diameter, so r= 2 meters. The height his 6 meters. Therefore, the

volume of the tank is:

V=π×(2)2×6 = 24πm3

Step 2: Calculate the weight of water in the tank. The weight of water in

the tank is equal to the mass of the water times the acceleration due to gravity,

g= 9.8 m/s2. The density of water is approximately 1000 kg/m3. Therefore,

the mass of the water is:

mass = density ×volume = 1000 ×24π= 24000πkg

And the weight of the water is:

weight = mass ×g= 24000π×9.8 N

Step 3: Calculate the weight of the tank. The weight of the tank is equal

to the mass of the tank times the acceleration due to gravity. To simplify the

calculation, let’s assume the density of the tank material is 8000 kg/m3. The

volume of the tank is 24πm3, so the mass of the tank is:

mass = density ×volume = 8000 ×24π= 192000πkg

And the weight of the tank is:

weight = mass ×g= 192000π×9.8 N

Step 4: Calculate the buoyant force. The buoyant force is equal to the weight

of the water displaced by the tank, which is equal to the weight of the water in

the tank. Therefore, the buoyant force acting on the tank is:

Buoyant force = weight of water = 24000π×9.8 N

Question 22

Question

A cylindrical tank filled with water has a radius of 2 meters and a height of 3

meters. A solid sphere with a radius of 0.5 meters and a density of 800 kg/m3is

submerged in the water. Calculate the buoyant force acting on the sphere and

determine whether the sphere will sink or float.

Given: Density of water = 1000 kg/m3, Acceleration due to gravity, g=

9.81 m/s2.

Solution

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

the formula:

V=4

3πr3

Substitute r= 0.5 meters into the formula to find the volume of the sphere:

V=4

3π(0.5)3=4

3π(0.125) = 1

6πm3

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

found using the formula:

mass = density ×volume

Given that the density of the sphere is 800 kg/m3, substitute the values of

density and volume into the formula:

mass = 800 ×1

6π=400

3πkg

Step 3: Calculate the weight of the sphere. The weight of the sphere is given

by the formula:

weight = mass ×g

Substitute the mass of the sphere and the acceleration due to gravity into the

formula:

weight = 400

3π×9.81 = 3920

3πN

Step 4: Calculate the weight of the water displaced by the sphere (buoyant

force). The weight of the water displaced is equal to the weight of the sphere

when it is fully submerged. The volume of water displaced is equal to the volume

of the sphere.

Weight of water displaced = 400

3π×9.81 = 3920

3πN

Step 5: Determine whether the sphere will sink or float. Since the weight of

the water displaced is equal to the weight of the sphere, the sphere will float in

the water.

Question 23

Question

A rectangular block of wood with a density of 0.6 g/cm3and dimensions 10 cm

x 5 cm x 3 cm is floating in water. What is the minimum mass of lead that

needs to be placed on top of the block so that it sinks completely in water? The

density of lead is 11.3 g/cm3.

Solution

Step 1: Calculate the volume of the wood block.

The volume of the wood block is given by:

Vwood = length ×width ×height

Vwood = 10 cm ×5 cm ×3 cm = 150 cm3

Step 2: Determine the volume of water displaced by the wood block when

it is floating.

Since the block is floating, the weight of the water displaced by the block is

equal to the weight of the block. The weight of the block can be calculated

using its volume and density:

Weight of wood block = Vwood ×density of wood

Weight of wood block = 150 cm3×0.6 g/cm3= 90 g

Step 3: Calculate the minimum mass of lead required to sink the wood block

completely.

To sink the wood block completely, the total weight of the system (wood block

+ lead) must be greater than the weight of the water displaced by the wood

block. Let m be the mass of the lead needed.

m×density of lead = Weight of wood block + m×1 g/cm3

m×11.3 = 90 + m

10.3m= 90

m=90

10.3≈8.74 g

Therefore, the minimum mass of lead that needs to be placed on top of the

block so that it sinks completely in water is approximately 8.74 g.

Question 24

Question

A cylindrical object with a height of 12 cm and a radius of 3 cm is placed in a

container of water. The object floats with 4 cm of its height above the water

surface. Calculate the density of the cylindrical object.

