PHYS 101 - ELEMENTS OF PHYSICS Buoyant forces and Archimedes' principle Question Bank Set 7

PHYS 101 - ELEMENTS OF PHYSICS

- Buoyant forces and Archimedes’

principle

Question Bank - Set 7

Liberty University

Question 1

Question

A 500 kg block of iron is placed on a boat floating on a lake. If the block of

iron is submerged in the water, what is the minimum volume of water displaced

needed to keep the boat afloat with the block of iron submerged? Take the

density of iron to be 7,860 kg/m3and the density of water to be 1,000 kg/m3.

Solution

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

block of iron is 500 kg, we can calculate its weight using the formula:

Weight = mass ×gravity

Weight = 500 kg ×9.81 m/s2= 4,905 N

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

force is equal to the weight of the water displaced by the block of iron. Let V

be the volume of water displaced by the block of iron. The buoyant force can

be calculated as:

Buoyant force = density of water ×V×gravity

Buoyant force = 1000 kg/m3×Vm3×9.81 m/s2

Step 3: Use Archimedes’ principle to determine the minimum volume of

water displaced. According to Archimedes’ principle, the buoyant force acting

on an object is equal to the weight of the fluid it displaces. Therefore, the

buoyant force should be equal to the weight of the block of iron when submerged:

1000 kg/m3×Vm3×9.81 m/s2= 4905 N

Step 4: Solve for the volume of water displaced, V.

1000 kg/m3×Vm3×9.81 m/s2= 4905 N

V=4905 N

1000 kg/m3×9.81 m/s2

Step 5: Calculate the minimum volume of water displaced.

V=4905

1000 ×9.81 =4905

9810 ≈0.5 m3

Therefore, the minimum volume of water displaced needed to keep the boat

afloat with the block of iron submerged is approximately 0.5 m3.

Question 2

Question

A cube of side length 10 cm and density 800 kg/m3is submerged in water.

Calculate the buoyant force acting on the cube. (Density of water = 1000

kg/m3and acceleration due to gravity = 9.8 m/s2)

Solution

Step 1: First, we need to determine the volume and mass of the cube. Given:

Side length of cube, a= 10 cm = 0.1 m

Density of cube, ρcube = 800 kg/m3

Density of water, ρwater = 1000 kg/m3

The volume of the cube is given by V=a3= 0.13m3. The mass of the cube

can be calculated using the equation mcube =ρcube ·V.

Step 2: Calculate the mass of the cube.

mcube = 800 ×0.13kg = 8 ×10−3kg

Step 3: Now, we need to determine the weight of the 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.

Step 4: Calculate the weight of the cube.

Wcube =mcube ×g= 8 ×10−3×9.8 N

Step 5: According to Archimedes’ principle, the buoyant force acting on the

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

force can be calculated using the equation Fbuoyant =ρwater ·V·g.

Step 6: Calculate the buoyant force acting on the cube.

Fbuoyant = 1000 ×0.13×9.8 N

Step 7: Simplify the expression to find the final answer for the buoyant force.

Fbuoyant = 100 ×10−3×9.8 N = 0.98 N

Therefore, the buoyant force acting on the cube is 0.98 N.

Question 3

Question

A block of wood with a density of 0.6 g/cm3and a volume of 100 cm3is floating

in a container of water. What is the minimum mass of lead that must be placed

on top of the block to submerge it completely in water? The density of lead is

11.3 g/cm3.

Solution

Step 1: First, let’s calculate the mass of the wooden block. Given: Density

of wood, ρwood = 0.6 g/cm3Volume of wood, Vwood = 100 cm3Let’s use the

formula:

mwood =ρwood ×Vwood

mwood = 0.6×100

mwood = 60 g

Step 2: Next, let’s find the volume of water displaced by the wooden block.

Since the block is floating, the weight of the block is equal to the weight of the

water displaced, which is equal to the buoyant force acting on the block. Let’s

calculate the volume of water displaced:

Vwater =Vwood

Vwater = 100 cm3

Step 3: Now, let’s calculate the mass of the water displaced by the wooden

block. Given: Density of water, ρwater = 1 g/cm3Using the formula:

mwater =ρwater ×Vwater

mwater = 1 ×100

mwater = 100 g

Step 4: The total mass that must be placed on top of the block to submerge

it completely in water is the sum of the mass of the water displaced and the

mass of the wooden block. Let’s calculate:

Total mass = mwood +mwater

Total mass = 60 + 100

Total mass = 160 g

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

block to submerge it completely in water is 160 g.

Question 4

Question

A cube of density ρ1and side length ais floating in a liquid of density ρ2. The

cube is such that one third of it is submerged in the liquid. Calculate the density

of the cube.

Solution

Step 1: Define the variables and known values.

Let’s represent: - Density of the cube as ρ(what we want to find) - Density of the

liquid as ρ2- Volume of the cube as V=a3- Volume of the submerged part of

the cube as V′=1

3×a2×a=a3

3- Weight of the cube as W=m×g=ρ1×V×g

- Buoyant force as Fb=ρ2×V′×g

Step 2: Apply Archimedes’ principle.

According to Archimedes’ principle, the buoyant force acting on the cube must

be equal to the weight of the cube for it to float. In equation form:

Fb=W

Step 3: Express the equations in terms of the variables.

Substitute the expressions for Wand Fbinto the equation:

ρ2×a3

3×g=ρ1×a3×g

Step 4: Solve for the density of the cube.

Simplify the equation and solve for ρ:

ρ2

3=ρ1

ρ1=ρ2

Therefore, the density of the cube is ρ2

Question 5

Question

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

in water. What is the height of the cube that is above the water surface?

Solution

Step 1: Determine the density of water Given that the cube is floating in water,

the density of water (ρw) is equal to the density of the cube, which is 0.6 g/cm3.

Step 2: Determine the volume of the cube submerged in water Let’s assume

the height of the cube submerged in the water is h, then the volume of the cube

submerged in water is given by:

Vsubmerged = (Area of base)(h) = (10 cm ×10 cm)(h) = 100hcm3

Step 3: Set up the equation using Archimedes’ principle The buoyant force

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

This can be represented by the equation:

(Buoyant force) = (Weight of water displaced)

Step 4: Calculate the buoyant force The buoyant force is given by the for-

mula:

Buoyant force = ρw·g·Vsubmerged

Substitute the values of ρw,g, and Vsubmerged into the equation.

Step 5: Calculate the weight of the cube The weight of the cube is given by

the formula:

Weight of the cube = ρc·g·V

where ρcis the density of the cube and Vis the volume of the cube.

