PHYS 101 - ELEMENTS OF PHYSICS Buoyant forces and Archimedes' principle Question Bank Set 6

PHYS 101 - ELEMENTS OF PHYSICS

- Buoyant forces and Archimedes’

principle

Question Bank - Set 6

Liberty University

Question 1

Question

A student conducts an experiment in which a metal object of mass 500 g and

volume 400 cm3is hung from a string and completely submerged in a container

of water. The student observes that the object appears to float with 60

Given: Density of water = 1000 kg/m3. Acceleration due to gravity, g= 9.8

m/s2.

Solution

Step 1: Find the volume of the object submerged in water. Let Vbe the total

volume of the object and Vsub be the volume submerged under water. Since

60Given that the volume of the object is 400 cm3,

V= 400 cm3= 0.0004 m3

Thus,

Vsub = 0.6×0.0004 = 0.00024 m3

Step 2: Calculate the density of the metal object. The density of an object

is given by:

Density = Mass

Volume

Given mass = 500 g = 0.5 kg and volume = 0.0004 m3,

Density = 0.5

0.0004 = 1250 kg/m3

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

acting on the object is given by:

Buoyant force = Weight of water displaced

The weight of the water displaced is equal to the weight of the water that would

fill the volume of the submerged fraction of the object.

Weight of water displaced = Volumesubmerged ×Density of water ×g

= 0.00024 ×1000 ×9.8

= 2.352 N

Question 2

Question

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

liquid with a density of 1.2 g/cm3. If the cube has 80

Solution

Step 1: First, let’s calculate the density of the cube using the given information.

The density of the cube can be calculated using the formula:

Density = mass

volume

Given that the density of the cube is 0.6 g/cm3and the volume submerged

is 80Let Vbe the volume of the cube. Since 80

mass = Density ×Volumesubmerged = 0.6 g/cm3×0.8V

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

the submerged portion of the cube. Let Vliquid be the volume of the liquid

displaced. Since the cube is floating, the volume of the liquid displaced is equal

to the volume of the submerged portion of the cube. So, we have:

Vliquid = 0.8V

Step 3: Now, we can calculate the buoyant force acting on the cube using

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

displaced by the cube. The weight of the liquid displaced by the cube is:

Weight = Densityliquid ×g×Vliquid

where gis the acceleration due to gravity and Densityliquid is the density of

the liquid. Given that the density of the liquid is 1.2 g/cm3, we can substitute

all the known values into the formula to find the buoyant force.

Question 3

Question

A spherical balloon filled with helium gas has a radius of 2 meters. If the density

of helium is 0.1785 kg/m3and the density of air is 1.225 kg/m3, calculate the

maximum mass the balloon can lift off the ground. Assume atmospheric pressure

is 101.3 kPa and the temperature is 25

C. (Hint: The buoyant force is equal to

the weight of the displaced air.)

Solution

Step 1: Calculate the volume of the balloon using the formula for the volume

of a sphere:

V=4

3πr3

Step 2: Substitute the given radius r= 2 meters into the formula to find the

volume V.

Step 3: Calculate the mass of the displaced air using the formula:

mair =ρair ×V

where ρair is the density of air.

Step 4: Calculate the weight Wair of the displaced air using Wair =mair ×g,

where g= 9.81 m/s2is the acceleration due to gravity.

Step 5: Calculate the buoyant force Fbuoyant acting on the balloon: Fbuoyant =

Wair

Step 6: Use Archimedes’ principle to find the maximum mass the balloon

can lift off the ground: Mmax =Fbuoyant

Step 7: Substitute the calculated values to find the maximum mass the

balloon can lift off the ground.

Question 4

Question

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

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

Solution

Step 1: Calculate the density of water as 1000 kg/m3. Step 2: Identify the

forces acting on the cube. The weight of the cube acting downwards is given by

W=ρwood ·Vcube ·g, where ρwood is the density of wood, Vcube is the volume

of the cube, and gis the acceleration due to gravity. The upthrust or buoyant

force acting upwards is given by B=ρwater ·Vsubmerged ·g, where Vsubmerged

is the volume of the cube submerged in water. Step 3: Apply Archimedes’

principle, which states that the upthrust or buoyant force on an object in a fluid

is equal to the weight of the fluid displaced by the object. Therefore, B=W.

Step 4: Set the weight of the cube equal to the upthrust: ρwood ·Vcube ·g=

ρwater ·Vsubmerged ·g. Step 5: Substitute the given values and solve for the depth

hto which the cube is submerged: ρwood ·A·h·g=ρwater ·A·h·gwhere A

is the area of one face of the cube. Step 6: Simplify the expression to find the

depth h:h=ρwood

ρwater

·h=600

1000 ·0.1=0.06 m.

Therefore, the cube is submerged to a depth of 0.06 meters in water.

Question 5

Question

A rectangular block of wood with dimensions 10 cm ×10 cm ×20 cm is floating

in water. The density of water is 1000 kg/m3and the density of wood is 800

kg/m3. Calculate the fraction of the volume of the block that is submerged in

water.

Solution

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

to the weight of the water displaced by the block. We can calculate this using

Archimedes’ principle.

Weight of block = Weight of water displaced

ρwater ×Vsubmerged ×g=ρwood ×Vblock ×g

where ρwater is the density of water, Vsubmerged is the volume submerged, ρwood

is the density of wood, Vblock is the total volume of the block, and gis the

acceleration due to gravity.

Step 2: Calculate the volume submerged. Since the block is floating, the

weight of the block is equal to the weight of the water displaced, as calculated

in Step 1.

ρwater ×Vsubmerged ×g=ρwood ×Vblock ×g

Vsubmerged =ρwood

ρwater ×Vblock

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

volume submerged is the ratio of the volume submerged to the total volume of

the block.

Fraction submerged = Vsubmerged

Vblock

Now, substitute the values and calculate the fraction submerged.

Question 6

Question

A cube of metal with a density of 8000 kg/m3and a volume of 0.04 m3is

placed in a container of water, with the top of the cube at the water’s surface.

Determine: (a) the weight of the cube, (b) the buoyant force acting on the cube,

or remain at rest.

Solution

(a) To calculate the weight of the cube, we use the formula W=mg, where

mis the mass of the cube and gis the acceleration due to gravity (9.81 m/s2).

Given that density ρ= 8000 kg/m3and volume V= 0.04 m3, we can find the

mass of the cube using the formula m=ρV . Let’s substitute the values into

the formula:

m= 8000 kg/m3×0.04 m3

m= 320 kg

Therefore, the weight of the cube is:

W= 320 kg ×9.81 m/s2

W= 3139.2 N

(b) The buoyant force acting on the cube is equal to the weight of the water

displaced by the cube. Using Archimedes’ principle, Fbuoyant =ρwaterVdisplacedg.

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

the volume of the cube under water, Vdisplaced =h×A, where his the height

submerged and Ais the base area of the cube. Given that Vdisplaced = 0.02 m3

(half the volume of the cube), we can now calculate the buoyant force:

Fbuoyant = 1000 kg/m3×0.02 m3×9.81 m/s2

Fbuoyant = 196.2 N

the cube and the buoyant force:

Fnet =W−Fbuoyant

Fnet = 3139.2 N −196.2 N

Fnet = 2943 N

(d) Since the net force acting on the cube is downward (positive), the cube

will sink in the water.

Question 7

Question

A cube of wood with a density of 500 kg/m3and side length 0.2 m is floating

in water. If a 2 kg weight is placed on top of the cube, how much of the cube’s

volume is now submerged in the water?

Solution

Step 1: We first need to calculate the density of water. Since the cube is floating,

the average density of the cube (ρavg) is equal to the density of water (ρwater).

We can use this relationship to find the density of water.

ρavg =ρwater

500 kg/m3=ρwater

Thus, ρwater = 500 kg/m3.

