PHYS 101 - ELEMENTS OF PHYSICS Buoyant forces and Archimedes' principle Question Bank Set 2

PHYS 101 - ELEMENTS OF PHYSICS

- Buoyant forces and Archimedes’

principle

Question Bank - Set 2

Liberty University

Question 1

Question

A cube of side length 20 cm and density 800 kg/m3is submerged in a container

of water. Find 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=s3, where sis the side length. Substituting s= 0.20 m, we have

V= (0.20 m)3= 0.008 m3

Step 2: Calculate 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. The mass can be calculated as M=ρV , where ρis the density of the

cube. Substituting ρ= 800 kg/m3and V= 0.008 m3, we get

M= 800 kg/m3×0.008 m3= 6.4 kg

Now, substituting M= 6.4 kg and g= 9.81 m/s2, we get

W= 6.4 kg ×9.81 m/s2= 62.784 N

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

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

fluid displaced by the object. The weight of the fluid displaced can be found

using ρwaterV g, where ρwater = 1000 kg/m3is the density of water. Substituting

ρwater = 1000 kg/m3and V= 0.008 m3, we get

Weight of water displaced = 1000 kg/m3×0.008 m3×9.81 m/s2= 78.48 N

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

Question 2

Question

A spherical balloon with a radius of 5 m is filled with helium at a density of 0.18

kg/m3. The balloon is tethered to the ground with a string. Find the tension

in the string when the balloon is released and begins to rise.

Solution

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

the weight of the fluid displaced by the object. In this case, the buoyant force

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

Step 2: The volume of the balloon can be calculated using the formula for

the volume of a sphere: V=4

3πr3, where ris the radius of the sphere.

Step 3: Substitute the given radius r= 5 m into the formula to find the

volume of the balloon.

Step 4: The weight of the air displaced by the balloon is equal to the mass

of the displaced air times the acceleration due to gravity. The mass of the

displaced air can be calculated using the density of air (ρair = 1.2 kg/m3) and

the volume of the balloon.

Step 5: The buoyant force is then given by Fb=ρair ·V·g, where gis the

acceleration due to gravity.

Step 6: The tension in the string can be found using Newton’s second law:

T−Fb=m·a, where Tis the tension in the string, Fbis the buoyant force, m

is the mass of the balloon (which is equal to the mass of the helium filling it),

and ais the acceleration of the balloon.

Step 7: Since the balloon is in equilibrium just as it is released, the acceler-

ation ais equal to zero. Thus, T=Fb, and we can substitute in the calculated

value for the buoyant force to find the tension in the string.

Question 3

Question

A cube of side length 10 cm and mass 2 kg is completely submerged in water. If

the cube is released from rest, what is the acceleration of the cube? (Assume the

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

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 of the cube is 10

cm, we convert it to meters: s= 0.1 m. So, V= (0.1 m)3= 0.001 m3.

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

W=mg, where mis the mass and gis the acceleration due to gravity. Given

that the mass of the cube is 2 kg, the weight is W= 2 kg ×9.81 m/s2= 19.62 N.

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

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

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

so the buoyant force is Fb=ρ·g·V, where ρis the density of water. Substitute

the known values: Fb= 1000 kg/m3×9.81 m/s2×0.001 m3= 9.81 N.

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

cube is the difference between the buoyant force and the weight of the cube, so

Fnet =Fb−W= 9.81 N −19.62 N = −9.81 N. The negative sign indicates that

the net force is acting in the upward direction.

Step 5: Calculate the acceleration of the cube. Using Newton’s second law,

Fnet =ma, where ais the acceleration. Substitute the known values: −9.81 N =

2 kg ×a. Therefore, the acceleration of the cube is a=−9.81 N

2 kg =−4.905 m/s2.

Question 4

Question

A cube of iron with a side length of 10 cm and a density of 7.8 g/cm3is sub-

merged in a container of water. Given that the density of water is 1 g/cm3,

calculate the buoyant force acting on the cube.

Solution

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

by V=a3, where ais the side length. In this case, a= 10 cm. Thus, V=

103= 1000 cm3.

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

by m= density ×V. For the iron cube, m= 7.8 g/cm3×1000 cm3= 7800 g.

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

given by w=m×g, where gis the acceleration due to gravity (approximately

9.81 m/s2). Converting the mass to kg, we have m= 7800 g ×0.001 kg/g = 7.8

kg. Therefore, w= 7.8 kg ×9.81 m/s2= 76.638 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.

Since the iron cube displaces water, the buoyant force is given by the weight of

the displaced water: Fb= density of water ×g×Vdisplaced.

Step 5: Calculate the volume of water displaced by the iron cube. The

volume of water displaced is equal to the volume of the iron cube, which is 1000

cm3.