Step 6: Equate the buoyant force to the weight of the cube Set the buoyant

force equal to the weight of the cube and solve for h, the height of the cube

above the water surface.

Step 7: Calculate the height of the cube above the water surface Substitute

the necessary values into the equation from Step 6 and solve for h.

Question 6

Question

A cube of side length aand density ρcube is completely submerged in a liquid

of density ρliquid. The cube is initially at rest, but when released, it accelerates

upwards with a constant velocity. Find the ratio of the density of the cube to

the density of the liquid.

Solution

Step 1: The force due to buoyancy acting on the cube is equal to the weight

of the liquid displaced by the cube. Let the volume of the cube be Vcube =a3,

then the weight of the cube is mg =ρcubeVcubegand the weight of the liquid

displaced is ρliquidVcubeg. So, equating these forces:

ρcubeVcubeg=ρliquidVcubeg

ρcube =ρliquid

Step 2: Since the cube is accelerating upwards with a constant velocity, the

net force acting on the cube is zero. The forces acting on the cube are its weight

W=ρcubeVcubegdownwards and the buoyant force B=ρliquidVcubegupwards.

Therefore, W−B= 0:

ρcubeVcubeg−ρliquidVcubeg= 0

⇒ρcube −ρliquid = 0

⇒ρcube =ρliquid

Step 3: The ratio of the density of the cube to the density of the liquid is:

Density ratio = ρcube

ρliquid

=ρliquid

ρliquid

= 1

Question 7

Question

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

density of helium is 0.178 kg/m3and the density of air is 1.29 kg/m3. What is

the maximum mass that the balloon can lift off the ground?

Solution

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

the formula V=4

3πr3, where ris the radius of the sphere. Substitute r= 2 m

into the formula:

V=4

3π(2 m)3

V=32

3πm3

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

force is given by the formula Fb=ρfluidV g, where ρfluid is the density of the

fluid, Vis the volume of the object submerged, and gis the acceleration due to

gravity. Substitute ρfluid = 1.29 kg/m3,V=4

3π(2 m)3, and g= 9.81 m/s2into

the formula:

Fb= 1.29 ×32

3π×9.81 N

Step 3: Calculate the gravitational force acting on the balloon. The gravita-

tional force is given by the formula Fg=mballoong, where mballoon is the mass

of the balloon and gis the acceleration due to gravity.

Step 4: Set up the equilibrium condition. For the balloon to lift off the

ground, the buoyant force must be greater than or equal to the gravitational

force. Therefore,

Fb≥Fg

1.29 ×32

3π×9.81 ≥mballoon ×9.81

Step 5: Solve for the maximum mass that the balloon can lift off the ground.

mballoon ≤1.29 ×32

3πkg

Question 8

Question

A cylindrical container with a radius of 0.5 meters and a height of 1 meter is

completely filled with water. Determine the buoyant force acting on a solid

iron sphere with a radius of 0.2 meters and a density of 7800 kg/m3that is

submerged in the water inside the container.

Solution

Step 1: First, let’s calculate the volume of the iron sphere using the formula for

the volume of a sphere: V=4

3πr3, where ris the radius of the iron sphere. The

radius of the iron sphere, r= 0.2 meters. Plugging in the values, we get:

V=4

3π(0.2)3

Step 2: Now, calculate the volume of water displaced by the iron sphere.

This volume is equal to the volume of the sphere. So, Vdisplaced =4

3π(0.2)3.

Step 3: Calculate the mass of the iron sphere using the formula m= density×

volume. The density of iron, ρiron = 7800 kg/m3. So, miron = 7800 ×4

3π(0.2)3.

Step 4: Calculate the weight of the iron sphere using the formula Wiron =

miron ×g, where gis the acceleration due to gravity (9.8m/s2).

Wiron = 7800 ×4

3π(0.2)3×9.8

Step 5: Now, let’s calculate the buoyant force acting on the iron sphere.

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

the water displaced by the iron sphere. Therefore, the buoyant force is equal to

the weight of the water displaced. So, Fbuoyant =ρwater ×Vdisplaced ×g. The

density of water, ρwater = 1000 kg/m3. Therefore, Fbuoyant = 1000 ×4

3π(0.2)3×

9.8.

Step 6: Finally, the buoyant force acting on the iron sphere is the difference

between the weight of the water displaced and the weight of the iron sphere:

Fbuoyant = 1000 ×4

3π(0.2)3×9.8−7800 ×4

3π(0.2)3×9.8

Question 9

Question

A cylinder with a cross-sectional area of 0.05 m2and a height of 2.0 m is floating

vertically in a pool of water. A mass of 100 kg is placed on top of the cylinder.

If the density of water is 1000 kg/m3, what is the height of the cylinder that

remains above the water surface?

Solution

Step 1: Determine the volume of water displaced by the cylinder. The mass

of the cylinder can be calculated using the density of water and the volume of

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

cylinder beneath the water surface.

Given: Cross-sectional area, A= 0.05m2Height of the cylinder, h= 2.0m

Density of water, ρwater = 1000kg/m3

The volume of water displaced is:

Vwater =A×h

Substitute the given values:

Vwater = 0.05 ×2.0

Vwater = 0.1m3

Step 2: Calculate the mass of the water displaced. The mass of the water

displaced can be calculated using the volume of water displaced and the density

of water.

Given: Density of water, ρwater = 1000kg/m3

The mass of the water displaced is:

mwater =ρwater ×Vwater

Substitute the values:

mwater = 1000 ×0.1

mwater = 100kg

Step 3: Determine the buoyant force on the cylinder. The buoyant force

acting on the cylinder is equal to the weight of the water displaced.

The buoyant force is given by:

Fbuoyant =mwater ×g

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

Substitute the values:

Fbuoyant = 100 ×9.81

Fbuoyant = 981N

Step 4: Calculate the weight of the added mass. The weight of the added

mass is equal to the mass multiplied by the acceleration due to gravity. Given:

Mass of the added mass, madded = 100kg

The weight of the added mass is:

Wadded =madded ×g

Wadded = 100 ×9.81

Wadded = 981N

Step 5: Determine the total downward force acting on the cylinder. The

total downward force acting on the cylinder is the sum of the weight of the

added mass and the weight of the water displaced.

Fdownward =Wadded +Fbuoyant

Fdownward = 981 + 981

Fdownward = 1962N

Step 6: Calculate the height of the cylinder above the water surface. The

height of the cylinder above the water surface, habove, can be found by equating

the total downward force to the weight of the entire cylinder.