Step 2: Next, we need to account for the additional weight placed on top of

the cube. The weight of the cube, Wcube, is equal to the weight of the water

displaced by the cube, and the weight of the added weight, Wadd, is equal to

the weight of the volume of water displaced by the added weight. Therefore,

the total weight of the system, Wtotal, is given by:

Wtotal =Wcube +Wadd

Wtotal =ρwater ·Vsubmerged ·g+madd ·g

where Vsubmerged is the volume of the cube submerged and gis the acceleration

due to gravity.

Step 3: We also know that the buoyant force acting on the system is equal to

the weight of the water displaced by the total volume of the system. Therefore,

the buoyant force, Fbuoyant, is given by:

Fbuoyant = (ρwater ·Vtotal)·g

where Vtotal is the total volume of the system.

Step 4: Using Archimedes’ principle, we know that the buoyant force is equal

to the total weight of the system. Thus, we have:

Fbuoyant =Wtotal

Step 5: By setting Fbuoyant equal to Wtotal and solving for Vsubmerged, we can

find the volume of water displaced by the cube and the added weight. Finally,

the ratio of Vsubmerged to the total volume of the cube will give us the proportion

of the cube that is now submerged in water.

Question 8

Question

A solid cube of side length 10 cm and density 1.2 g/cm3is placed in a container

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

water = 1 g/cm3)

Solution

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

V=s3, where sis the side length. Given that the side length s= 10 cm, the

volume of the cube is:

V= 103cm3= 1000 cm3

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 and Vis the

volume. Given that the density ρ= 1.2 g/cm3, the mass of the cube is:

m= 1.2×1000 = 1200 g = 1.2 kg

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

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

acceleration due to gravity (g= 9.81 m/s2). Therefore, the weight of the cube

is:

w= 1.2×9.81 = 11.772 N

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

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

Since the cube is fully submerged in water, the buoyant force is equal to the

weight of the water displaced by the cube. The volume of water displaced by

the cube is equal to the volume of the cube, which is 1000 cm3or 0.001 m3.

The weight of the water displaced is given by mwater =ρwaterV g, where ρwater

is the density of water (1 g/cm3or 1000 kg/m3). Therefore, the buoyant force

is:

Fbuoyant =ρwaterV g = 1000 ×0.001 ×9.81 = 9.81 N

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

Question 9

Question

A large wooden cube with a density of 600 kg/m3and side length 5 m is floating

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

Given: Density of water = 1000 kg/m3, acceleration due to gravity = 9.8

m/s2.

Solution

Step 1: Calculate the buoyant force on the wooden cube. The buoyant force

(Fb) on an object is equal to the weight of the fluid displaced by the object. In

this case, the weight of the fluid displaced is equal to the weight of the water

displaced by the cube.

Fb=ρwater ·Vdisplaced ·g

where ρwater is the density of water, Vdisplaced is the volume of water displaced

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

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

water displaced by the cube is equal to the volume of the cube that is submerged

in water. Let’s denote the depth of immersion as h.

Vdisplaced = (5 m)2·h

Step 3: Equate the weight of the cube to the buoyant force. The weight of

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

ρcube ·Vcube ·g=ρwater ·Vdisplaced ·g

Step 4: Solve for the depth of immersion h. Plug in the given values and

density of the cube:

600 kg/m3·(5 m)3·g= 1000 kg/m3·(5 m)2·h·g

53·600 = 52·1000 ·h

5h=53·600

1000

h=5·5·5·600

1000

h= 15 m

Therefore, the depth of immersion of the wooden cube is 15 meters.

Question 10

Question

A cylindrical object with a radius of 8 cm and a height of 12 cm is floating upright

in water. The top of the object is 4 cm above the water surface. Determine the

density of the material the object is made of. (Density of water = 1000 kg/m3)

Solution

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

can be calculated using the formula:

V=πr2h

where ris the radius and his the height of the cylinder. Substituting in the

given values:

V=π×(0.08 m)2×0.12 m

V= 0.03072 m3

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

of water displaced is equal to the volume of the part of the object submerged

in water, which is equal to the volume of a cylinder with a radius of 8 cm and

a height of 4 cm:

Vdisplaced water =π×(0.08 m)2×0.04 m

Vdisplaced water = 0.008064 m3

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

force is equal to the weight of the water displaced by the object, and can be

calculated using the formula:

Fb=ρwater ×g×Vdisplaced water

where ρwater is the density of water and gis the acceleration due to gravity.

Substituting in the given values:

Fb= 1000 kg/m3×9.81 m/s2×0.008064 m3

Fb≈79.22 N

Step 4: Calculate the weight of the object. The weight of the object is equal

to the weight of the water displaced by the part of the object submerged in

water:

Wobject =ρmaterial ×g×Vdisplaced water

where ρmaterial is the density of the material the object is made of. Since the

object floats, the buoyant force is equal to the weight of the object:

Wobject =Fb

ρmaterial ×g×0.008064 m3= 79.22 N

ρmaterial =79.22 N

9.81 m/s2×0.008064 m3

ρmaterial ≈9816.45 kg/m3

Therefore, the density of the material the object is made of is approximately

9816.45 kg/m3.

Question 11

Question

A cube of steel with side length 0.1 m and density 7850 kg/m3is placed in a

tank of oil. The density of the oil is 800 kg/m3. Calculate the buoyant force

acting on the cube of steel.

Solution

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

by V= side length3. Therefore, the volume of the steel cube is:

V= (0.1 m)3= 0.001 m3

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

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

due to gravity. Since m=ρV , where ρis the density of the object and Vis the

volume, we have:

msteel =ρsteelV= 7850 kg/m3×0.001 m3= 7.85 kg

Wsteel =msteel ·g= 7.85 kg ·9.81 m/s2= 77.1485 N

Step 3: Calculate the volume of the displaced oil. Since the cube is fully

submerged in the oil, the volume of oil displaced is equal to the volume of the

cube, which is 0.001 m3.

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

oil is given by Woil =ρoilVdisplaced ·g, where Vdisplaced is the volume of oil

displaced.

Woil = 800 kg/m3×0.001 m3×9.81 m/s2= 7.848 N

Step 5: Calculate the buoyant force. The buoyant force is equal to the weight

of the displaced fluid, which is 7.848 N. Therefore, the buoyant force acting on

the steel cube is 7.848 N.

Question 12

Question

A cube of iron with dimensions 10 cm ×10 cm ×10 cm is submerged in water.

If the density of iron is 7.87 g/cm3and the density of water is 1.0 g/cm3,

determine the buoyant force acting on the iron cube.

Solution

Step 1: Calculate the volume of the iron cube. Given that the dimensions of

the cube are 10 cm ×10 cm ×10 cm, the volume Vis given by:

V= 10 cm ×10 cm ×10 cm = 1000 cm3

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

can be found by multiplying its volume by its density:

m=V×density = 1000 cm3×7.87 g/cm3= 7870 g

Step 3: Calculate the volume of water displaced by the iron cube. Since the

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

of the cube, which is 1000 cm3.

Step 4: 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

cube.

weight = density×volume×gravity = 1.0 g/cm3×1000 cm3×9.81 m/s2= 9810 g

Step 5: Calculate the buoyant force. The buoyant force is equal to the weight

of the water displaced, which is also equal to the weight of the fluid the object

displaces. Therefore, the buoyant force acting on the iron cube is 9810 g or 98.1

Question 13

Question

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

The cube floats with a fraction fof its volume submerged in the liquid.

If the volume of the cube is Vcube =a3, determine an expression for the

fraction fin terms of the densities ρcube and ρliquid.

Solution

Let’s denote the volume of the submerged part of the cube as Vsubmerged. Since

the cube is floating, the buoyant force acting on the cube must be equal to the

weight of the submerged part.