Solution

Step 1: First, let’s determine the volume of the cylindrical object submerged in

water. The total volume of the cylinder can be calculated using the formula for

the volume of a cylinder: Vcylinder =πr2h, where ris the radius and his the

height. Given r= 3 cm and h= 12 cm, we have:

Vcylinder =π×(3 cm)2×12 cm

Vcylinder = 108πcm3

Step 2: Since 4 cm of the cylinder is above the water surface, the submerged

volume can be calculated as Vsubmerged =πr2×submerged height. Given the

submerged height is 8 cm, we have:

Vsubmerged =π×(3 cm)2×8 cm

Vsubmerged = 72πcm3

Step 3: The buoyant force acting on the cylinder is equal to the weight

of the water displaced by the submerged volume of the cylinder. This can be

calculated using Archimedes’ principle, Fbuoyant =ρ×Vsubmerged ×g, where ρ

is the density of water and gis the acceleration due to gravity. The weight of

the cylinder can be calculated using its density ρcylinder, volume Vcylinder, and

acceleration due to gravity g, as Wcylinder =ρcylinder ×Vcylinder ×g.

Step 4: Since the cylinder is floating, the buoyant force Fbuoyant must equal

the weight of the cylinder Wcylinder.T hisgivesustheequation :ρ×Vsubmerged ×g=

ρcylinder ×Vcylinder ×g.

Step 5: Substitute the known values into the equation:

ρ×72π×g=ρcylinder ×108π×g

Step 6: Simplify the equation by canceling out gand π:

72ρ= 108ρcylinder

Step 7: Solve for the density of the cylinder, ρcylinder:

ρcylinder =72ρ

108

ρcylinder =2

3ρ

Therefore, the density of the cylindrical object is 2

3times the density of

water.

Question 25

Question

A cylindrical container is filled with water up to a height of 4 meters. Inside the

container, there is a solid aluminum sphere with a radius of 0.2 meters. If the

aluminum sphere is completely submerged in the water, what is the buoyant

force acting on the sphere? (Density of aluminum = 2700 kg/m3, density of

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

Solution

Step 1: Calculate the volume of the aluminum sphere using the formula for the

volume of a sphere:

V=4

3πr3

where r= 0.2 m.

Step 2: Substitute the radius into the formula to find the volume:

V=4

3×π×(0.2)3

Step 3: Calculate the volume:

V≈0.03351 m3

Step 4: Determine the mass of the aluminum sphere using the formula:

mass = density ×volume

The density of aluminum is 2700 kg/m3.

Step 5: Substitute the density and volume to find the mass of the aluminum

sphere:

mass = 2700 ×0.03351

Step 6: Calculate the mass of the aluminum sphere:

mass ≈90.327 kg

Step 7: Determine the weight of the aluminum sphere using the formula:

weight = mass ×acceleration due to gravity

Given that the acceleration due to gravity is 9.81 m/s2.

Step 8: Substitute the mass to find the weight of the aluminum sphere:

weight = 90.327 ×9.81

Step 9: Calculate the weight of the aluminum sphere:

weight ≈886.91 N

The weight of the aluminum sphere is 886.91 N.

Step 10: Determine the buoyant force acting on the sphere using Archimedes’

principle, which states that the buoyant force is equal to the weight of the fluid

displaced by the object. The buoyant force can be calculated using the formula:

buoyant force = weight of water displaced

Step 11: Calculate the weight of the water displaced by the aluminum sphere

when submerged completely. The volume of water displaced is equal to the

volume of the aluminum sphere.

Step 12: Determine the weight of the water displaced using the formula:

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

The density of water is 1000 kg/m3.

Step 13: Substitute the density of water and volume of water displaced:

weight of water displaced = 1000 ×0.03351

Step 14: Calculate the weight of the water displaced:

weight of water displaced = 33.51 N

Step 15: Hence, the buoyant force acting on the aluminum sphere is equal

to the weight of the water displaced:

buoyant force = 33.51 N

Question 26

Question

A spherical balloon with a radius of 2 meters is filled with helium gas. The

density of helium gas is 0.178 kg/m3, while the density of air is 1.29 kg/m3.

Determine the maximum mass that the balloon can carry without sinking.