Step 6: Substitute the values into the equation for buoyant force. Fb=

1 g/cm3×9.81 m/s2×1000 cm3= 9810 N.

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

Question 5

Question

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

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

Solution

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

approximately 1 g/cm3, we can use this value to calculate the buoyant force

acting on the cube.

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

calculated using the formula:

Weight = density ×volume ×acceleration due to gravity

Weight = 0.8 g/cm3×(10 cm)3×9.81 m/s2

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

can be calculated using the formula:

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

Let’s assume the depth to which the cube is immersed is dcm. Then the volume

immersed is 10 cm ×10 cm ×dcm. Thus,

Buoyant force = 1 g/cm3×10 cm ×10 cm ×dcm ×9.81 m/s2

Step 4: Set up the equilibrium condition. For the cube to float, the buoyant

force must equal the weight of the cube. Therefore, set the weight equal to the

buoyant force and solve for d.

Weight = Buoyant force

0.8 g/cm3×(10 cm)3×9.81 m/s2= 1 g/cm3×10 cm ×10 cm ×dcm ×9.81 m/s2

Question 6

Question

A cube of aluminum with sides of length 10 cm and a density of 2.7 g/cm3is

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

aluminum cube.

Solution

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

given by V=s3, where sis the length of a side. For this cube, s= 10 cm, so

the volume is:

V= 103cm3= 1000 cm3

Step 2: Convert the density of the aluminum cube to kg/m3. The density

of the aluminum is 2.7 g/cm3, which can be converted to kg/m3by multiplying

by 1000:

2.7 g/cm3= 2.7×10−3kg/cm3= 2700 kg/m3

Step 3: Calculate the mass of the aluminum cube. The mass of the aluminum

can be calculated using the formula m=ρV , where ρis the density and Vis

the volume. Substituting the known values:

m= 2700 kg/m3×0.001 m3= 2.7 kg

Step 4: Calculate the weight of the aluminum cube. The weight of an object

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

gravity (9.81 m/s2):

W= 2.7 kg ×9.81 m/s2= 26.487 N

Step 5: Calculate the buoyant force acting on the aluminum cube. According

to Archimedes’ principle, the buoyant force acting on an object submerged in a

fluid is equal to the weight of the fluid displaced by the object. Since the cube is

fully submerged, the buoyant force is equal to the weight of the water displaced

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

Vdisplaced = 1000 cm3= 0.001 m3

The density of water is 1000 kg/m3, so the weight of the water displaced is:

Wwater =ρwaterVdisplacedg= 1000 kg/m3×0.001 m3×9.81 m/s2= 9.81 N

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

Question 7

Question

A cube of wood with a density of 700 kg/m3is floating in a container filled with

water. If the cube has a side length of 10 cm, what is the depth to which the

cube is submerged in the water?

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

m/s2.

Solution

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

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

acceleration due to gravity. First, calculate the volume of the cube using the

formula: V= side length3= (0.10 m)3. Then, calculate the mass of the cube

using the density formula: m= density ×V. Finally, calculate the weight of

the cube.

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

is given by the formula: Fb=ρ·Vsub ·g, where ρis the density of water, Vsub

is the volume of the cube submerged in the water, and gis the acceleration due

to gravity. Since the cube is floating, the weight of the cube is equal to the

buoyant force. Therefore, W=Fb.

Step 3: Calculate the volume of the cube submerged in water. Let hbe

the depth to which the cube is submerged in water. The volume of the cube

submerged is equal to the area of the base of the cube multiplied by h:Vsub =

side length2×h. Substitute this expression for Vsub into the equation W=Fb.

Step 4: Solve for the depth h. From step 1, we have the value of V. From

step 2, we can calculate Fb. Substitute these values into the equation W=Fb

and solve for h.

Question 8

Question

A cylindrical object of height hand radius ris fully submerged in a liquid

of density ρliquid. The object has a density ρobject such that ρobject > ρliquid.

Determine the expression for the buoyant force acting on the object, in terms

of h,r,ρliquid, and ρobject.

Solution

Step 1: Find the volume of the object submerged in the liquid. The volume of

the object submerged in the liquid is equal to the volume of the liquid displaced

by the object. The volume of the object is given by the formula: Vobject =πr2h.

Therefore, the volume of the liquid displaced by the object is also πr2h.

Step 2: Find the mass of the liquid displaced by the object. The mass of

the liquid displaced by the object can be calculated using the formula mliquid =

Vliquid ×ρliquid, where Vliquid is the volume of the liquid displaced. Therefore,

the mass of the liquid displaced is mliquid =πr2h×ρliquid.