Step 1: Write an expression for the buoyant force. The buoyant force Fb

acting on the cube is given by:

Fb=ρliquid ·Vsubmerged ·g

Step 2: Write an expression for the weight of the submerged part. The

weight Wsubmerged of the submerged part is given by:

Wsubmerged =ρcube ·Vsubmerged ·g

Step 3: Equate the buoyant force to the weight of the submerged part. Since

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

part:

ρliquid ·Vsubmerged ·g=ρcube ·Vsubmerged ·g

Step 4: Solve for the fraction f. Since the volume of the cube is Vcube =a3:

Vsubmerged =f·Vcube =f·a3

Substitute the expression for Vsubmerged into the equation from Step 3:

ρliquid ·f·a3=ρcube ·f·a3

Divide by ρcubea3on both sides to solve for f:

f=ρliquid

ρcube

Therefore, the fraction fof the cube submerged in the liquid is f=ρliquid

ρcube

Question 14

Question

A wooden cube with a side length of 0.5 meters and a density of 600 kg/m3is

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

water.

Solution

Step 1: First, let’s determine the density of water. The density of water is 1000

kg/m3.

Step 2: The buoyant force acting on the cube is equal to the weight of the

water displaced by the cube. Using Archimedes’ Principle, the buoyant force

(Fb) is given by the formula:

Fb=ρwater ·Vsubmerged ·g

where ρwater is the density of water, Vsubmerged is the volume of the cube sub-

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

Step 3: The weight of the cube is equal to the force of gravity acting on it.

The weight of the cube (W) is given by:

W=ρcube ·Vcube ·g

where ρcube is the density of the cube, Vcube is the total volume of the cube,

and gis the acceleration due to gravity.

Step 4: Since the cube is floating, the buoyant force must be equal to the

weight of the cube. Therefore, we have:

Fb=W

ρwater ·Vsubmerged ·g=ρcube ·Vcube ·g

Step 5: Substitute the given values into the equation:

1000 ·Vsubmerged = 600 ·0.53

Step 6: Solve for Vsubmerged:

Vsubmerged =600 ·0.53

1000 = 0.15 m3

Step 7: The depth to which the cube is submerged in the water is equal to

the height of the cube submerged in water, which is equal to the side length of

the cube:

Depth submerged = 0.5 m

Question 15

Question

A solid sphere of radius Rand density ρ1is placed in a fluid of density ρ2

such that it floats with half of its volume submerged. Calculate the ratio of the

density of the sphere to the density of the fluid.

Solution

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

fluid as Vs. Given that the sphere floats with half of its volume submerged, we

have:

Vs=1

2×4

3πR3

Step 2: The weight of the fluid displaced by the submerged volume Vswill

be equal to the weight of the sphere. Using Archimedes’ principle, we have:

ρ2·Vs·g=ρ1·4

3πR3·g

Step 3: Substitute the expression for Vsinto the equation from Step 2 and

simplify the equation.

ρ2·1

2×4

3πR3·g=ρ1·4

3πR3·g

Step 4: Cancel out common terms and solve for the ratio of the densities.

2ρ2=ρ1

ρ1

ρ2

Step 5: Therefore, the ratio of the density of the sphere to the density of the

fluid is 1

Question 16

Question

A cylindrical metal object of height 20 cm and radius 5 cm is floating in water.

If the density of the metal is 8000 kg/m3and the density of water is 1000 kg/m3,

determine the fraction of the object that is submerged in water.

Solution

Step 1: We first calculate the volume of the metal object. The volume of a

cylinder is given by:

Vcylinder =πr2h

where ris the radius and his the height of the cylinder. Plugging in the values:

Vcylinder =π(0.05 m)2·0.20 m

Vcylinder =π·0.0025 m2·0.20 m

Vcylinder = 0.00157 m3

Step 2: The weight of the object must equal the buoyant force acting on the

object when it is floating. The weight of the object is given by:

Wobject =ρmetal ·Vcylinder ·g

where ρmetal is the density of the metal, Vcylinder is the volume of the cylinder,

and gis the acceleration due to gravity.

Wobject = 8000 kg/m3·0.00157 m3·9.81 m/s2

Wobject = 124.1 N

Step 3: The buoyant force is given by:

Fbuoyant =ρwater ·Vsubmerged ·g

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

submerged, and gis the acceleration due to gravity.

Step 4: Since the object is floating, the weight of the object is equal to the

buoyant force:

Wobject =Fbuoyant

ρmetal ·Vcylinder ·g=ρwater ·Vsubmerged ·g

Step 5: Solving for the fraction of the object submerged:

Vsubmerged

Vcylinder

=ρmetal

ρwater

Vsubmerged

0.00157 =8000

1000

Vsubmerged = 0.01256 m3

Step 6: Finally, the fraction of the object submerged is:

Vsubmerged

Vcylinder

=0.01256

0.00157 = 8

Therefore, 8/20 or 40% of the object is submerged in water.

Question 17

Question

A cylindrical tank is filled with water to a height of 3 meters. The tank has a

diameter of 4 meters. Calculate the total force exerted on the base of the tank

by the water. Take the density of water to be 1000 kg/m3and the acceleration

due to gravity to be 9.81 m/s2.

Solution

Step 1: Calculate the volume of water in the tank. The volume of water in the

tank can be calculated using the formula for the volume of a cylinder:

V=πr2h

where ris the radius of the tank and his the height of the water in the tank.

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

diameter: r= 2 m. Substitute r= 2 m and h= 3 m into the formula:

V=π×(2 m)2×3 m

V= 12πm3

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

can be calculated using the formula:

m= density ×volume

Given that the density of water is 1000 kg/m3:

m= 1000 kg/m3×12πm3

m= 12000πkg

Step 3: Calculate the weight of the water in the tank. The weight of the

water can be calculated using the formula:

W=mg

where g= 9.81 m/s2is the acceleration due to gravity.

W= 12000πkg ×9.81 m/s2

W≈372360πN

Step 4: Calculate the total force exerted on the base of the tank by the

water. The total force exerted on the base of the tank is equal to the weight of

the water, which is the same as the gravitational force:

Total Force = 372360πN

Question 18

Question

A cube of ice with sides of length 5 cm floats in a glass of water. The density

of ice is 0.92 g/cm3and the density of water is 1.00 g/cm3. What is the length

of the side of the cube that is submerged in the water?

Solution

Step 1: The buoyant force is equal to the weight of the water displaced by the

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

cube. Step 2: The weight of the cube is equal to the mass of the cube times

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

volume and density. Step 3: Let’s calculate the volume of the cube. The volume

of a cube is given by V=s3, where sis the length of a side of the cube. Step 4:

Submerge a distance of xof the cube in water, the volume of water displaced

can be represented as Vdisplaced =s2·x. Step 5: Since the cube is floating, the

weight of the cube is balanced by the weight of the displaced water. Hence,

mcube ·g=ρwater ·Vdisplaced ·g. Step 6: We have: s3·ρice ·g=s2·x·ρwater ·g.

Step 7: Solving for x, we find x=s·ρice

ρwater . Step 8: Substituting in the given

values, x= 5 cm ·0.92 g/cm3

1.00 g/cm3. Step 9: Therefore, the length of the side of the

cube that is submerged in the water is 4.6 cm.

Question 19

Question

A cube of wood with side length 15 cm and density 0.8 g/cm3is floating in

water. What is the depth to which the cube is submerged in water? (Density

of water = 1 g/cm3)

Solution

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

V=s3, where sis the side length of the cube. In this case, s= 15 cm.

V= 153cm3= 3375 cm3

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

calculated using the formula m= density ×volume. In this case, the density of

the wood cube is 0.8 g/cm3.

m= 0.8×3375 g = 2700 g

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

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

displaced by the object. The weight of the fluid displaced is given by mg, where

mis the mass of the fluid displaced and gis the acceleration due to gravity

(9.81 m/s2). Since the cube is floating, the buoyant force is equal to the weight

of the cube. Therefore,

buoyant force = mg = 2700 ×9.81 N = 26487 N

Step 4: Determine the volume of water displaced by the cube. The volume

of water displaced by the cube is equal to the volume of the submerged part of

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

Volume of water displaced = 15 ×15 ×hcm3= 3375 cm3

Step 5: Determine the mass of the water displaced by the cube. The density

of water is 1 g/cm3. The mass of the water displaced is given by mwater =

density ×volume.

mwater = 1 ×15 ×15 ×hg = 225hg

Step 6: Equate the mass of the water displaced to the mass of the cube.