Given: Mass of the entire cylinder, mcylinder = 100kg

The weight of the entire cylinder is:

Wcylinder =mcylinder ×g

Wcylinder = 100 ×9.81

Wcylinder = 981N

Since the cylinder is in equilibrium, the total downward force is equal to the

weight of the cylinder:

Fdownward =Wcylinder

1962 = 100 ×9.81

1962 = 981 ×habove

habove =1962

981

habove = 2.0m

Therefore, the height of the cylinder that remains above the water surface

is 2.0 meters.

Question 10

Question

A rectangular block of metal measures 6 m in length, 3 m in width, and 2 m

in height. The block has a density of 5000 kg/m3. What is the buoyant force

acting on the block when it is submerged in water? (Assume the density of

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

Solution

Step 1: Begin by calculating the volume of the block. The volume of a rectan-

gular block is given by V=l×w×h, where lis the length, wis the width,

and his the height of the block. Step 2: Substitute the given values into the

formula to find the volume. Step 3: The volume of the block is:

V= 6 m ×3 m ×2 m = 36 m3

Step 4: Next, calculate the weight of the block. The weight of an object is

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

due to gravity (9.81 m/s2). Step 5: The mass of the block can be found using

the formula m=ρ×V, where ρis the density of the block and Vis the volume

of the block. Step 6: Substitute the given density of the block and the volume

of the block into the formula to find the mass. Step 7: The mass of the block

is:

m= 5000 kg/m3×36 m3= 180000 kg

Step 8: Calculate the weight of the block by multiplying the mass by the

acceleration due to gravity. Step 9: The weight of the block is:

W= 180000 kg ×9.81 m/s2= 1765800 N

Step 10: According to Archimedes’ principle, the buoyant force acting on

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

the object. The buoyant force is given by 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 11: The

volume of the block submerged in water is equal to the volume of the block.

Step 12: Substitute the given density of water, the volume of the block, and the

acceleration due to gravity into the formula to find the buoyant force. Step 13:

The buoyant force acting on the block when submerged in water is:

Fbuoyant = 1000 kg/m3×36 m3×9.81 m/s2= 353160 N

Therefore, the buoyant force acting on the block when submerged in water

is 353160 N.

Question 11

Question

A cube of side length 0.5 m and density 1000 kg/m3is submerged in a liquid

with a density of 800 kg/m3. Calculate the buoyant force acting on the cube.

(Take acceleration due to gravity to be 9.81 m/s2)

Solution

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

cube is 0.5 m, the volume of the cube is given by:

Volume (V) = (side length)3= (0.5 m)3= 0.125 m3

Step 2: Calculate the mass of the cube. The mass of the cube is given by:

Mass (m) = Density ×Volume = 1000 kg/m3×0.125 m3= 125 kg

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

by:

Weight = Mass ×Acceleration due to gravity = 125 kg ×9.81 m/s2= 1226.25 N

Step 4: Calculate the buoyant force. The buoyant force acting on an object

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

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

The buoyant force is given by:

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

Buoyant force = 800 kg/m3×0.125 m3×9.81 m/s2= 981 N

Therefore, the buoyant force acting on the cube is 981 N.

Question 12

Question

A cylindrical container filled with water has a volume of 2.5 liters. A metal

cube with a density of 8000 kg/m3is submerged in the water. The dimensions

of the cube are 5 cm ×5 cm ×5 cm. Calculate the buoyant force acting on the

cube and determine if it floats or sinks.

Solution

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

the formula:

Vcube = side length3= (0.05 m)3= 0.000125 m3

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

using the formula:

mass = density ×volume = 8000 kg/m3×0.000125 m3= 1 kg

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

by:

weight = mass ×acceleration due to gravity = 1 kg ×9.8 m/s2= 9.8 N

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

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

weight of the water with a volume equal to the volume of the submerged cube.

The volume of water displaced is equal to the volume of the cube:

Vwater displaced =Vcube = 0.000125 m3

The weight of the water displaced is:

weight displaced = mass×acceleration due to gravity = 1000 kg/m3×0.000125 m3×9.8 m/s2= 1.225 N

Step 5: Determine if the cube floats or sinks. The buoyant force acting on

the cube is equal to the weight of the water displaced, which is 1.225 N. Since

the weight of the cube is 9.8 N, which is greater than the buoyant force, the

cube sinks in water.

Question 13

Question

A solid cube of steel with side length 0.2 m is immersed in a container of water.

Given that the density of steel is 7,800 kg/m3and the density of water is 1,000

kg/m3, determine the buoyant force acting on the steel cube.

Solution

Step 1: Calculate the weight of the steel cube. The weight Wof the steel cube

can be calculated by:

W=m·g

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

(9.81 m/s2). The mass mcan be calculated using the formula:

m=ρsteel ·V

where ρsteel is the density of steel and Vis the volume of the cube. Given that

the side length of the cube is 0.2 m, the volume Vis:

V= side length3= (0.2 m)3= 0.008 m3

Substitute the values to find m:

m= 7,800 kg/m3×0.008 m3= 62.4 kg

Then, calculate the weight W:

W= 62.4 kg ×9.81 m/s2= 612.14 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. According to Archimedes’ principle, the magnitude of the buoyant force

is given by:

Fb=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. The volume

Vsubmerged of the cube submerged in water can be calculated as Vsubmerged =V.

Substitute the values for water density and volume of the cube:

Fb= 1,000 kg/m3×0.008 m3×9.81 m/s2= 78.48 N

Therefore, the buoyant force acting on the steel cube is 78.48 N.

Question 14

Question

A solid cylinder of height hand base radius Ris floating in a liquid of density

ρl. Given that the density of the cylinder is ρc, determine the fraction of the

cylinder that is submerged in the liquid.

Solution

Step 1: Let’s denote the fraction of the cylinder that is submerged in the liquid

as f. We are looking for the value of f.

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

the liquid displaced. The weight of the liquid displaced is equal to the weight of

the submerged part of the cylinder. Therefore, the buoyant force, Fb, is equal

to ρl·g·πR2·f·h, where gis the acceleration due to gravity.

Step 3: The weight of the cylinder is equal to the weight of the submerged

part plus the weight of the part above the liquid. Therefore, the weight of the

cylinder, Wc, is equal to ρc·g·πR2·h.

Step 4: The cylinder is in equilibrium, so the buoyant force is equal to the

weight of the cylinder. We can set Fbequal to Wcand solve for f.

Step 5: Setting the buoyant force equal to the weight of the cylinder, we

have: ρl·g·πR2·f·h=ρc·g·πR2·h

Step 6: Simplifying the equation, we find: ρl·f=ρc

Step 7: Finally, the fraction of the cylinder submerged in the liquid is given

by:

f=ρc

ρl

Question 15

Question

A spherical balloon with a radius of 5 meters is filled with helium at a density of

0.178 kg/m3. The density of air is 1.29 kg/m3. Determine the maximum mass

that the balloon can carry without sinking in the air.