Step 3: Find the weight of the liquid displaced by the object. The weight

of the liquid displaced by the object is equal to the buoyant force acting on the

object. The weight of the liquid displaced can be calculated using the formula

Fbuoyant =mliquid ×g, where gis the acceleration due to gravity. Therefore,

Fbuoyant =πr2h×ρliquid ×g.

Step 4: Express the buoyant force in terms of h,r,ρliquid, and ρobject. Since

the object is fully submerged, the buoyant force acting on the object is equal to

the weight of the liquid displaced. Therefore, Fbuoyant =πr2h×ρliquid ×g.

Therefore, the expression for the buoyant force acting on the object is

πr2h×ρliquid ×g.

Question 9

Question

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

The cube is initially completely submerged in the liquid. Determine the mini-

mum value of ρcube such that the cube will float at the liquid’s surface.

Solution

Let’s denote the acceleration due to gravity as g. We can start by considering

the forces acting on the cube when it is completely submerged:

Step 1: Calculate the weight of the cube submerged in the liquid. The

weight of the cube submerged in the liquid is given by the formula:

Wcube =ρcube ·g·Vcube

where Vcube is the volume of the cube.

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

acting on the cube is given by the formula:

Fbuoyant =ρliquid ·g·Vcube

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

liquid’s surface, the weight of the cube submerged in the liquid must be equal

to the buoyant force acting on it:

ρcube ·g·Vcube =ρliquid ·g·Vcube

Step 4: Solve for the minimum value of ρcube. Cancelling out Vcube from

both sides of the equation, we get:

ρcube =ρliquid

Therefore, the minimum value of ρcube such that the cube will float at the

liquid’s surface is ρliquid.

Question 10

Question

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

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

Solution

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

ρwater = 1000 kg/m3.

Step 2: Apply Archimedes’ Principle to calculate the buoyant force acting

on the cube. The buoyant force Fbis equal to the weight of the water displaced

by the cube. Therefore,

Fb=ρwater ×Vsubmerged ×g,

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

due to gravity.

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

cube is Vcube = (0.1 m)3. Let the depth to which the cube is submerged be d.

Then the volume submerged is

Vsubmerged = (0.1 m)2×d.

Step 4: Equate the buoyant force to the weight of the cube. Since the cube

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

weight of the cube is Wcube =ρwood ×Vcube ×g, where ρwood is the density of

the wood.

Step 5: Solve for the depth dsubmerged. Set the buoyant force equal to

the weight of the cube and solve for d. This gives us

ρwater ×(0.1 m ×d)×g=ρwood ×(0.1 m)3×g.

Step 6: Calculate the depth of submersion. Solve the equation from the

previous step for dto find the depth to which the cube is submerged in water.

Question 11

Question

A rectangular block of wood with a volume of 0.02 m3and a density of 600

kg/m3is floating in a tub of water. What is the minimum mass of lead that can

be placed on top of the block of wood to submerge it completely in the water?

Solution

Step 1: First, we need to find the weight of the block of wood in order to

determine the minimum mass of lead needed to submerge it completely. The

weight of the block of wood is equal to the buoyant force acting on it, which

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

floating, the weight of the block of wood must be exactly equal to the weight of

the water displaced. Given that the density of water is 1000 kg/m3, the weight

of the water displaced by the block is:

Wwater displaced = densitywater ×volumeblock ×g

Wwater displaced = 1000 kg/m3×0.02 m3×9.8 m/s2

Wwater displaced = 196 N

Step 2: The minimum mass of lead needed to completely submerge the block

of wood is equal to the weight of the water displaced plus the weight of the block

of wood. The total weight required to completely submerge the block of wood

is:

Wtotal =Wwater displaced +Wblock

Wtotal = 196 N + (densityblock ×volumeblock ×g)

Wtotal = 196 N + 600 kg/m3×0.02 m3×9.8 m/s2

Wtotal = 196 N + 117.6 N

Wtotal = 313.6 N

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

of wood to submerge it completely in the water is:

313.6 N

9.8 m/s2= 32 kg

Question 12

Question

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

in water. Calculate the volume of the cube that is submerged in water.

Solution

Step 1: Determine the density of water. Given that the density of pure water

is 1 g/cm3, we can calculate that the density of water in kg/m3is 1000 kg/m3.

Step 2: Determine the volume of the cube submerged in water. Let the

volume of the cube submerged be V. The weight of the cube must be equal to

the buoyant force acting on it. The weight of the cube is given by:

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

Weight of cube = 0.6 g/cm3×Vcm3×9.81 m/s2

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

is given by:

Buoyant force = Density of water ×V×Acceleration due to gravity

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

Step 4: Equate the weight of the cube to the buoyant force. Setting the

weight of the cube equal to the buoyant force gives:

0.6 g/cm3×Vcm3×9.81 m/s2= 1000 kg/m3×Vm3×9.81 m/s2

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

0.6×V×10−6×9.81 = 1000 ×V

0.006 ×V= 1000 ×V

999.994 ×V= 0

V≈0 m3

Therefore, the volume of the cube that is submerged in water is approxi-

mately 0 m3.