Since the cube is floating, the mass of the water displaced is equal to the mass

of the cube.

225h= 2700

h=2700

225 = 12 cm

Therefore, the cube is submerged to a depth of 12 cm in water.

Question 20

Question

A solid sphere of radius Rand density ρsis placed in a liquid of density ρl. The

sphere is released from rest and allowed to sink in the liquid. Determine the

expression for the acceleration of the sphere as it sinks.

Solution

Step 1: Firstly, let’s consider the forces acting on the sphere. When the sphere

is submerged in the liquid, it experiences the force of gravity (Fgravity) acting

downward and the buoyant force (Fbuoyant) acting upward. The net force acting

on the sphere is given by the difference between these two forces:

Fnet =Fbuoyant −Fgravity

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

the liquid displaced by the sphere. This can be determined using Archimedes’

principle:

Fbuoyant =ρlVsphereg

where Vsphere is the volume of the sphere.

Step 3: The weight of the sphere can be calculated as:

Fgravity =ρsVsphereg

Step 4: Substituting these expressions back into the net force equation gives:

Fnet =ρlVsphereg−ρsVsphereg

Step 5: Simplifying, we find the net force acting on the sphere as:

Fnet = (ρl−ρs)Vsphereg

Step 6: The mass of the sphere can be written in terms of its density and

volume:

m=ρsVsphere

Step 7: Finally, we can write the net force in terms of the mass of the sphere:

Fnet = (ρl−ρs)mg

The acceleration of the sphere can then be determined using Newton’s second

law, F=ma:

a=(ρl−ρs)g

Question 21

Question

A solid metal cube of side length 10 cm and density 8000 kg/m3is placed in a

container of water. Calculate the buoyant force acting on the cube when it is

fully submerged in the water.

Solution

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

V= side length3= (0.1 m)3= 0.001 m3.

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

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

to gravity. Since density ρ=m

V, we can rewrite the weight as W=ρ·V·g=

(8000 kg/m3)·(0.001 m3)·(9.81 m/s2) = 78.48 N.

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

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

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

of the cube when it is fully submerged. Therefore, Fb= 78.48 N.

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

in water is 78.48 N.

Question 22

Question

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

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

Solution

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

V= (edge length)3. Substituting the given edge length of 0.1 m, we find:

V= (0.1 m)3= 0.001 m3

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

calculated using the formula m= density ×volume. Substituting the given

density of 800 kg/m3and the volume calculated in step 1, we get:

m= 800 kg/m3×0.001 m3= 0.8 kg

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

W=mg, where gis the acceleration due to gravity (approximately 9.81 m/s2).

Substituting the mass calculated in step 2, we find:

W= 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 cube. The buoyant force can be calculated using the formula

Fbuoyant =ρwater ×Vsubmerged ×g, where ρwater is the density of water (1000

kg/m3), Vsubmerged is the volume of the cube submerged in water, and gis the

acceleration due to gravity. Since the cube is completely submerged, Vsubmerged

is equal to the volume of the cube, which is 0.001 m3. Substituting these values,

we get:

Fbuoyant = 1000 kg/m3×0.001 m3×9.81 m/s2= 9.81 N

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

Question 23

Question

A cube of iron with sides of length 10 cm is submerged in a fluid. The density

of iron is 7.87 g/cm3and the density of the fluid is 1.20 g/cm3. Determine the

depth to which the cube sinks in the fluid.

Solution

Step 1: Calculate the weight of the iron cube. Step 2: Calculate the buoyant

force on the iron cube. Step 3: Use Archimedes’ principle to find the depth to

which the cube sinks.

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

given by Vcube = (10 cm)3= 1000 cm3= 1000 ×10−6m3= 0.001 m3. The mass

of the iron cube is miron =ρiron ×Vcube, where ρiron = 7.87 g/cm3= 7870 kg/m3.

Therefore, miron = 7870 ×0.001 = 7.87 kg. The weight of the iron cube is given

by Wiron =miron ×g, where g≈9.81 m/s2. Thus, Wiron ≈7.87 ×9.81 = 77.3 N.

Step 2: Calculate the buoyant force on the iron cube. The volume of fluid

displaced by the cube is equal to the volume of the cube under the fluid. The

buoyant force Fbon the cube is given by Fb=ρfluid ×Vcube ×g, where ρfluid =

1.20 g/cm3= 1200 kg/m3. Therefore, Fb= 1200 ×0.001 ×9.81 = 11.76 N.

Step 3: Use Archimedes’ principle to find the depth to which the cube sinks.

The net force acting on the cube when it is submerged is the difference between

the weight of the cube and the buoyant force. Thus, Net force = Wiron −Fb=

77.3−11.76 = 65.54 N. The depth to which the cube sinks is determined by this

net force. Using the equation Net force = ρfluid ×g×Vsubmerged, where Vsubmerged

is the volume of the cube submerged in the fluid. Rearranging the equation, we

find that Vsubmerged =Net force

ρfluid×g. Therefore, Vsubmerged =65.54

1200×9.81 ≈0.0054 m3.

Since the cube is submerged with only one face on the bottom, the depth to

which it sinks is equal to the side length of the cube minus the length of the side

submerged. Hence, the depth to which the cube sinks is 10 cm −Vsubmerged

0.001 ≈

5.46 cm.

Question 24

Question

A large block of ice with a uniform density of 900 kg/m3is floating in a fresh-

water lake. The ice block has a volume of 2.5 m3. Calculate the fraction of the

ice block that is submerged in the water.

Solution

Step 1: Determine the buoyant force acting on the ice block.

The buoyant force Fbacting on an object immersed in a fluid is given by

Archimedes’ principle:

Fb=ρf·Vsub ·g

Where: - ρfis the density of the fluid, - Vsub is the volume of fluid displaced by

the immersed object, - gis the acceleration due to gravity.

Given that the ice block is floating in freshwater with a density of 1000 kg/m3

and that the entire volume of the ice block is 2.5 m3, the buoyant force can be

calculated as:

Fb= 1000 kg/m3·2.5 m3·9.81 m/s2

Step 2: Determine the weight of the ice block.

The weight Wof the ice block can be calculated using the formula:

W=ρice ·Vice ·g

Where: - ρice is the density of the ice block, - Vice is the total volume of the ice

block.

Substitute the given values into the formula to find the weight of the ice

block.

Step 3: Use the fact that the ice block is in equilibrium to find the fraction

submerged.

In equilibrium, the weight of the ice block is equal to the buoyant force:

W=Fb

Set the weight of the ice block equal to the buoyant force and find the fraction

of the ice block submerged.

Question 25

Question

A solid sphere of radius Rand density ρsolid is floating in a liquid of density

ρliquid. Determine the fraction of the sphere that is submerged in the liquid.

Solution

Step 1: Let’s consider the forces acting on the sphere. The buoyant force FB

is equal to the weight of the liquid displaced by the sphere, which is also the

weight of the sphere when it is floating.

FB=mg

Step 2: The weight of the sphere is given by mg =ρsolidVsphereg, where Vsphere

is the volume of the sphere. Step 3: The volume of the sphere Vsphere can be

expressed in terms of the radius Ras Vsphere =4

3πR3. Step 4: The buoyant

force FBcan also be expressed as FB=ρliquidVsubmergedg, where Vsubmerged is

the volume of the sphere submerged in the liquid. Step 5: Since the sphere is

floating, the weight of the sphere must be balanced by the buoyant force.

ρsolidVsphereg=ρliquidVsubmergedg

Step 6: Plugging in the expressions for Vsphere and Vsubmerged, we get

ρsolid

3πR3=ρliquid

3πR3h

where his the fraction of the sphere submerged in the liquid. Step 7: Simplifying

the equation gives us the fraction of the sphere that is submerged in the liquid:

h=ρsolid

ρliquid

Therefore, the fraction of the sphere submerged in the liquid is ρsolid

ρliquid .