Solution

Step 1: The buoyant force acting on the balloon can be calculated using Archimedes’

principle:

Fbuoyant =ρair ·Vballoon ·g

where ρair = 1.29 kg/m3is the density of air, Vballoon =4

3πr3is the volume of

the balloon, and g= 9.8 m/s2is the acceleration due to gravity.

Step 2: Substituting the given values into the equation, we get:

Vballoon =4

3π(5)3=500

3π= 523.6 m3

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

Fbuoyant = 1.29 ×523.6×9.8≈6500 N

Step 4: For the balloon to float in the air, the buoyant force must be greater

than or equal to the weight of the balloon (mass carried by the balloon):

Fbuoyant ≥mballoon ·g

Step 5: We can rearrange the equation to solve for the maximum mass that

the balloon can carry:

mballoon ≤Fbuoyant

Step 6: Substituting the calculated value of Fbuoyant into the equation:

mballoon ≤6500

9.8≈663.3 kg

Therefore, the maximum mass that the balloon can carry without sinking in

the air is approximately 663.3 kg.

Question 16

Question

A cube of wood with dimensions 10 cm ×10 cm ×10 cm floats in water with

5 cm of its height above the water level. Determine the density of the wood.

Solution

Let’s denote the density of the wood as ρwood and the density of water as ρwater.

We can start solving the problem by considering the forces acting on the cube.

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

of the cube submerged in water is given by the product of the base area and

the height submerged:

Vsubmerged = 10 cm ×10 cm ×5 cm = 500 cm3

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

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

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

Fbuoyant =ρwater ·g·Vsubmerged

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

to the weight of the volume of wood submerged in water:

Fwood =ρwood ·g·Vsubmerged

Step 4: Apply equilibrium condition. Since the cube is floating in water,

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

Fwood =Fbuoyant

Step 5: Solve for the density of the wood. Setting the expressions for Fwood

and Fbuoyant equal:

ρwood ·g·Vsubmerged =ρwater ·g·Vsubmerged

ρwood =ρwater

Therefore, the density of the wood is equal to the density of water.

Question 17

Question

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

submerged in water. The object has a density of 800 kg/m3. Calculate the

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

Solution

Step 1: Calculate the volume of the cylindrical object.

Volume = πr2h

Volume = π(0.05 m)2×0.20 m

Volume = 0.000157 m3

Step 2: Calculate the weight of the object.

Weight = Mass ×Gravity

Weight = Volume ×Density ×Gravity

Weight = 0.000157 m3×800 kg/m3×9.81 m/s2

Weight ≈1.24 N

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

Buoyant Force = Volume ×Density of Water ×Gravity

Buoyant Force = 0.000157 m3×1000 kg/m3×9.81 m/s2

Buoyant Force ≈1.54 N

Step 4: Determine whether the object will sink or float. Since the weight of

the object is less than the buoyant force acting on it, the object will float.

Question 18

Question

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

water. The density of wood is 0.8 g/cm3and the density of water is 1 g/cm3.

Determine the depth to which the block of wood is submerged in water.

Solution

Step 1: First, let’s calculate the mass of the block of wood using its volume and

density. Given: Density of wood, ρwood = 0.8 g/cm3Dimensions of the block:

10 cm ×5 cm ×3 cm

The volume of the block of wood is given by:

Vwood = 10 cm ×5 cm ×3 cm

Vwood = 150 cm3

The mass of the block of wood can be calculated using its volume and density:

mwood =ρwood ×Vwood

mwood = 0.8 g/cm3×150 cm3

mwood = 120 g

Step 2: Next, let’s determine the buoyant force acting on the block of wood.

The buoyant force can be calculated using the volume of water displaced by the

block and the density of water. The volume of water displaced by the block is

equal to the volume of the block submerged in water.

Let xbe the depth to which the block of wood is submerged in water. Then

the submerged volume of the block is given by:

Vsubmerged = 10 cm ×5 cm ×x

Vsubmerged = 50xcm3

The mass of water displaced by the block is equal to the mass of the block

of wood (since the block is in equilibrium):

mwater =mwood = 120 g

The volume of water displaced by the block is equal to the submerged volume

of the block:

Vwater =Vsubmerged = 50xcm3

The mass of water displaced by the block can also be calculated using the

density of water:

mwater =ρwater ×Vwater

120 g = 1 g/cm3×50xcm3

x=120

50 = 2.4 cm

Therefore, the block of wood is submerged to a depth of 2.4 cm in water.

Question 19

Question

A rectangular block of wood with dimensions 10 cm ×4 cm ×6 cm floats in a

container of water. The weight of the block is measured to be 1.2 N. Calculate

the density of the wood. (Density of water is 1000 kg/m3and acceleration due

to gravity is 9.81 m/s2)

Solution

Step 1: Calculate the volume of the wood block. The volume of the wood block

can be calculated using the formula:

Volume = length ×width ×height

Given that the dimensions of the wood block are 10 cm×4 cm×6 cm, the volume

is:

Volume = 10 ×4×6 cm3= 240 cm3

Step 2: Convert the volume to cubic meters. To convert from cubic centime-

ters to cubic meters, we need to divide by 106:

Volume = 240 cm3= 240 ×10−6m3= 0.00024 m3

Step 3: Calculate the buoyant force acting on the wood block. The buoyant

force acting on the wood block is equal to the weight of the water displaced by

the block. This can be calculated using Archimedes’ principle:

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

Given that the density of water is 1000 kg/m3and the volume of the water

displaced equals the volume of the wood block, the buoyant force is:

Buoyant force = 1000 ×0.00024 ×9.81 N = 2.35 N

Step 4: Calculate the density of the wood. Since the wood block is float-

ing, the weight of the block must be equal to the buoyant force acting on it.

Therefore, the density of the wood can be calculated as:

Density of the wood = Weight of the wood

Volume of the wood ×Acceleration due to gravity

Plugging in the values we have:

Density of the wood = 1.2

0.00024 ×9.81 = 520.83 kg/m3

Therefore, the density of the wood block is 520.83 kg/m3.

Question 20

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 whether the cube will sink

or float.

Solution

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

due to gravity as g= 9.8 m/s2.

Step 1: Calculate the weight of the cube. The weight of the cube can be

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

the acceleration due to gravity. Given that the density ρis related to mass by

ρ=m

V, where Vis the volume of the cube V=s3with sbeing the side length.