Question 13

Question

A steel block of mass 500 kg is submerged in water. Calculate the buoyant force

acting on the block and determine if the block will sink or float. Assume that

the density of steel is 7800 kg/m3and the density of water is 1000 kg/m3.

Solution

Step 1: Calculate the volume of the steel block using the formula V=m

ρ, where

mis the mass of the steel block and ρis the density of steel. Given: m= 500 kg,

ρsteel = 7800 kg/m3

Substitute the values into the formula: V=500 kg

7800 kg/m3= 0.064 m3

Step 2: Calculate the weight of the steel block using the formula W=mg,

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

Given: m= 500 kg, g= 9.81 m/s2

Substitute the values into the formula: W= 500 kg ×9.81 m/s2= 4905 N

Step 3: Calculate the buoyant force acting on the steel block using Archimedes’

principle, which states that the buoyant force equals the weight of the fluid

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

block, which is 0.064 m3. The buoyant force can be calculated using the formula

Fbuoyant =ρwater ×Vsubmerged ×g, where ρwater is the density of water and gis

the acceleration due to gravity.

Given: ρwater = 1000 kg/m3

Substitute the values into the formula: Fbuoyant = 1000 kg/m3×0.064 m3×

9.81 m/s2= 629.76 N

Step 4: Compare the buoyant force with the weight of the steel block to

determine if the block will float or sink. The buoyant force is 629.76 N and the

weight of the block is 4905 N. Since the weight is greater than the buoyant force,

the block will sink.

Question 14

Question

A cylindrical object with a radius of 5 cm and a height of 12 cm is floating in

water with 4 cm of its height submerged. If the density of water is 1000 kg/m3,

find the density of the material from which the object is made.

Solution

Step 1: To solve this problem, we need to consider the forces acting on the

cylindrical object. The buoyant force, FB, acting upward is equal to the weight

of the water displaced by the object. The weight of the object, Fg, acts down-

ward and is equal to the weight of the object. Since the object is floating, these

forces are equal in magnitude. Step 2: The volume of the portion of the ob-

ject submerged in water can be calculated using the formula for the volume of a

cylinder: V=πr2h. Substituting the given values, we find that the volume sub-

merged, Vsub, is π(0.05 m)2(0.04 m). Step 3: The weight of the water displaced

by the submerged portion of the object is equal to the buoyant force. This

weight can be calculated using the formula FB= density of water ×g×Vsub,

where gis the acceleration due to gravity. Substituting the given values, we get

FB= 1000 kg/m3×9.8 m/s2×π(0.05 m)2(0.04 m). Step 4: The weight of the ob-

ject can be calculated using the formula Fg= density of object×g×Vtotal, where

Vtotal is the total volume of the object. Substituting the given values, we get

Fg= density of object ×9.8 m/s2×π(0.05 m)2(0.12 m). Step 5: Since the buoy-

ant force and the weight of the object are equal, we can set FB=Fgand solve

for the density of the object. This gives us the equation: 1000 kg/m3×9.8 m/s2×

π(0.05 m)2(0.04 m) = density of object ×9.8 m/s2×π(0.05 m)2(0.12 m). Step 6:

Solving the equation from Step 5 for the density of the object, we find that the

density of the material from which the object is made is 2500 kg/m3.

Question 15

Question

A cylindrical container filled with water has a density of ρ= 1000 kg/m3. A solid

sphere with a radius of 0.1 m and a density of ρs= 2000 kg/m3is submerged

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

Solution

Step 1: Firstly, calculate the volume of the sphere using the formula V=4

3πr3,

where ris the radius of the sphere.

Volume of sphere, V=4

3π(0.1)3=4

3π(0.001) = 0.004188 m3

Step 2: Next, calculate the weight of the sphere using the formula W=m·g,

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

(g= 9.8 m/s2).

Mass of sphere, m=ρs·V= 2000 ×0.004188 = 8.376 kg

W= 8.376 ×9.8 = 82.2528 N

Step 3: Determine the weight of the water displaced by the sphere. This is

equal to the weight of the water that would occupy the volume of the sphere

when submerged.

Wwater =ρ·V·g= 1000 ×0.004188 ×9.8 = 41.0232 N

Step 4: Finally, calculate the buoyant force acting on the sphere, which is

equal to the weight of the water displaced by the sphere.