Question 26

Question

A cube of side length 0.2 m and density 800 kg/m3is immersed partially in oil

of density 920 kg/m3and partially in water of density 1000 kg/m3as shown in

the figure. The depth of the oil layer is 0.05 m and the depth of the water layer

is 0.1 m. Determine the total buoyant force acting on the cube.

Cube

Oil (ρ= 920 kg/m3)

Water (ρ= 1000 kg/m3)

Solution

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

The volume of the cube is Vcube = side length3= 0.23m3= 0.008 m3.

The volume of the cube submerged in oil is Voil = area of cube base ×

height of submerged portion = 0.22×0.05 m3= 0.002 m3.

The volume of the cube submerged in water is Vwater = area of cube base ×

height of submerged portion = 0.22×0.1 m3= 0.004 m3.

Step 2: Calculate the weight of the cube, oil, and water.

The weight of the cube is Wcube =ρcube ×Vcube ×g= 800 ×0.008 ×9.8 N =

62.72 N.

The weight of the oil displaced by the cube is Woil =ρoil ×Voil ×g=

920 ×0.002 ×9.8 N = 17.92 N.

The weight of the water displaced by the cube is Wwater =ρwater×Vwater×g=

1000 ×0.004 ×9.8 N = 39.2 N.

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

The total buoyant force is the sum of the buoyant forces from the oil and water,

which is equal to the weight of the oil and water displaced by the cube. There-

fore, the total buoyant force is Fbuoy =Woil +Wwater = 17.92 + 39.2 = 57.12 N.

Thus, the total buoyant force acting on the cube is 57.12 N.

Question 27

Question

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

The top face of the cube is at a distance hbelow the liquid surface. Calculate

the fraction of the volume of the cube that is submerged in the liquid.

Solution

Let’s assume the cube has a mass mand its volume is V=a3. We are asked

to find the fraction of the volume that is submerged, which is the volume of the

cube below the liquid surface divided by the total volume of the cube.

Step 1: Find the equation of motion in terms of the forces acting on the

cube. The forces acting on the cube are: - The weight of the cube mg =ρcubeV g.

- The buoyant force Fbuoyant =V ρliquidg. Since the cube is in equilibrium, the

net force on the cube is zero, so we have:

ρcubeV g =V ρliquidg

Step 2: Solve for the depth of the cube submerged h. The volume of the

cube submerged in the liquid is Vsub =a2h. Thus, ρcubea2hg =a3ρliquidg.

Solving for hgives:

h=aρcube

ρliquid

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

the volume submerged is given by:

Vsub

V=a2h

a3=h

a=ρcube

ρliquid

Therefore, the fraction of the volume of the cube that is submerged in the

liquid is ρcube

ρliquid .

Question 28

Question

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

in water with part of the block submerged. The block has a mass of 150 g.

What is the density of the wood if the density of water is 1000 kg/m3?

Solution

Step 1: Find the volume of the block. The volume of the block is given by:

V=l×w×h

where lis the length, wis the width, and his the height of the block. Converting

the dimensions to meters:

V= 0.1 m ×0.05 m ×0.03 m

V= 0.00015 m3

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

is equal to the weight of the water displaced by the block. The weight of the

water displaced is given by:

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

Weight of water displaced = V×1000 kg/m3×9.8 m/s2

Weight of water displaced = 0.00015 m3×1000 kg/m3×9.8 m/s2

Weight of water displaced = 1.47 N

Step 3: Find the weight of the block. The weight of the block is equal to its

mass times the acceleration due to gravity:

Weight of block = Mass of block ×Acceleration due to gravity

Weight of block = 0.15 kg ×9.8 m/s2

Weight of block = 1.47 N

Step 4: Apply Archimedes’ principle. Since the block is floating, the buoyant

force is equal to the weight of the block:

Weight of water displaced = Weight of block

1.47 N = 1.47 N

Step 5: Calculate the density of wood. The density of the wood can be

calculated using the formula:

Density of wood = Mass of block

Volume of block

Density of wood = 0.15 kg

0.00015 m3

Density of wood = 1000 kg/m3

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

Question 29

Question

A cube of side length 0.1 m and density 1000 kg/m3is floating in water. What

is the depth of the cube below the water surface? (Assume the density of water

is 1000 kg/m3and acceleration due to gravity is 9.81 m/s2)

Solution

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

displaced by the cube. This buoyant force can be calculated using Archimedes’

principle:

Buoyant force = Weight of water displaced = Density of water×Volume of cube×g

Buoyant force = 1000 kg/m3×0.13m3×9.81 m/s2

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

Weight of cube = Density of cube ×Volume of cube ×g

Weight of cube = 1000 kg/m3×0.13m3×9.81 m/s2

Step 3: From the condition that the cube is floating, the buoyant force must

equal the weight of the cube. Therefore,

Buoyant force = Weight of cube

Step 4: Setting the expressions for buoyant force and weight of the cube

equal to each other gives us:

1000 ×0.13×9.81 = 1000 ×0.13×9.81

Step 5: Solving for the depth of the cube below the water surface:

Depth = Side length of cube = 0.1 m

Question 30

Question

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

Determine the buoyant force acting on the cube and the depth to which the

cube is submerged in water.

Solution

Let’s denote the density of water as ρw= 1000 kg/m3and the acceleration due

to gravity as g= 9.81 m/s2.

Step 1: Calculate 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,

which can be calculated using Archimedes’ principle:

Buoyant force = Weight of water displaced = ρw·g·Vsubmerged

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

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

cube is given by Vcube = (0.5 m)3. Since the density of the cube is 800 kg/m3

and it is submerged in water (density 1000 kg/m3), the volume of the cube that

is submerged can be calculated as:

Vsubmerged =Density of cube

Density of water ·Vcube

Thus,

Vsubmerged =800 kg/m3

1000 kg/m3·(0.5 m)3

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

Vsubmerged into the expression for buoyant force:

Buoyant force = 1000 kg/m3·9.81 m/s2· 800 kg/m3

1000 kg/m3·0.5 m!3

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

buoyant force is equal to the weight of the cube, we can calculate the depth h

(submerged) of the cube using the buoyant force and the weight of the cube:

Weight of cube = Buoyant force = ρcube ·g·Vcube

The weight of the cube can be calculated as:

Weight of cube = 800 kg/m3·g·(0.5 m)3

Now, equate the weight of the cube to the buoyant force and solve for h.

Question 31

Question

A cube of aluminum with a mass of 200 g and a volume of 100 cm3is submerged

in a container of water. Calculate the buoyant force acting on the cube and

determine if the cube will sink or float in water.

Solution

Step 1: Calculate the density of aluminum using the mass and volume informa-

tion.

Density = mass

volume

Density of aluminum = 200 g

100 cm3= 2 g/cm3

Step 2: Determine the density of water. The density of water at 25

C is

approximately 1 g/cm3.

Step 3: Compare the density of aluminum and water to determine if the

cube will sink or float. Since the density of aluminum (2 g/cm3) is greater than

the density of water (1 g/cm3), the cube will sink as it is denser than water.

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

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

Buoyant force = density of water×volume of cube×acceleration due to gravity

Buoyant force = 1 g/cm3×100 cm3×9.81 m/s2= 981 dyne

Therefore, the buoyant force acting on the cube is 981 dyne, and the cube

will sink in water.

Question 32

Question

A metal sphere with a radius of 0.1 m and a density of 8000 kg/m

is placed in

a container of water. The sphere floats such that 30

Solution

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

the sphere submerged can be calculated using the formula for the volume of a

sphere:

Vsubmerged =4

3πr3×30

100

Step 2: Substitute the radius of the sphere into the formula.

Vsubmerged =4

3π(0.1)3×0.30

Step 3: Calculate the volume of the sphere submerged.

Vsubmerged =4

3×1

103π×0.001 ×0.30

Step 4: Simplify the expression for the volume of the sphere submerged.

Vsubmerged =4

3000π

Step 5: Calculate the mass of the water displaced by the submerged sphere.

The mass of the water displaced can be calculated using the formula:

Massdisplaced = Volumesubmerged ×Densitywater

Step 6: Substitute the known values into the formula.