Therefore, m=ρcube ·V. Substitute in the given values to find the weight of

the cube.

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

acting on the cube is given by Archimedes’ principle: Fbuoyant =ρwater ·V·g.

Substitute in the values to calculate the buoyant force.

Step 3: Analyze whether the cube will sink or float. If the buoyant force is

greater than or equal to the weight of the cube, the cube will float. Otherwise,

it will sink. Compare the two forces to determine the fate of the cube.

Question 21

Question

A cylindrical object of radius rand height his floating in a liquid of density

ρL. The density of the object is ρO. If the cylinder is pushed down slightly and

released, it oscillates up and down in simple harmonic motion. Determine the

period of oscillation in terms of the given parameters.

Solution

Step 1: Determine the buoyant force on the cylinder when it is floating at

equilibrium.

The buoyant force Fbacting on the cylinder is equal in magnitude to the

weight of the liquid displaced by the cylinder. The weight of the displaced liquid

is given by mliquidg=ρLVcylinderg, where Vcylinder is the volume of the cylinder.

Using the formula for the volume of a cylinder (V=πr2h), we find that

Vcylinder =πr2h.

Therefore, the buoyant force when the cylinder is floating at equilibrium is:

Fb=ρLπr2hg

Step 2: Write the equilibrium condition for the floating cylinder.

For the cylinder to be floating at equilibrium, the weight of the cylinder

Wcylinder must be equal to the buoyant force Fb:

Wcylinder =Fb

The weight of the cylinder is given by Wcylinder =ρOVcylinderg, where Vcylinder

is the volume of the cylinder.

Substitute the expression for Vcylinder and the weight of the cylinder into the

equilibrium condition:

ρOπr2hg =ρLπr2hg

Step 3: Determine the period of oscillation of the cylinder.

The period of oscillation of a floating object is given by:

T= 2πrm

where mis the effective mass of the object in the liquid and Fbis the buoyant

force.

The effective mass of the object in the liquid is the sum of the actual mass

of the object and the mass of the liquid displaced by the object:

m=ρOVcylinder +ρLVcylinder =ρOπr2h+ρLπr2h

Substitute the expressions for the effective mass and the buoyant force into

the equation for the period of oscillation:

T= 2πsρOπr2h+ρLπr2h

ρLπr2h

Simplify the expression to find the period of oscillation in terms of the given

parameters.

Question 22

Question

A metal block of mass 500 kg and volume 0.2 m3is lowered into a container of

water. If the density of water is 1000 kg/m3, determine: a) The buoyant force

acting on the block when it is completely submerged in the water. b) Whether

the block will sink or float in the water.

Solution

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

500 kg, we can calculate the weight using the formula W=mg, where mis the

mass and gis the acceleration due to gravity. With m= 500 kg and g= 9.81

m/s2, we get:

W= (500 kg)(9.81 m/s2) = 4905 N

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 volume of the

block is 0.2 m3, so the volume of water displaced is also 0.2 m3. The weight

of this volume of water is given by Wwater = densitywater ×volumewater ×g.

Substituting in densitywater = 1000 kg/m3and volumewater = 0.2 m3, we get:

Wwater = (1000 kg/m3)(0.2 m3)(9.81 m/s2) = 1962 N

Therefore, the buoyant force acting on the block when completely submerged

in water is 1962 N.

Step 3: Determine if the block will sink or float in the water. Since the weight

of the block (4905 N) is greater than the buoyant force acting on it (1962 N),

the block will sink in the water.

Question 23

Question

A cube with a volume of 0.5 m3and a density of 800 kg/m3is completely

submerged in a liquid of density 1000 kg/m3. Calculate the buoyant force

acting on the cube.

Solution

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

weight of the cube is given by:

W=mg

where: - mis the mass of the cube, - gis the acceleration due to gravity.

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

m=ρV

where: - ρis the density of the cube, and - Vis the volume of the cube.

Substitute the values, we get:

m= 800 kg/m3×0.5 m3

Step 3: Calculate the mass of the cube:

m= 400 kg

Step 4: Now, calculate the weight of the cube:

W= 400 kg ×9.8 m/s2

Step 5: Determine the upward buoyant force, which is equal in magnitude

to the weight of the liquid displaced by the cube. The buoyant force is given

by:

Fbuoyant =ρliquidV g

where: - ρliquid is the density of the liquid.

Step 6: Substitute the values, we get:

Fbuoyant = 1000 kg/m3×0.5 m3×9.8 m/s2

Step 7: Calculate the buoyant force:

Fbuoyant = 4900 N

Step 8: Therefore, the buoyant force acting on the cube is 4900 N.

Question 24

Question

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

water. What is the volume of the cube that is submerged in the water?

Solution

Step 1: The density of water is 1000 kg/m3. Let’s denote the volume of the

submerged part of the cube as Vs, the volume of the cube as V, and the total

weight of the cube as W. The buoyant force Fbacting on the cube is equal to

the weight of the water displaced by the submerged part of the cube:

Fb=ρw·Vs·g

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

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

plus the weight of the submerged cube:

W=ρ·V·g=ρw·Vs·g+ρ·Vs·g

Substitute the given values into the equation:

600 ·0.13·g= 1000 ·Vs·g+ 600 ·Vs·g

60 = 1000 ·Vs+ 600 ·Vs

60 = 1600 ·Vs

Vs=60

1600 = 0.0375 m3

Step 3: Therefore, the volume of the cube that is submerged in water is

0.0375 m3.

Question 25

Question

A cube of side length 0.2 m and density 800 kg/m3is submerged in water.

Calculate 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=a3, where ais the length of a side of the cube. In this case, the side length

a= 0.2 m. Thus, the volume of the cube is:

V= 0.23m3= 0.008 m3

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

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

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

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

volume.

m= 800 ×0.008 = 6.4 kg

Now, calculate the weight of the cube:

W= 6.4×9.81 = 62.784 N

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

weight of the water displaced by the cube. Using Archimedes’ principle, the

buoyant force can be calculated as the density of water times the volume of

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

displaced is equal to the volume of the cube: 0.008 m3.

Fbuoyant =ρwaterVdispg= 1000 ×0.008 ×9.81

Fbuoyant = 78.48 N

Therefore, the buoyant force acting on the cube is 78.48 N.

Question 26

Question

A cube made of solid gold with sides of length 5 cm is placed in a container of

water. The density of gold is 19.3 g/cm3and the density of water is 1 g/cm3.

What percentage of the cube’s volume will be submerged in the water?