Solution

Step 1: Let’s calculate the volume of the spherical balloon using the formula for

the volume of a sphere: V=4

3πr3, where ris the radius. Given that the radius

of the balloon is 2 meters, we have:

V=4

3π(2 m)3=32

3πm3

Step 2: Now, let’s calculate the weight of the air displaced by the balloon.

The weight of the displaced air is equal to the buoyant force acting on the bal-

loon. Buoyant force, Fb= Weight of the displaced air = Volume of air displaced×

Density of air ×g

Fb=V×Density of air ×g=32

3πm3×1.29 kg/m3×9.81 m/s2

Step 3: Next, let’s calculate the buoyant force acting on the balloon. The

buoyant force is also equal to the weight of the air that the balloon displaces.

Fb= Volume of balloon×Density of helium×g=V×Density of helium×g=32

3πm3×0.178 kg/m3×9.81 m/s2

Step 4: The maximum mass that the balloon can carry without sinking is

when the buoyant force equals the weight of the load. Let mbe the maximum

mass the balloon can carry.

m×g=Fb=32

3πm3×0.178 kg/m3×9.81 m/s2

3π×0.178 ×9.81

gkg

Step 5: Substituting the value of g= 9.81 m/s2and evaluating the expression

gives us the maximum mass that the balloon can carry.

3π×0.178 ×9.81

9.81 =32

3π×0.178 kg ≈5.96 kg

Therefore, the maximum mass that the balloon can carry without

sinking is approximately 5.96 kg.

Question 27

Question

A cylindrical container with a radius of 0.5 m and a height of 1.0 m is filled

with oil. The density of oil is 800 kg/m3. A solid sphere with a radius of 0.3

m and a density of 1200 kg/m3is carefully lowered into the oil until it is fully

submerged and at rest. Calculate the buoyant force acting on the sphere.

Solution

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

of a sphere is given by the formula:

V=4

3πr3

where ris the radius of the sphere. Substituting r= 0.3 m, we get:

V=4

3π(0.3)3= 0.1131 m3

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

the fluid displaced by the sphere. The weight of the fluid displaced is given by:

Weight = density ×volume ×acceleration due to gravity

Step 3: The weight of the fluid displaced by the submerged sphere is equal

to the weight of the oil that would occupy the volume of the sphere when it is

removed. The density of the oil is 800 kg/m3, so the weight of the displaced oil

is:

Weight = 800 ×0.1131 ×9.81 = 885.67 N

Step 4: The weight of the sphere in the oil is equal to the weight of the

sphere in the air minus the buoyant force acting on it. The weight of the sphere

in the air is given by:

Weightsphere = densitysphere ×volumesphere ×acceleration due to gravity

Weightsphere = 1200 ×4

3π(0.3)3×9.81 = 1065.5 N

Step 5: Finally, the buoyant force acting on the sphere is the weight of the

oil displaced, which is equal to:

Buoyant force = Weightsphere−Weight of displaced oil = 1065.5−885.67 = 179.83 N

Thus, the buoyant force acting on the sphere is 179.83 N.

Question 28

Question

A metal block with a volume of 0.05 m3and a density of 8000 kg/m3is immersed

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

will sink or float.

Solution

Step 1: Calculate the weight of the block The weight of the block is given by the

formula: W=m·g, where mis the mass of the block and gis the acceleration

due to gravity. Since density ρ=m

V, we can solve for mto find the mass of the

block.

m=ρ·V= 8000 kg/m3×0.05 m3= 400 kg

Therefore, the weight of the block is:

W=m·g= 400 kg ×9.81 m/s2= 3924 N

Step 2: Calculate the buoyant force The buoyant force acting on an object

immersed in a fluid is equal to the weight of the fluid displaced by the object.

The density of water is 1000 kg/m3. Since the volume of the block is 0.05 m3,

the weight of the water displaced by the block is:

Weight of displaced water = density of water ×volume of block ×g

= 1000 kg/m3×0.05 m3×9.81 m/s2= 490.5 N

Therefore, the buoyant force acting on the block is 490.5 N.