Buoyant force = Wwater = 41.0232 N

Question 16

Question

A metal cube of side length 10 cm and density 8000 kg/m3is submerged in

a container filled with oil of density 900 kg/m3. Determine 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= side length3. Therefore, the volume of the cube is:

V= (0.1 m)3= 0.001 m3

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

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

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

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

m= 8000 kg/m3×0.001 m3= 8 kg

Therefore, the weight of the cube is:

W= 8 kg ×9.81 m/s2= 78.48 N

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

submerged in the oil is equal to the weight of the oil displaced by the cube. This

can be calculated using Archimedes’ principle, which states that the buoyant

force is equal to the weight of the fluid displaced. The volume of oil displaced

by the cube is equal to the volume of the cube, so the weight of the oil displaced

is given by:

Woil =ρoilV g = 900 kg/m3×0.001 m3×9.81 m/s2= 8.82 N

Therefore, the buoyant force acting on the cube is 8.82 N.

Question 17

Question

A rectangular block of wood measuring 10 cm by 6 cm by 4 cm floats in a tub

of water. Determine the density of the wood. (Density of water = 1000 kg/m3)

Solution

Step 1: Calculate the volume of the wood block. Given the dimensions of the

wood block, the volume can be calculated as:

Volume of wood block = 10 cm ×6 cm ×4 cm

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

meters, we need to convert from cubic centimeters to cubic meters.

1 cm3= 10−6m3

Step 3: Calculate the mass of the water displaced by the wood block. Using

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

displaced by the wood block.

Buoyant force = Weight of water displaced = Density of water×Volume of wood block×g

Step 4: Calculate the weight of the wood block. The weight of the wood

block is equal to its mass times the acceleration due to gravity.

Weight of wood block = Density of wood ×Volume of wood block ×g

Step 5: Set up the equilibrium condition. For the block of wood to float, the

weight of the wood block must equal the buoyant force.

Density of wood×Volume of wood block×g= Density of water×Volume of wood block×g

Step 6: Solve for the density of the wood. By canceling the volume and

acceleration due to gravity from both sides of the equation, we find:

Density of wood = Density of water

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

Question 18

Question

A cylindrical tank with a radius of 2 meters is completely filled with water. A

solid sphere with a radius of 1 meter and a density of 1000 kg/m3is submerged

in the water inside the tank. Calculate the buoyant force acting on the sphere.

Solution

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

a sphere:

Volume of sphere = 4

3πr3

Volume of sphere = 4

3π(1 m)3

Volume of sphere = 4

3πm3

Volume of sphere ≈4.19 m3

Step 2: Calculate the weight of the sphere using the formula for the weight

of an object:

Weight of sphere = mass ×acceleration due to gravity

Weight of sphere = density ×volume ×acceleration due to gravity

Weight of sphere = 1000 kg/m3×4.19 m3×9.81 m/s2

Weight of sphere ≈41129.59 N

Step 3: Calculate the buoyant force acting on the sphere using Archimedes’

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

displaced by the object, in this case, water:

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

Buoyant force = 1000 kg/m3×4.19 m3×9.81 m/s2

Buoyant force ≈41129.59 N

Therefore, the buoyant force acting on the sphere is approximately 41129.59

Question 19

Question

A cube with sides of length 10 cm and a density of 800 kg/m3is placed in a

container of water. If the cube is completely submerged in the water, what is

the buoyant force acting on it? (Note: The density of water is 1000 kg/m3and

the acceleration due to gravity is 9.81 m/s2)

Solution

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

formula:

Volume of cube = side length3

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

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

calculated using the formula:

Mass = Density ×Volume

Mass = 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 the formula:

Weight = Mass ×Acceleration due to gravity

Weight = 0.8 kg ×9.81 m/s2= 7.848 N

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

can be calculated using Archimedes’ principle which states that the buoyant

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

of water displaced by the cube is equal to the volume of the cube itself, as the

cube is completely submerged. The density of water is 1000 kg/m3, so the mass

of the water displaced is:

Mass of water displaced = 1000 kg/m3×0.001 m3= 1 kg

Therefore, the buoyant force is equal to the weight of the water displaced,

which is:

Buoyant force = Mass of water displaced ×Acceleration due to gravity

Buoyant force = 1 kg ×9.81 m/s2= 9.81 N

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

Question 20

Question

A cube of wood with a density of 0.7 g/cm3and side length of 5 cm is floating

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

water. (Density of water = 1 g/cm3)

Solution

Step 1: Determine the weight of the cube. The weight of an object can be

calculated using the formula:

Weight = Density ×Volume ×Acceleration due to gravity

The density of wood is 0.7 g/cm3, and the volume of the cube can be calcu-

lated as (side length)3. The acceleration due to gravity is 9.8 m/s2. Therefore,

the weight of the cube is:

Weight = 0.7 g/cm3×(5 cm)3×9.8 m/s2

Step 2: Determine the buoyant force acting on the cube. According to

Archimedes’ principle, the buoyant force acting on an object submerged in a

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

of water displaced by the cube is the volume of the cube that is submerged.