Solution

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

its mass times the acceleration due to gravity (g= 9.81 m/s2). Given that the

density of gold is 19.3 g/cm3, the volume of the cube is (5 cm)3= 125 cm3.

Therefore, the mass of the cube is:

Volume ×Density = 125 cm3×19.3 g/cm3= 2412.5 g

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

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

displaced is the volume of the submerged portion of the cube. Let Vdbe the

volume submerged in water. Given that the density of water is 1 g/cm3, the

mass of the water displaced is:

Densitywater ×Vd= 1 g/cm3×Vd

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

principle states that the buoyant force is equal to the weight of the water dis-

placed by the object. Therefore:

Buoyant force = Densitywater ×Vd×g

Step 4: Calculate the volume of the cube that is submerged in water. Since

the cube is a cube, the submerged volume is equal to Vd= (side length)2×

(depth submerged). Given that the side length is 5 cm, and the density of

water is 1 g/cm3, the volume submerged can be calculated by:

Vd= 5 cm ×5 cm ×(Densitygold

Densitywater

×side length)

Step 5: Calculate the percentage of the cube’s volume submerged in wa-

ter. Finally, the percentage of the cube’s volume submerged in water can be

calculated by:

Percentage submerged = Vd

Total volume ×100%

Question 27

Question

A cube of side length 20 cm and density 800 kg/m3is floating at the surface of

water. What is the depth of the cube below the surface of the water? (Density

of water = 1000 kg/m3, acceleration due to gravity = 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. Step 2: The volume of water displaced by the cube is

equal to the volume of the cube submerged in water. Step 3: Let the depth of

the cube below the surface of the water be d. The volume submerged in water

can be expressed as (20 cm)2×d= 400 cm2×d. Step 4: The weight of the water

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

is given by mg, where mis the mass of the cube and gis the acceleration due

to gravity. Step 5: The mass of the cube can be calculated using the density of

the cube and the volume submerged in water. m=ρV , where ρis the density

of the cube and Vis the volume submerged. Step 6: Equate the weight of the

water displaced to the weight of the cube to find the depth d. Solve for din

terms of the given quantities. Step 7: Substitute the values of the densities and

accelerations to find the depth of the cube below the surface of the water.

Question 28

Question

A solid object with a volume of 0.03 m3and a density of 800 kg/m3is submerged

in water. Determine the buoyant force acting on the object.

Solution

Step 1: Calculate the weight of the object. Given that the density of the object

is 800 kg/m3and the volume is 0.03 m3, we can calculate the weight using the

formula:

weight = mass ×acceleration due to gravity

weight = density ×volume ×acceleration due to gravity

weight = 800 kg/m3×0.03 m3×9.81 m/s2

weight = 235.44 N

Step 2: Calculate the buoyant force. The buoyant force acting on the ob-

ject is equal to the weight of the water displaced by the object. According to

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

displaced. The density of water is 1000 kg/m3.

buoyant force = density of water×volume of object submerged×acceleration due to gravity

Now, we determine the volume of the object submerged in water. Since the

object is completely submerged, the volume of water displaced is equal to the

volume of the object.

volume of object submerged = 0.03 m3

buoyant force = 1000 kg/m3×0.03 m3×9.81 m/s2

buoyant force = 294.3 N

Therefore, the buoyant force acting on the object is 294.3 N.

Question 29

Question

A spherical balloon with a radius of 5 meters is filled with helium gas at a

density of 0.18 kg/m3. The balloon itself has a mass of 15 kg. Calculate the

maximum mass of the load that the balloon can carry to just barely float in the

air. Take the density of air to be 1.29 kg/m3.

Solution

Step 1: Find the volume of the balloon using the formula V=4

3πr3where ris

the radius of the balloon.

Given radius, r= 5 m

V=4

3π(5 m)3

V=4

3π125

V=500

3πm3

Step 2: Find the weight of the displaced air (buoyant force) using the formula

Fbuoyant = densityair ×g×Vwhere gis the acceleration due to gravity and V

is the volume of the balloon.

Given density of air, densityair = 1.29 kg/m3

Given volume of balloon, V=500

3πm3

Fbuoyant = 1.29 ×9.8×500

3π

Fbuoyant = 1.29 ×9.8×500

3π

Step 3: Find the weight of the helium gas in the balloon using the formula

Whelium = densityhelium ×g×Vwhere densityhelium is the density of helium gas

and Vis the volume of the balloon.

Given density of helium, densityhelium = 0.18 kg/m3

Given volume of balloon, V=500

3πm3

Whelium = 0.18 ×9.8×500

3π

Step 4: Set up the equilibrium condition where the weight of the balloon

and the load is equal to the sum of the weight of the helium gas and the buoyant

force.

Weight of balloon + Weight of load = Whelium+Fbuoyant+Weight of the helium

15 + Weight of load = 0.18 ×9.8×500

3π+ 1.29 ×9.8×500

3π+ 15

Step 5: Solve for the weight of the load.

Weight of load = 0.18 ×9.8×500

3π+ 1.29 ×9.8×500

3π

Therefore, the maximum mass of the load that the balloon can carry to just

barely float in the air is Weight of load.

Question 30

Question

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

The cube is completely submerged and is floating stationary. Determine the

ratio of the densities of the cube and the liquid, ρ1

ρ2.

Solution

Step 1: We start by considering the forces acting on the cube. The forces are the

weight of the cube acting downward and the buoyant force acting upward. Since

the cube is floating stationary, these two forces must be equal in magnitude.

Step 2: The weight of the cube is given by W=m·g=ρ1V g, where mis

the mass of the cube, gis the acceleration due to gravity, Vis the volume

of the cube, and ρ1is the density of the cube. Step 3: The buoyant force is

given by Fb=ρ2V g, where ρ2is the density of the liquid. Step 4: Since the

cube is floating stationary, the weight of the cube equals the buoyant force:

ρ1V g =ρ2V g. Step 5: Simplifying the equation above, we find ρ1=ρ2. Step

6: Therefore, the ratio of the densities of the cube and the liquid is ρ1

ρ2= 1 .

Question 31

Question

A spherical balloon filled with helium gas has a radius of 1 meter. The density

of helium gas is 0.178 kg/m3and the density of air is 1.25 kg/m3. Calculate the

maximum mass of payload that the balloon can carry without sinking in air.

Assume that the weight of the balloon itself is negligible.

Solution

Step 1: Calculate the buoyant force acting on the balloon. The buoyant force

is given by the formula:

Fb=ρair ·Vdisplaced ·g

where ρair is the density of air, Vdisplaced is the volume of air displaced by the

balloon, and gis the acceleration due to gravity.