Step 3: Analyze the forces Since the weight of the block is 3924 N and the

buoyant force is 490.5 N, the net force acting on the block is:

3924 N −490.5 N = 3433.5 N

The net force is greater than zero, so the block will sink in water.

Question 29

Question

A spherical metal ball with a radius of 10 cm and a density of 8000 kg/m3is

submerged in a liquid that exerts an upwards buoyant force of 200 N on the

ball. Determine the density of the liquid.

Solution

Step 1: Recall that the buoyant force (Fbuoyant) acting on an object submerged

in a fluid is given by the formula:

Fbuoyant =ρfluid ·Vsubmerged ·g

where: - ρfluid is the density of the fluid, - Vsubmerged is the volume of the

object submerged in the fluid, and - gis the acceleration due to gravity.

Step 2: The volume of a sphere is given by the formula:

V=4

3πr3

where ris the radius of the sphere.

Step 3: Substitute the given information into the equations. The volume of

the spherical ball can be calculated as:

V=4

3π(0.1 m)3=4

3π(0.001 m3) = 0.004189 m3

Step 4: Since the entire ball is submerged, Vsubmerged =V. The buoyant

force is given as 200 N.

200 = ρfluid ·0.004189 ·9.81

Step 5: Solve for ρfluid:

ρfluid =200

0.004189 ·9.81 ≈4792 kg/m3

Therefore, the density of the liquid is approximately 4792 kg/m3.

Question 30

Question

A cube of side length sand density ρis floating in a liquid of density σ. If the

cube is submerged to a depth h, determine the expression for the buoyant force

acting on the cube.

Solution

Step 1: Find the volume of the cube submerged in the liquid.

The volume of the cube is V=s3. The volume of the cube submerged in the

liquid is Vsubmerged =A·h, where Ais the area of the cube face in contact with

the liquid. Since the cube is floating, the submerged volume must be equal to

the volume of liquid displaced. Therefore, Vsubmerged =σAh.

Step 2: Determine the mass of the cube and the mass of the liquid displaced.

The mass of the cube is mcube =ρV , and the mass of the liquid displaced is

mliquid =σVsubmerged. Substituting the expressions for Vand Vsubmerged, we

have mcube =ρs3and mliquid =σσAh.

Step 3: Apply Archimedes’ principle to find the buoyant force.

Archimedes’ principle states that the buoyant force acting on a submerged object

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

Fbuoyant is given by the difference in the weight of the liquid before and after

the cube was submerged:

Fbuoyant = (σAh)g−(ρs3)g

Fbuoyant =σAhg −ρs3g

Therefore, the expression for the buoyant force acting on the cube floating

in the liquid to a depth his Fbuoyant =σAhg −ρs3g.

Question 31

Question

A solid cube of side length 20 cm and density 800 kg/m3is floating in a liquid

of density 1000 kg/m3. What is the volume of the cube that is submerged in

the liquid?

Solution

Step 1: Determine the volume of the cube.

Volume of cube = side length3

= (0.20 m)3

= 0.008 m3

Step 2: Calculate the weight of the cube.

Weight of cube = Volume ×Density ×Acceleration due to gravity

= 0.008 m3×800 kg/m3×9.81 m/s2

≈62.79 N

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

Buoyant force = Weight of displaced liquid

= Volume submerged ×Density of liquid ×Acceleration due to gravity

Step 4: Use Archimedes’ principle to find the volume submerged.

Buoyant force = Weight of cube

Volume submerged ×1000 kg/m3×9.81 m/s2= 62.79 N

Volume submerged = 62.79 N

9810 N/m3

≈0.0064 m3

Therefore, the volume of the cube that is submerged in the liquid is approx-

imately 0.0064 m3.

Question 32

Question

A solid cube of side length 0.5 m and density 800 kg/m3is floating in a liquid

of density 1000 kg/m3. What fraction of the cube’s volume is submerged in the

liquid?

Solution

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

given by the formula Fb=ρfluid ·Vsub ·g, where: - ρfluid is the density of the fluid,

-Vsub is the volume of the cube submerged in the fluid, - gis the acceleration

due to gravity.

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

calculated using the formula W=m·g, where: - mis the mass of the cube, -

gis the acceleration due to gravity.