Let’s denote the depth to which the cube is submerged as h. The submerged

volume is the area of one face of the cube (5 cm ×5 cm) multiplied by the depth

h. Therefore, the buoyant force is:

Buoyant force = Density of water×Submerged volume×Acceleration due to gravity

Step 3: Set up the equilibrium condition. Since the cube is floating, the

weight of the cube is balanced by the buoyant force acting on it:

Weight of the cube = Buoyant force

Step 4: Solve for the depth h. Set the expressions for the weight of the cube

and the buoyant force equal to each other, and solve for the depth hto which

the cube is submerged.

Question 21

Question

A cube with a density of 1000 kg/m3and a side length of 0.1 m is submerged in

water. Calculate the buoyant force acting on the cube and determine whether

the cube will sink or float in water. The density of water is 1000 kg/m3and the

acceleration due to gravity is 9.81 m/s2.

Solution

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

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

gis the acceleration due to gravity. Given that the density of the cube is 1000

kg/m3and the side length is 0.1 m, the mass mof the cube can be calculated

as:

m= Density×Volume = 1000 kg/m3×(0.1 m)3= 1000 kg/m3×0.001 m3= 1 kg

Therefore, the weight of the cube is:

W= 1 kg ×9.81 m/s2= 9.81 N

Step 2: Calculate the buoyant force acting on the cube. According to

Archimedes’ principle, the buoyant force Fbacting on the cube 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. Therefore, the buoyant force can

be calculated as:

Fb= Density of water×Volume submerged×g= 1000 kg/m3×0.001 m3×9.81 m/s2= 9.81 N

Step 3: Compare the buoyant force and weight of the cube. Since the buoyant

force (9.81 N) is equal to the weight of the cube (9.81 N), the cube will neither

sink nor float in water. It will remain at its current position in the water.

Question 22

Question

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

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

water.

Solution

Step 1: We first need to determine the density of water, which is 1 g/cm3. Since

the cube is floating, the buoyant force acting on the cube is equal to the weight

of the water displaced by the cube.

Step 2: The volume of the cube can be calculated as Vcube = (15 cm)3=

3375 cm3. The weight of the cube can be calculated as Wcube =mcube ·g=

ρcube ·Vcube ·g, where g= 9.81 m/s2is the acceleration due to gravity.

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

buoyant force, which is B=ρwater ·Vdisp ·g, where Vdisp is the volume of water

displaced by the cube.

Step 4: Setting the weight of the cube equal to the buoyant force gives

ρcube ·Vcube ·g=ρwater ·Vdisp ·g.

Step 5: Solving for Vdisp, we find Vdisp =ρcube ·Vcube

ρwater

. Substituting the

known values into the equation, we get Vdisp =0.8 g/cm3·3375 cm3

1 g/cm3.

Step 6: Calculating the volume of water displaced gives Vdisp = 2700 cm3.

Since the cube is floating, this volume of water displaced corresponds to the

volume submerged by the cube.

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

using the formula h=Vdisp

A, where Ais the area of the cube’s face in contact

with the water. Since the cube is square, A= (side length)2.

Step 8: Substituting the values Vdisp = 2700 cm3,A= (15 cm)2= 225 cm2,

we find h=2700 cm3

225 cm2= 12 cm.

Step 9: Therefore, the cube is submerged to a depth of 12 cm in the water.

Question 23

Question

A solid weighs 8 N in air and 5 N when completely submerged in water. Calcu-

late the volume of the solid. (Assume g= 10 m/s2and the density of water is

1000 kg/m3)

Solution

Step 1: Identify the given values and unknowns. Let the volume of the solid be

represented by V. The weight of the solid in air, Wair, is 8 N and the weight of

the solid when completely submerged in water, Wwater, is 5 N. We are asked to

find the volume V.

Step 2: Understand the weight loss in water. The difference in weight be-

tween the solid in air and in water is due to the buoyant force acting on the

solid when it is submerged. The weight of the water displaced by the solid is

equal to the buoyant force acting on the solid.