Massdisplaced =4

3000π×1000 ×800

Step 7: Calculate the mass of the water displaced.

Massdisplaced =3200

3π

Step 8: Calculate the weight of the water displaced. The weight of the water

displaced can be calculated using the formula:

Weightdisplaced = Massdisplaced ×g

Step 9: Substitute the acceleration due to gravity into the formula.

Weightdisplaced =3200

3π×9.81

Step 10: Calculate the weight of the water displaced.

Weightdisplaced ≈31624.23 N

Therefore, the buoyant force acting on the metal sphere is approximately

31624.23 N.

Question 33

Question

A solid metal cube with sides of length 10 cm and a density of 8000 kg/m3is

submerged in a container of water. If the cube floats with 1 cm of its height

above the water surface, determine the depth of the water.

Solution

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

by V=s3, where sis the side length of the cube. In this case, s= 0.1 m.

Therefore,

V= (0.1)3= 0.001 m3

Step 2: Since the cube is floating, the buoyant force acting on it is equal to

the weight of the water the cube displaces. The buoyant force can be calculated

using Archimedes’ principle:

Buoyant force = Weight of water displaced = ρwater ·g·V

where ρwater is the density of water (1000 kg/m3) and gis the acceleration due

to gravity (9.81 m/s2). Substituting the given values,

Buoyant force = 1000 ·9.81 ·0.001 = 9.81 N

Step 3: The weight of the cube can be calculated using its mass and gravi-

tational acceleration:

Weight of cube = mass ·g=ρcube ·V·g

Substitute the given values, we have

Weight of cube = 8000 ·0.001 ·9.81 = 78.48 N

Step 4: The water pushes upwards with a force equal to the buoyant force,

and gravity pulls the cube downwards with a force equal to its weight. Since

the cube floats, the net force must be zero. Therefore,

Buoyant force = Weight of cube

9.81 = 78.48

This equation implies that the cube is in equilibrium.

Step 5: The depth of the water is the level where the buoyant force can

balance the weight of the cube. Since 1 cm of the cube is above the water

surface, the depth of the water must be such that the cube displaces enough

water to counteract the weight of the part of the cube submerged. This implies

that the depth of the water is equal to the height of the submerged part of the

cube. Therefore, the depth of the water is 9 cm.

Question 34

Question

A balloon filled with helium has a volume of 3.0 m3and a density of 0.18 kg/m3.

The balloon is released underwater at a depth of 50 m. What is the net buoyant

force acting on the balloon?

Solution

Step 1: Calculate the buoyant force acting on the balloon at the depth of 50 m

by using Archimedes’ principle. Archimedes’ principle states that the buoyant

force is equal to the weight of the fluid displaced by the object. Given: - Volume

of the balloon, V= 3.0 m3- Density of the fluid (water), ρfluid = 1000 kg/m3

- Acceleration due to gravity, g= 9.81 m/s2- Depth of the balloon, d= 50

m The volume of fluid displaced by the balloon at depth dis Vfluid =V+Ad,

where Ais the area of the balloon.

Vfluid = 3.0 m3+A×50 m2

Step 2: Calculate the area of the balloon using the formula for the surface

area of a sphere, A= 4πr2, where ris the radius of the balloon. Given: -

Volume of the balloon, V=4

3πr3Solving for r:

3πr3= 3.0

r=3

4π×3.01

r≈0.87 m

The area of the balloon is:

A= 4π×(0.87)2

Step 3: Calculate the total volume of fluid displaced by the balloon:

Vfluid = 3.0 m3+ 4π×(0.87)2×50 m3

Step 4: Calculate the mass of the fluid displaced:

mfluid =ρfluid ×Vfluid

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

Fbuoyant =mfluid ×g

Step 6: Finally, substitute the values and solve for the net buoyant force

acting on the balloon at a depth of 50 m.

Question 35

Question

A wooden block with a density of 700 kg/m3and a volume of 0.05 m3floats in a

pool of water. Determine the buoyant force acting on the block and the fraction

of the block that is submerged in the water.

Solution

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

of wooden block, ρwood = 700 kg/m3Volume of wooden block, Vwood = 0.05 m3

Acceleration due to gravity, g= 9.81 m/s2

The weight of the wooden block is given by:

Wwood =ρwood ·Vwood ·g

Wwood = 700 kg/m3·0.05 m3·9.81 m/s2

Wwood = 343.35 N

Step 2: Next, we consider the buoyant force acting on the wooden block.

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

can be calculated using Archimedes’ principle:

Fbuoyant =ρwater ·Vsubmerged ·g

where Density of water, ρwater = 1000 kg/m3

Step 3: We also know that the weight of the water displaced is equal to the

weight of the wooden block.

Wwater =Wwood

Step 4: Setting the weight of the water equal to the buoyant force:

ρwater ·Vsubmerged ·g=Wwood

1000 kg/m3·Vsubmerged ·9.81 m/s2= 343.35 N

Vsubmerged =343.35 N

1000 kg/m3·9.81 m/s2

Vsubmerged = 0.035 m3

Step 5: Finally, we find the fraction of the block submerged in the water.

Fraction submerged = Vsubmerged

Vwood

Fraction submerged = 0.035 m3

0.05 m3

Fraction submerged = 0.7 or 70%

Therefore, the buoyant force acting on the block is 343.35 N and 70

Question 3

Question

A spherical balloon filled with helium gas has a radius of 2 meters. If the density

of helium is 0.1785 kg/m3and the density of air is 1.225 kg/m3, calculate the

maximum mass the balloon can lift off the ground. Assume atmospheric pressure

is 101.3 kPa and the temperature is 25

C. (Hint: The buoyant force is equal to

the weight of the displaced air.)

Solution

Step 1: Calculate the volume of the balloon using the formula for the volume

of a sphere:

V=4

3πr3

Step 2: Substitute the given radius r= 2 meters into the formula to find the

volume V.

Step 3: Calculate the mass of the displaced air using the formula:

mair =ρair ×V

where ρair is the density of air.

Step 4: Calculate the weight Wair of the displaced air using Wair =mair ×g,

where g= 9.81 m/s2is the acceleration due to gravity.

Step 5: Calculate the buoyant force Fbuoyant acting on the balloon: Fbuoyant =

Wair

Step 6: Use Archimedes’ principle to find the maximum mass the balloon

can lift off the ground: Mmax =Fbuoyant

Step 7: Substitute the calculated values to find the maximum mass the

balloon can lift off the ground.

Question 4

Question

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

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

Solution

Step 1: Calculate the density of water as 1000 kg/m3. Step 2: Identify the

forces acting on the cube. The weight of the cube acting downwards is given by

W=ρwood ·Vcube ·g, where ρwood is the density of wood, Vcube is the volume

of the cube, and gis the acceleration due to gravity. The upthrust or buoyant

force acting upwards is given by B=ρwater ·Vsubmerged ·g, where Vsubmerged

is the volume of the cube submerged in water. Step 3: Apply Archimedes’

principle, which states that the upthrust or buoyant force on an object in a fluid

is equal to the weight of the fluid displaced by the object. Therefore, B=W.

Step 4: Set the weight of the cube equal to the upthrust: ρwood ·Vcube ·g=

ρwater ·Vsubmerged ·g. Step 5: Substitute the given values and solve for the depth

hto which the cube is submerged: ρwood ·A·h·g=ρwater ·A·h·gwhere A

is the area of one face of the cube. Step 6: Simplify the expression to find the

depth h:h=ρwood

ρwater

·h=600

1000 ·0.1=0.06 m.

Therefore, the cube is submerged to a depth of 0.06 meters in water.

Question 5

Question

A rectangular block of wood with dimensions 10 cm ×10 cm ×20 cm is floating

in water. The density of water is 1000 kg/m3and the density of wood is 800

kg/m3. Calculate the fraction of the volume of the block that is submerged in

water.

Solution

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

to the weight of the water displaced by the block. We can calculate this using

Archimedes’ principle.