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

of air displaced by the balloon is equal to the volume of the balloon itself. For

a sphere, the volume formula is:

V=4

3πr3

Substitute r= 1 m into the formula to find V.

Step 3: Substitute the given values into the formula for the buoyant force to

find Fb.

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

air is equal to the mass of the displaced air times g. The mass of the displaced

air can be found using the formula:

mdisplaced =ρair ·Vdisplaced

where ρair is the density of air and Vdisplaced is the volume of air displaced by

the balloon.

Step 5: Calculate the maximum mass of the payload. The maximum mass

of the payload the balloon can carry without sinking is equal to the weight of

the displaced air minus the weight of the balloon itself. Since the weight of the

balloon is negligible, this is equal to the weight of the displaced air.

Step 6: Substitute the calculated values for mdisplaced and ginto the formula

for the maximum mass of the payload to find the final answer.

Question 32

Question

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

water. A solid iron sphere with a radius of 0.3 m is submerged in the water

inside the container. Given that the density of iron is 7800 kg/m3and the

density of water is 1000 kg/m3, what is the buoyant force on the iron sphere?

Solution

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

of a sphere V=4

3πr3, where ris the radius of the sphere. The volume of the

iron sphere is:

V=4

3π(0.3)3≈0.1131 m3

Step 2: Calculate the mass of the iron sphere using its volume and density.

The mass of the iron sphere is:

mass = density ×volume = 7800 kg/m3×0.1131 m3≈884.13 kg

Step 3: Calculate the volume of water displaced by the iron sphere. The

displaced volume is equal to the volume of the sphere. So, the volume of water

displaced is approximately 0.1131 m3.

Step 4: Calculate the buoyant force acting on the iron sphere using Archimedes’

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

displaced. The buoyant force can be calculated as:

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

Buoyant force = 1000 kg/m3×0.1131 m3×9.81 m/s2≈1104.78 N

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

1104.78 N.

Question 33

Question

A rectangular block of wood with a density of 800 kg/m3and dimensions 0.1 m

×0.2 m ×0.3 m is floating in a container of water. What is the minimum mass

that needs to be added on top of the block of wood to make it sink completely

in the water?

Given: Density of water = 1000 kg/m3, acceleration due to gravity = 9.81

m/s2.

Solution

Step 1: First, let’s calculate the volume of the wood block: The volume of the

wood block can be calculated using the formula:

Volume of block = length ×width ×height

Volume of block = 0.1 m ×0.2 m ×0.3 m

Volume of block = 0.006 m3

Step 2: Next, we can calculate the weight of the block of wood: The weight

of the block of wood is given by:

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

Weight of block = 800 kg/m3×0.006 m3×9.81 m/s2

Weight of block = 47.88 N

Step 3: Now, in order to make the block sink completely, the added mass

must create a total weight equal to the weight of the water displaced by the

block. Therefore, we need to find the weight of the water displaced by the

block: The weight of the water displaced by the block is given by:

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

Weight of water displaced = 1000 kg/m3×0.006 m3×9.81 m/s2

Weight of water displaced = 58.86 N

Step 4: The minimum mass that needs to be added to the block is the

difference between the weight of water displaced and the weight of the block:

Minimum mass added = Weight of water displaced −Weight of block

Minimum mass added = 58.86 N −47.88 N

Minimum mass added = 10.98 N

Therefore, the minimum mass that needs to be added on top of the block of

wood to make it sink completely in the water is 10.98 N.

Question 34

Question

A cubical block of wood with a density of 0.7 g/cm3is floating in a container of

water. The block has sides of length 10 cm. If the block is pushed down under

the water until it is completely submerged, what will be the apparent weight of

the block in water? (Density of water = 1 g/cm3)

Solution

Step 1: Calculate the volume of the wooden block. The volume of a cube is

given by V=L3, where Lis the length of the side. Substituting L= 10 cm,

we find:

V= 103cm3= 1000 cm3

Step 2: Calculate the mass of the wooden block. The mass of the wooden

block is given by m=ρ·V, where ρis the density. Substituting ρ= 0.7 g/cm3

and V= 1000 cm3, we find:

m= 0.7×1000 = 700 g

Step 3: Calculate the weight of the wooden block in air. The weight in air is

given by Wair =m·g, where gis the acceleration due to gravity (approximately

9.81 m/s2). Converting grams to kilograms, we get:

Wair = 0.7 kg ×9.81 m/s2= 6.87 N

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

of the water displaced is equal to the buoyant force acting on the block. The

buoyant force is given by Fbuoyant =ρwater ·Vdisplaced ·g, where ρwater is the

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

block, so Vdisplaced = 1000 cm3. Substituting ρwater = 1 g/cm3, we get:

Fbuoyant = 1 ×1000 ×9.81 = 9810 N

Step 5: Calculate the apparent weight of the block in water. When sub-

merged in water, the buoyant force reduces the effective weight of the block in

water. The apparent weight in water is given by:

Wapparent =Wair −Fbuoyant = 6.87 N −9.81 N = −2.94 N

Therefore, the apparent weight of the block in water is 2.94 N upward.

Question 35

Question

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

a reservoir of water. Calculate the depth to which the cube is immersed in the

water.

Question 5

Question

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

in water. What is the height of the cube that is above the water surface?

Solution

Step 1: Determine the density of water Given that the cube is floating in water,

the density of water (ρw) is equal to the density of the cube, which is 0.6 g/cm3.

Step 2: Determine the volume of the cube submerged in water Let’s assume

the height of the cube submerged in the water is h, then the volume of the cube

submerged in water is given by:

Vsubmerged = (Area of base)(h) = (10 cm ×10 cm)(h) = 100hcm3

Step 3: Set up the equation using Archimedes’ principle The buoyant force

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

This can be represented by the equation:

(Buoyant force) = (Weight of water displaced)

Step 4: Calculate the buoyant force The buoyant force is given by the for-

mula:

Buoyant force = ρw·g·Vsubmerged

Substitute the values of ρw,g, and Vsubmerged into the equation.

Step 5: Calculate the weight of the cube The weight of the cube is given by

the formula:

Weight of the cube = ρc·g·V

where ρcis the density of the cube and Vis the volume of the cube.

Step 6: Equate the buoyant force to the weight of the cube Set the buoyant

force equal to the weight of the cube and solve for h, the height of the cube

above the water surface.

Step 7: Calculate the height of the cube above the water surface Substitute

the necessary values into the equation from Step 6 and solve for h.

Question 6

Question

A cube of side length aand density ρcube is completely submerged in a liquid

of density ρliquid. The cube is initially at rest, but when released, it accelerates

upwards with a constant velocity. Find the ratio of the density of the cube to

the density of the liquid.