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

displaced by the solid is the difference between the weight of the solid in air and

in water:

Weight of water displaced = Wair −Wwater

Step 4: Calculate the volume of the solid. Using Archimedes’ principle, the

weight of the water displaced is equal to the weight of the solid in air minus the

weight of the solid in water. This can be expressed as:

ρwater ·V·g=Wair −Wwater

Substitute the given values into the equation:

1000 ·V·10 = 8 −5

Step 5: Solve for the volume of the solid. Solving the equation for V:

10000V= 3

V=3

10000 = 0.0003 m3

Therefore, the volume of the solid is 0.0003 m3.

Question 24

Question

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

Calculate the buoyant force acting on the cube.

Solution

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

V=s3, where sis the side length. Substituting s= 0.10 m into the formula

gives:

V= (0.10 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=ρV , where ρis the density. Substituting

ρ= 800 kg/m3and V= 0.001 m3into the formula gives:

m= 800 kg/m3×0.001 m3= 0.8 kg

Step 3: Determine the weight of the cube. The weight of an object can be

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

due to gravity (g= 9.81 m/s2). Substituting m= 0.8 kg into the formula gives:

W= 0.8 kg ×9.81 m/s2= 7.848 N

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

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

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

formula Fbuoy =ρfluidVdisplacedg, where ρfluid is the density of the fluid, Vdisplaced

is the volume of the fluid displaced, and gis the acceleration due to gravity.

For an object submerged in water, ρfluid = 1000 kg/m3. The volume of water

displaced by the cube is equal to the volume of the cube, so Vdisplaced = 0.001

m3. Substituting these values into the formula gives:

Fbuoy = 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 25

Question

A spherical balloon filled with helium has a radius of 1.5 meters. The density of

helium is 0.179 kg/m3and the density of air is 1.23 kg/m3. Calculate the max-

imum mass the balloon can carry without sinking in air. Take the acceleration

due to gravity as 9.81 m/s2.

Solution

Step 1: Determine the buoyant force on the balloon. The buoyant force on the

balloon can be calculated using Archimedes’ principle:

Fbuoyant =ρfluid ·Vdisplaced ·g

where - ρfluid = 1.23 kg/m3is the density of air, - Vdisplaced is the volume of air

displaced by the balloon, and - g= 9.81 m/s2is 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, which can

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

Vballoon =4

3πr3

where r= 1.5 m is the radius of the balloon.

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

the values into the formula for the buoyant force:

Fbuoyant = 1.23 kg/m3·4

3π(1.5 m)3·9.81 m/s2

Step 4: Determine the weight of the air displaced by the balloon. The weight

of the air displaced by the balloon is equal to the mass of the air displaced times

the acceleration due to gravity:

Wair =ρfluid ·Vdisplaced ·g

Step 5: Set up the equilibrium condition. For the balloon to float in air, the

buoyant force must be equal to the weight of the air displaced:

Fbuoyant =Wair

Step 6: Solve for the mass the balloon can carry. Equating the buoyant force

and the weight of the air displaced, we have:

1.23 ·4

3π(1.5)3·9.81 = 0.179 ·M·9.81

Solving for M, the maximum mass the balloon can carry without sinking in air,

we find:

M=1.23 ·4

3π(1.5)3

0.179

Question 26

Question

A block of wood with a density of 700 kg/m3and a volume of 0.02 m3is floating

in a tub of water. The block has a 0.005 m thick layer of oil (density 800 kg/m3)

on its top surface. Determine the buoyant force acting on the block of wood.

Solution

Step 1: First, we find the total weight of the block of wood. The weight of an

object is given by the formula:

Weight = Mass ×Acceleration due to gravity

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

Mass = Density ×Volume

Mass = 700 kg/m3×0.02 m3= 14 kg

Therefore, the weight of the block is:

Weight = 14 kg ×9.8 m/s2= 137.2 N

Step 2: Next, we calculate the weight of the oil layer on top of the block.

The volume of the oil layer is given by:

Volumeoil = 0.02 m3×0.005 m = 0.0001 m3

The mass of the oil layer is:

Massoil = Densityoil ×Volumeoil = 800 kg/m3×0.0001 m3= 0.08 kg

Therefore, the weight of the oil layer is:

Weightoil = 0.08 kg ×9.8 m/s2= 0.784 N

Step 3: Finally, we can calculate the total buoyant force acting on the block

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

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

block is equal to the weight of the block (including the oil layer).

Buoyant force = Weight + Weightoil = 137.2 N + 0.784 N = 137.984 N

Therefore, the buoyant force acting on the block of wood is 137.984 N.

Question 27

Question

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

The cube is submerged to a depth of hin the liquid. Determine the expression

for the buoyant force acting on the cube in terms of a,ρcube,ρliquid, and h.

Solution

Step 1: Let’s first determine the volume of the cube submerged in the liquid.

Since the cube is submerged to a depth of h, the submerged volume of the cube

is a2h.