Weight of block = Weight of water displaced

ρwater ×Vsubmerged ×g=ρwood ×Vblock ×g

where ρwater is the density of water, Vsubmerged is the volume submerged, ρwood

is the density of wood, Vblock is the total volume of the block, and gis the

acceleration due to gravity.

Step 2: Calculate the volume submerged. Since the block is floating, the

weight of the block is equal to the weight of the water displaced, as calculated

in Step 1.

ρwater ×Vsubmerged ×g=ρwood ×Vblock ×g

Vsubmerged =ρwood

ρwater ×Vblock

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

volume submerged is the ratio of the volume submerged to the total volume of

the block.

Fraction submerged = Vsubmerged

Vblock

Now, substitute the values and calculate the fraction submerged.

Question 6

Question

A cube of metal with a density of 8000 kg/m3and a volume of 0.04 m3is

placed in a container of water, with the top of the cube at the water’s surface.

Determine: (a) the weight of the cube, (b) the buoyant force acting on the cube,

or remain at rest.

Solution

(a) To calculate the weight of the cube, we use the formula W=mg, where

mis the mass of the cube and gis the acceleration due to gravity (9.81 m/s2).

Given that density ρ= 8000 kg/m3and volume V= 0.04 m3, we can find the

mass of the cube using the formula m=ρV . Let’s substitute the values into

the formula:

m= 8000 kg/m3×0.04 m3

m= 320 kg

Therefore, the weight of the cube is:

W= 320 kg ×9.81 m/s2

W= 3139.2 N

(b) The buoyant force acting on the cube is equal to the weight of the water

displaced by the cube. Using Archimedes’ principle, Fbuoyant =ρwaterVdisplacedg.

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

the volume of the cube under water, Vdisplaced =h×A, where his the height

submerged and Ais the base area of the cube. Given that Vdisplaced = 0.02 m3

(half the volume of the cube), we can now calculate the buoyant force:

Fbuoyant = 1000 kg/m3×0.02 m3×9.81 m/s2

Fbuoyant = 196.2 N

the cube and the buoyant force:

Fnet =W−Fbuoyant

Fnet = 3139.2 N −196.2 N

Fnet = 2943 N

(d) Since the net force acting on the cube is downward (positive), the cube

will sink in the water.

Question 7

Question

A cube of wood with a density of 500 kg/m3and side length 0.2 m is floating

in water. If a 2 kg weight is placed on top of the cube, how much of the cube’s

volume is now submerged in the water?

Solution

Step 1: We first need to calculate the density of water. Since the cube is floating,

the average density of the cube (ρavg) is equal to the density of water (ρwater).

We can use this relationship to find the density of water.

ρavg =ρwater

500 kg/m3=ρwater

Thus, ρwater = 500 kg/m3.

Step 2: Next, we need to account for the additional weight placed on top of

the cube. The weight of the cube, Wcube, is equal to the weight of the water

displaced by the cube, and the weight of the added weight, Wadd, is equal to

the weight of the volume of water displaced by the added weight. Therefore,

the total weight of the system, Wtotal, is given by:

Wtotal =Wcube +Wadd

Wtotal =ρwater ·Vsubmerged ·g+madd ·g

where Vsubmerged is the volume of the cube submerged and gis the acceleration

due to gravity.

Step 3: We also know that the buoyant force acting on the system is equal to

the weight of the water displaced by the total volume of the system. Therefore,

the buoyant force, Fbuoyant, is given by:

Fbuoyant = (ρwater ·Vtotal)·g

where Vtotal is the total volume of the system.

Step 4: Using Archimedes’ principle, we know that the buoyant force is equal

to the total weight of the system. Thus, we have:

Fbuoyant =Wtotal

Step 5: By setting Fbuoyant equal to Wtotal and solving for Vsubmerged, we can

find the volume of water displaced by the cube and the added weight. Finally,

the ratio of Vsubmerged to the total volume of the cube will give us the proportion

of the cube that is now submerged in water.

Question 8

Question

A solid cube of side length 10 cm and density 1.2 g/cm3is placed in a container

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

water = 1 g/cm3)

Solution

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

V=s3, where sis the side length. Given that the side length s= 10 cm, the

volume of the cube is:

V= 103cm3= 1000 cm3

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 and Vis the

volume. Given that the density ρ= 1.2 g/cm3, the mass of the cube is:

m= 1.2×1000 = 1200 g = 1.2 kg

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

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

acceleration due to gravity (g= 9.81 m/s2). Therefore, the weight of the cube

is:

w= 1.2×9.81 = 11.772 N

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

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

Since the cube is fully submerged in water, the buoyant force is equal to the

weight of the water displaced by the cube. The volume of water displaced by

the cube is equal to the volume of the cube, which is 1000 cm3or 0.001 m3.

The weight of the water displaced is given by mwater =ρwaterV g, where ρwater

is the density of water (1 g/cm3or 1000 kg/m3). Therefore, the buoyant force

is:

Fbuoyant =ρwaterV g = 1000 ×0.001 ×9.81 = 9.81 N

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

Question 9

Question

A large wooden cube with a density of 600 kg/m3and side length 5 m is floating

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

Given: Density of water = 1000 kg/m3, acceleration due to gravity = 9.8

m/s2.

Solution

Step 1: Calculate the buoyant force on the wooden cube. The buoyant force

(Fb) on an object is equal to the weight of the fluid displaced by the object. In

this case, the weight of the fluid displaced is equal to the weight of the water

displaced by the cube.

Fb=ρwater ·Vdisplaced ·g

where ρwater is the density of water, Vdisplaced is the volume of water displaced

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

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

water displaced by the cube is equal to the volume of the cube that is submerged

in water. Let’s denote the depth of immersion as h.

Vdisplaced = (5 m)2·h

Step 3: Equate the weight of the cube to the buoyant force. The weight of

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

ρcube ·Vcube ·g=ρwater ·Vdisplaced ·g

Step 4: Solve for the depth of immersion h. Plug in the given values and

density of the cube:

600 kg/m3·(5 m)3·g= 1000 kg/m3·(5 m)2·h·g

53·600 = 52·1000 ·h

5h=53·600

1000

h=5·5·5·600

1000

h= 15 m

Therefore, the depth of immersion of the wooden cube is 15 meters.

Question 10

Question

A cylindrical object with a radius of 8 cm and a height of 12 cm is floating upright

in water. The top of the object is 4 cm above the water surface. Determine the

density of the material the object is made of. (Density of water = 1000 kg/m3)

Solution

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

can be calculated using the formula:

V=πr2h

where ris the radius and his the height of the cylinder. Substituting in the

given values:

V=π×(0.08 m)2×0.12 m

V= 0.03072 m3

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

of water displaced is equal to the volume of the part of the object submerged

in water, which is equal to the volume of a cylinder with a radius of 8 cm and

a height of 4 cm:

Vdisplaced water =π×(0.08 m)2×0.04 m

Vdisplaced water = 0.008064 m3

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

force is equal to the weight of the water displaced by the object, and can be

calculated using the formula:

Fb=ρwater ×g×Vdisplaced water

where ρwater is the density of water and gis the acceleration due to gravity.

Substituting in the given values:

Fb= 1000 kg/m3×9.81 m/s2×0.008064 m3

Fb≈79.22 N

Step 4: Calculate the weight of the object. The weight of the object is equal

to the weight of the water displaced by the part of the object submerged in

water:

Wobject =ρmaterial ×g×Vdisplaced water

where ρmaterial is the density of the material the object is made of. Since the

object floats, the buoyant force is equal to the weight of the object:

Wobject =Fb

ρmaterial ×g×0.008064 m3= 79.22 N

ρmaterial =79.22 N

9.81 m/s2×0.008064 m3

ρmaterial ≈9816.45 kg/m3

Therefore, the density of the material the object is made of is approximately

9816.45 kg/m3.

Question 11

Question

A cube of steel with side length 0.1 m and density 7850 kg/m3is placed in a

tank of oil. The density of the oil is 800 kg/m3. Calculate the buoyant force

acting on the cube of steel.