Step 3: Set up an equilibrium condition. For an object to float, the weight

of the object must be equal to the buoyant force acting on it. Therefore, we can

write: W=Fb

Step 4: Express the weight and buoyant force in terms of the volume sub-

merged. - The weight of the cube is W=ρcube ·Vcube ·g, where ρcube is the

density of the cube. - The volume of the cube is Vcube =a3, where ais the side

length of the cube.

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

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

Vcube ·g=ρfluid ·Vsub ·g

Solving for Vsub, we get: Vsub =ρcube

ρfluid

·Vcube

The fraction of the cube’s volume submerged is then given by: f=Vsub

Vcube =

ρcube

ρfluid

Substitute the values of ρcube and ρfluid to find the answer.

Question 33

Question

A metal block of density ρband volume Vis floating in a container filled with

water. The block is connected to a spring scale as shown in the figure. The

scale reads T0when the block is completely submerged.

If the block is slowly pulled out of the water and partially hangs out, what

will be the reading on the scale? Justify your answer.

images/buoyant_forces.png

Solution

1. When the block is completely submerged in water, the buoyant force acting

on the block is equal to the weight of the water displaced by the block. We

know that the weight of the water displaced is given by mg, where mis the

mass of the water displaced and gis the acceleration due to gravity. So, the

buoyant force Fbcan be expressed as:

Fb=mg

2. When the block is partially out of the water, the buoyant force F′

bacting

on the block is still equal to the weight of the water displaced. Since the block

is partially submerged, the weight of the water displaced is equal to the weight

of the volume of water equal to the volume of the submerged part of the block.

Let the volume of the submerged part of the block be V′. Then the weight of

the water displaced is given by ρwgV ′.

3. The weight of the block when it is partially submerged is given by ρbgV .

The apparent weight, which is the difference between the actual weight of the

block and the buoyant force, is then:

T=ρbgV −ρwgV ′

4. From Archimedes’ principle, the buoyant force is equal to the weight of

the fluid displaced, so F′

b=ρwgV ′. Therefore, the reading on the scale when

the block is partially submerged is:

T=ρbgV −ρwgV ′

5. Since ρbV=ρwV′(because the block is floating), the term ρbV−ρwV′

evaluates to zero and the scale simply reads the weight of the block T0. There-

fore, the reading on the scale remains T0when the block is pulled out of the

water and partially hangs out.

Question 34

Question

A cube of wood with a density of 600 kg/m3and dimensions 0.2 m ×0.2 m ×

0.2 m is floating in a liquid with a density of 800 kg/m3. Calculate the depth

to which the cube is submerged in the liquid.

Solution

Step 1: First, let’s find the weight of the cube. The weight of the cube can be

calculated using the formula:

Weight = Volume ×Density ×Acceleration due to gravity

Given that the density of the cube is 600 kg/m3, dimensions of the cube are 0.2

m×0.2 m ×0.2 m, and the acceleration due to gravity is 9.81 m/s2:

Volume = Side length3= 0.23= 0.008 m3

Weight = 0.008 m3×600 kg/m3×9.81 m/s2= 47.04 N

Step 2: Next, let’s find the buoyant force acting on the cube. The buoy-

ant force can be calculated using Archimedes’ principle which states that the

buoyant force is equal to the weight of the liquid displaced by the object:

Buoyant force = Volume submerged×Density of liquid×Acceleration due to gravity

Let hbe the depth to which the cube is submerged. The volume submerged is

the area of the base of the cube (0.2 m ×0.2 m) times h:

Volume submerged = 0.2×0.2×hm3= 0.04hm3

Given that the density of the liquid is 800 kg/m3and the acceleration due to

gravity is 9.81 m/s2:

Buoyant force = 0.04h×800 ×9.81 = 313.6hN

Step 3: Since the cube is floating, the weight of the cube is equal to the

buoyant force.

47.04 N = 313.6hN

Solving for h:

h=47.04

313.6= 0.15 m

Therefore, the cube is submerged to a depth of 0.15 meters in the liquid.

Question 35

Question

A metal cube of side length 10 cm and density 8000 kg/m3is floating in water.

What is the depth of the cube under the water surface? (Density of water =

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

Solution

Step 1: First, we need to determine the fraction of the cube’s volume submerged

in water.