Solution

Step 1: The force due to buoyancy acting on the cube is equal to the weight

of the liquid displaced by the cube. Let the volume of the cube be Vcube =a3,

then the weight of the cube is mg =ρcubeVcubegand the weight of the liquid

displaced is ρliquidVcubeg. So, equating these forces:

ρcubeVcubeg=ρliquidVcubeg

ρcube =ρliquid

Step 2: Since the cube is accelerating upwards with a constant velocity, the

net force acting on the cube is zero. The forces acting on the cube are its weight

W=ρcubeVcubegdownwards and the buoyant force B=ρliquidVcubegupwards.

Therefore, W−B= 0:

ρcubeVcubeg−ρliquidVcubeg= 0

⇒ρcube −ρliquid = 0

⇒ρcube =ρliquid

Step 3: The ratio of the density of the cube to the density of the liquid is:

Density ratio = ρcube

ρliquid

=ρliquid

ρliquid

= 1

Question 7

Question

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

density of helium is 0.178 kg/m3and the density of air is 1.29 kg/m3. What is

the maximum mass that the balloon can lift off the ground?

Solution

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

the formula V=4

3πr3, where ris the radius of the sphere. Substitute r= 2 m

into the formula:

V=4

3π(2 m)3

V=32

3πm3

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

force is given by the formula Fb=ρfluidV g, where ρfluid is the density of the

fluid, Vis the volume of the object submerged, and gis the acceleration due to

gravity. Substitute ρfluid = 1.29 kg/m3,V=4

3π(2 m)3, and g= 9.81 m/s2into

the formula:

Fb= 1.29 ×32

3π×9.81 N

Step 3: Calculate the gravitational force acting on the balloon. The gravita-

tional force is given by the formula Fg=mballoong, where mballoon is the mass

of the balloon and gis the acceleration due to gravity.

Step 4: Set up the equilibrium condition. For the balloon to lift off the

ground, the buoyant force must be greater than or equal to the gravitational

force. Therefore,

Fb≥Fg

1.29 ×32

3π×9.81 ≥mballoon ×9.81

Step 5: Solve for the maximum mass that the balloon can lift off the ground.

mballoon ≤1.29 ×32

3πkg

Question 8

Question

A cylindrical container with a radius of 0.5 meters and a height of 1 meter is

completely filled with water. Determine the buoyant force acting on a solid

iron sphere with a radius of 0.2 meters and a density of 7800 kg/m3that is

submerged in the water inside the container.

Solution

Step 1: First, let’s calculate the volume of the iron sphere using the formula for

the volume of a sphere: V=4

3πr3, where ris the radius of the iron sphere. The

radius of the iron sphere, r= 0.2 meters. Plugging in the values, we get:

V=4

3π(0.2)3

Step 2: Now, calculate the volume of water displaced by the iron sphere.

This volume is equal to the volume of the sphere. So, Vdisplaced =4

3π(0.2)3.

Step 3: Calculate the mass of the iron sphere using the formula m= density×

volume. The density of iron, ρiron = 7800 kg/m3. So, miron = 7800 ×4

3π(0.2)3.

Step 4: Calculate the weight of the iron sphere using the formula Wiron =

miron ×g, where gis the acceleration due to gravity (9.8m/s2).

Wiron = 7800 ×4

3π(0.2)3×9.8

Step 5: Now, let’s calculate the buoyant force acting on the iron sphere.

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

the water displaced by the iron sphere. Therefore, the buoyant force is equal to

the weight of the water displaced. So, Fbuoyant =ρwater ×Vdisplaced ×g. The

density of water, ρwater = 1000 kg/m3. Therefore, Fbuoyant = 1000 ×4

3π(0.2)3×

9.8.

Step 6: Finally, the buoyant force acting on the iron sphere is the difference

between the weight of the water displaced and the weight of the iron sphere:

Fbuoyant = 1000 ×4

3π(0.2)3×9.8−7800 ×4

3π(0.2)3×9.8

Question 9

Question

A cylinder with a cross-sectional area of 0.05 m2and a height of 2.0 m is floating

vertically in a pool of water. A mass of 100 kg is placed on top of the cylinder.

If the density of water is 1000 kg/m3, what is the height of the cylinder that

remains above the water surface?

Solution

Step 1: Determine the volume of water displaced by the cylinder. The mass

of the cylinder can be calculated using the density of water and the volume of

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

cylinder beneath the water surface.

Given: Cross-sectional area, A= 0.05m2Height of the cylinder, h= 2.0m

Density of water, ρwater = 1000kg/m3

The volume of water displaced is:

Vwater =A×h

Substitute the given values:

Vwater = 0.05 ×2.0

Vwater = 0.1m3

Step 2: Calculate the mass of the water displaced. The mass of the water

displaced can be calculated using the volume of water displaced and the density

of water.

Given: Density of water, ρwater = 1000kg/m3

The mass of the water displaced is:

mwater =ρwater ×Vwater

Substitute the values:

mwater = 1000 ×0.1

mwater = 100kg

Step 3: Determine the buoyant force on the cylinder. The buoyant force

acting on the cylinder is equal to the weight of the water displaced.

The buoyant force is given by:

Fbuoyant =mwater ×g

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

Substitute the values:

Fbuoyant = 100 ×9.81

Fbuoyant = 981N

Step 4: Calculate the weight of the added mass. The weight of the added

mass is equal to the mass multiplied by the acceleration due to gravity. Given:

Mass of the added mass, madded = 100kg

The weight of the added mass is:

Wadded =madded ×g

Wadded = 100 ×9.81

Wadded = 981N

Step 5: Determine the total downward force acting on the cylinder. The

total downward force acting on the cylinder is the sum of the weight of the

added mass and the weight of the water displaced.

Fdownward =Wadded +Fbuoyant

Fdownward = 981 + 981

Fdownward = 1962N

Step 6: Calculate the height of the cylinder above the water surface. The

height of the cylinder above the water surface, habove, can be found by equating

the total downward force to the weight of the entire cylinder.

Given: Mass of the entire cylinder, mcylinder = 100kg

The weight of the entire cylinder is:

Wcylinder =mcylinder ×g

Wcylinder = 100 ×9.81

Wcylinder = 981N

Since the cylinder is in equilibrium, the total downward force is equal to the

weight of the cylinder:

Fdownward =Wcylinder

1962 = 100 ×9.81

1962 = 981 ×habove

habove =1962

981

habove = 2.0m

Therefore, the height of the cylinder that remains above the water surface

is 2.0 meters.