Solution

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

by V= side length3. Therefore, the volume of the steel cube is:

V= (0.1 m)3= 0.001 m3

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

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

due to gravity. Since m=ρV , where ρis the density of the object and Vis the

volume, we have:

msteel =ρsteelV= 7850 kg/m3×0.001 m3= 7.85 kg

Wsteel =msteel ·g= 7.85 kg ·9.81 m/s2= 77.1485 N

Step 3: Calculate the volume of the displaced oil. Since the cube is fully

submerged in the oil, the volume of oil displaced is equal to the volume of the

cube, which is 0.001 m3.

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

oil is given by Woil =ρoilVdisplaced ·g, where Vdisplaced is the volume of oil

displaced.

Woil = 800 kg/m3×0.001 m3×9.81 m/s2= 7.848 N

Step 5: Calculate the buoyant force. The buoyant force is equal to the weight

of the displaced fluid, which is 7.848 N. Therefore, the buoyant force acting on

the steel cube is 7.848 N.

Question 12

Question

A cube of iron with dimensions 10 cm ×10 cm ×10 cm is submerged in water.

If the density of iron is 7.87 g/cm3and the density of water is 1.0 g/cm3,

determine the buoyant force acting on the iron cube.

Solution

Step 1: Calculate the volume of the iron cube. Given that the dimensions of

the cube are 10 cm ×10 cm ×10 cm, the volume Vis given by:

V= 10 cm ×10 cm ×10 cm = 1000 cm3

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

can be found by multiplying its volume by its density:

m=V×density = 1000 cm3×7.87 g/cm3= 7870 g

Step 3: Calculate the volume of water displaced by the iron cube. Since the

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

of the cube, which is 1000 cm3.

Step 4: 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

cube.

weight = density×volume×gravity = 1.0 g/cm3×1000 cm3×9.81 m/s2= 9810 g

Step 5: Calculate the buoyant force. The buoyant force is equal to the weight

of the water displaced, which is also equal to the weight of the fluid the object

displaces. Therefore, the buoyant force acting on the iron cube is 9810 g or 98.1

Question 13

Question

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

The cube floats with a fraction fof its volume submerged in the liquid.

If the volume of the cube is Vcube =a3, determine an expression for the

fraction fin terms of the densities ρcube and ρliquid.

Solution

Let’s denote the volume of the submerged part of the cube as Vsubmerged. Since

the cube is floating, the buoyant force acting on the cube must be equal to the

weight of the submerged part.

Step 1: Write an expression for the buoyant force. The buoyant force Fb

acting on the cube is given by:

Fb=ρliquid ·Vsubmerged ·g

Step 2: Write an expression for the weight of the submerged part. The

weight Wsubmerged of the submerged part is given by:

Wsubmerged =ρcube ·Vsubmerged ·g

Step 3: Equate the buoyant force to the weight of the submerged part. Since

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

part:

ρliquid ·Vsubmerged ·g=ρcube ·Vsubmerged ·g

Step 4: Solve for the fraction f. Since the volume of the cube is Vcube =a3:

Vsubmerged =f·Vcube =f·a3

Substitute the expression for Vsubmerged into the equation from Step 3:

ρliquid ·f·a3=ρcube ·f·a3

Divide by ρcubea3on both sides to solve for f:

f=ρliquid

ρcube

Therefore, the fraction fof the cube submerged in the liquid is f=ρliquid

ρcube

Question 14

Question

A wooden cube with a side length of 0.5 meters and a density of 600 kg/m3is

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

water.

Solution

Step 1: First, let’s determine the density of water. The density of water is 1000

kg/m3.

Step 2: The buoyant force acting on the cube is equal to the weight of the

water displaced by the cube. Using Archimedes’ Principle, the buoyant force

(Fb) is given by the formula:

Fb=ρwater ·Vsubmerged ·g

where ρwater is the density of water, Vsubmerged is the volume of the cube sub-

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

Step 3: The weight of the cube is equal to the force of gravity acting on it.

The weight of the cube (W) is given by:

W=ρcube ·Vcube ·g

where ρcube is the density of the cube, Vcube is the total volume of the cube,

and gis the acceleration due to gravity.

Step 4: Since the cube is floating, the buoyant force must be equal to the

weight of the cube. Therefore, we have:

Fb=W

ρwater ·Vsubmerged ·g=ρcube ·Vcube ·g

Step 5: Substitute the given values into the equation:

1000 ·Vsubmerged = 600 ·0.53

Step 6: Solve for Vsubmerged:

Vsubmerged =600 ·0.53

1000 = 0.15 m3

Step 7: The depth to which the cube is submerged in the water is equal to

the height of the cube submerged in water, which is equal to the side length of

the cube:

Depth submerged = 0.5 m

Question 15

Question

A solid sphere of radius Rand density ρ1is placed in a fluid of density ρ2

such that it floats with half of its volume submerged. Calculate the ratio of the

density of the sphere to the density of the fluid.

Solution

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

fluid as Vs. Given that the sphere floats with half of its volume submerged, we

have:

Vs=1

2×4

3πR3

Step 2: The weight of the fluid displaced by the submerged volume Vswill

be equal to the weight of the sphere. Using Archimedes’ principle, we have:

ρ2·Vs·g=ρ1·4

3πR3·g

Step 3: Substitute the expression for Vsinto the equation from Step 2 and

simplify the equation.

ρ2·1

2×4

3πR3·g=ρ1·4

3πR3·g

Step 4: Cancel out common terms and solve for the ratio of the densities.

2ρ2=ρ1

ρ1

ρ2

Step 5: Therefore, the ratio of the density of the sphere to the density of the

fluid is 1

Question 16

Question

A cylindrical metal object of height 20 cm and radius 5 cm is floating in water.

If the density of the metal is 8000 kg/m3and the density of water is 1000 kg/m3,

determine the fraction of the object that is submerged in water.

Solution

Step 1: We first calculate the volume of the metal object. The volume of a

cylinder is given by:

Vcylinder =πr2h

where ris the radius and his the height of the cylinder. Plugging in the values:

Vcylinder =π(0.05 m)2·0.20 m

Vcylinder =π·0.0025 m2·0.20 m

Vcylinder = 0.00157 m3

Step 2: The weight of the object must equal the buoyant force acting on the

object when it is floating. The weight of the object is given by:

Wobject =ρmetal ·Vcylinder ·g

where ρmetal is the density of the metal, Vcylinder is the volume of the cylinder,

and gis the acceleration due to gravity.

Wobject = 8000 kg/m3·0.00157 m3·9.81 m/s2

Wobject = 124.1 N

Step 3: The buoyant force is given by:

Fbuoyant =ρwater ·Vsubmerged ·g

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

submerged, and gis the acceleration due to gravity.

Step 4: Since the object is floating, the weight of the object is equal to the

buoyant force:

Wobject =Fbuoyant

ρmetal ·Vcylinder ·g=ρwater ·Vsubmerged ·g

Step 5: Solving for the fraction of the object submerged:

Vsubmerged

Vcylinder

=ρmetal

ρwater

Vsubmerged

0.00157 =8000

1000

Vsubmerged = 0.01256 m3

Step 6: Finally, the fraction of the object submerged is:

Vsubmerged

Vcylinder

=0.01256

0.00157 = 8

Therefore, 8/20 or 40% of the object is submerged in water.

Question 17

Question

A cylindrical tank is filled with water to a height of 3 meters. The tank has a

diameter of 4 meters. Calculate the total force exerted on the base of the tank

by the water. Take the density of water to be 1000 kg/m3and the acceleration

due to gravity to be 9.81 m/s2.

Solution

Step 1: Calculate the volume of water in the tank. The volume of water in the

tank can be calculated using the formula for the volume of a cylinder:

V=πr2h

where ris the radius of the tank and his the height of the water in the tank.

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

diameter: r= 2 m. Substitute r= 2 m and h= 3 m into the formula:

V=π×(2 m)2×3 m

V= 12πm3

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

can be calculated using the formula:

m= density ×volume

Given that the density of water is 1000 kg/m3:

m= 1000 kg/m3×12πm3