Density of cube = 8000 kg/m3

Density of water = 1000 kg/m3

Since the cube is floating, the buoyant force must be equal to the weight of the

cube:

Weight of cube = Buoyant force

Vcube ×Density of water ×g=Vsubmerged ×Density of cube ×g

where Vcube is the total volume of the cube, and Vsubmerged is the volume sub-

merged in water.

Step 2: Since the cube is a cube, the volume of the cube is given by Vcube =

(side length)3= (0.10 m)3= 0.001 m3. Let DV be the depth of the cube under

the water surface. Then Vsubmerged = 0.10 m ×0.10 m ×DV . Plugging in the

values into the equation above:

0.001 m3×1000 kg/m3= 0.01 m2×DV ×8000 kg/m3

Step 3: Solve for the depth DV :

DV =0.001 m3×1000 kg/m3

0.01 m2×8000 kg/m3= 0.000125 m = 0.125 mm

Therefore, the depth of the cube under the water surface is 0.125 mm.

Question 3

Question

A cylindrical metal object with a density of 8000 kg/m3and a volume of 0.04

m3is submerged in a container of water. If the metal object is suspended by a

rope and completely submerged in water, calculate the tension in the rope.

Solution

Step 1: First, we need to determine the weight of the metal object. The weight

of an object can be calculated using the formula W=mg, where mis the mass

of the object and gis the acceleration due to gravity (approximately 9.81 m/s2).

Given that the density of the metal object is 8000 kg/m3and its volume is 0.04

m3, we can calculate the mass of the object as follows:

m= Density ×Volume = 8000 kg/m3×0.04 m3= 320 kg

Therefore, the weight of the metal object is:

W=mg = 320 kg ×9.81 m/s2= 3139.2 N

Step 2: Now, we can determine the buoyant force acting on the metal object.

The buoyant force is equal to the weight of the fluid displaced by the object,

which can be calculated using the formula Fbuoyant =ρ·V·g, where ρis the

density of the fluid (1000 kg/m3for water), Vis the volume of the object, and

gis the acceleration due to gravity:

Fbuoyant = 1000 kg/m3×0.04 m3×9.81 m/s2= 392.4 N

Step 3: Since the metal object is suspended by a rope and completely sub-

merged in water, the tension in the rope is equal to the difference between the

weight of the object and the buoyant force:

Tension = W−Fbuoyant = 3139.2 N −392.4 N = 2746.8 N

Therefore, the tension in the rope when the metal object is completely sub-

merged in water is 2746.8 N.

Question 4

Question

A metal block of mass 500 kg and volume 0.5 m3is submerged in a container of

water. Calculate the buoyant force acting on the block and determine if it will

float or sink.

Solution

Step 1: Calculate the weight of the block. Given that the mass of the block is

500 kg and the acceleration due to gravity is 9.81 m/s2, the weight Wcan be

calculated using the formula W=m·g, where mis the mass of the block and

gis the acceleration due to gravity.

W= 500 kg ×9.81 m/s2

W= 4905 N

Step 2: Calculate the buoyant force. The buoyant force Fbacting on an

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

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

the block. Since the density of water is 1000 kg/m3, the buoyant force can be

calculated as:

Fb= density of water ×volume of water displaced ×g

Fb= 1000 kg/m3×0.5 m3×9.81 m/s2

Fb= 4905 N

Step 3: Determine if the block will float or sink. The block will float if the

buoyant force is greater than or equal to the weight of the block. In this case,

since 4905 N = 4905 N, the buoyant force is equal to the weight of the block.

Therefore, the block will remain submerged in the water.

Question 5

Question

A cube of wood with a side length of 0.1 m and a density of 800 kg/m3is floating

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

that the density of water is 1000 kg/m3.

Solution

Step 1: Let’s first determine the volume of the cube. The volume of a cube is

given by the formula V=s3, where sis the side length. So, for the cube in this

problem:

V= (0.1 m)3= 0.001 m3

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

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

the values:

m= 800 kg/m3×0.001 m3= 0.8 kg

Step 3: Using Archimedes’ principle, the buoyant force acting on the cube

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