Step 2: Next, we can find the volume of liquid displaced by the cube. This

is equal to the volume of the cube that is submerged, a2h.

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

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

Fb=ρliquid ·g·Vdisplaced

Fb=ρliquid ·g·a2h

Step 4: The weight of the cube is given by

W=ρcube ·g·Vsubmerged

W=ρcube ·g·a2h

Step 5: Since the cube is floating, the buoyant force Fbis equal in magnitude

to the weight Wof the cube but in the opposite direction. Therefore, the

expression for the buoyant force acting on the cube is:

Fb=ρliquid ·g·a2h

Question 28

Question

A cylindrical metal object with a density of 6000 kg/m3and a weight of 200 N

is placed in water. The object is found to sink to a depth where it experiences

a buoyant force of 100 N. Calculate the volume of the metal object submerged

in water.

Solution

Step 1: Determine the buoyant force on the object. The buoyant force is equal

to the weight of the water displaced by the object. Given that the buoyant force

is 100 N, this means the weight of the water displaced is also 100 N.

Step 2: Recall that the weight of an object is given by W=m·g, where m

is the mass of the object and gis the acceleration due to gravity. Given that the

weight of the object is 200 N and g= 9.81 m/s2, we can write 200 = m·9.81.

Solving for m, we find the mass of the object to be m=200

9.81 ≈20.37 kg.

Step 3: Using the definition of density, ρ=m

V, where ρis the density, mis

the mass, and Vis the volume, we can rewrite this equation as V=m

ρ. Given

that the density of the metal object is 6000 kg/m3, we can calculate the volume

of the submerged part of the object as V=20.37

6000 ≈0.0034 m3or 3.4 liters.

Therefore, the volume of the metal object submerged in water is approxi-

mately 0.0034 m3or 3.4 liters.

Question 29

Question

A heavy piece of metal is suspended by a string and completely immersed in a

beaker of water. The tension in the string is measured to be 10 N. The beaker

is then removed and the piece of metal is placed in a beaker of oil. The tension

in the string is now measured to be 6 N. Calculate the density of the metal.

Solution

Step 1: Calculate the apparent weight of the metal in water. In water, the

buoyant force equals the weight of the water displaced by the metal. Let Wwbe

the weight of the metal in air, Wapparent be the apparent weight of the metal in

water. The tension in the string is equal to the apparent weight in water minus

the buoyant force. Therefore,

Ww=Wapparent + buoyant force

10 N = Wapparent + buoyant force

Step 2: Calculate the buoyant force in water. The buoyant force can be

calculated using Archimedes’ principle. The buoyant force in water is equal

to the weight of the water displaced by the metal. The density of water is

ρw= 1000 kg/m3. Let Vbe the volume of the metal. The buoyant force is

given by

buoyant force = ρw·g·V

Step 3: Calculate the volume of the metal. The volume Vof the metal can

be calculated using the formula V=Wapparent

ρw

·g.

Step 4: Calculate the density of the metal. The density of the metal can be

calculated using the formula ρ=Ww

Now, we can substitute the given values and solve for the density of the

metal.

Question 30

Question

A cube of copper with side lengths of 10 cm is submerged in a container filled

with water. The cube has a mass of 8 kg. Calculate the buoyant force acting on

the cube and determine whether the cube will sink or float in the water. (Density

of copper = 8.96 g/cm3, Density of water = 1000 kg/m3,g= 9.81 m/s2)

Solution

Step 1: Calculate the volume of the copper cube.

Volume of cube = (side length)3= (0.1 m)3

Volume of cube = 0.001 m3

Step 2: Calculate the mass of the water displaced by the cube.

Density of water = mass of water

volume of water displaced

mass of water = Density of water×volume of cube = 1000 kg/m3×0.001 m3= 1 kg

Step 3: Calculate the weight of the cube.

Weight of cube = mass ×g= 8 kg ×9.81 m/s2

Weight of cube = 78.48 N

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

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

Buoyant force = mass of water displaced ×g= 1 kg ×9.81 m/s2

Buoyant force = 9.81 N

Step 5: Compare the weight of the cube and the buoyant force. Since the

weight of the cube is greater than the buoyant force, the cube will sink in the

water.

Question 31

Question

A rectangular object with a density of 800 kg/m3and dimensions 2 m×3 m×4 m

is partially submerged in water. Calculate the buoyant force acting on the

object.

Solution

Step 1: First, calculate the volume of the object. Given that the object has

dimensions 2 m ×3 m ×4 m, the volume of the object is given by

Vobject = length ×width ×height = 2 m ×3 m ×4 m = 24 m3.

Step 2: Next, calculate the weight of the object. The weight of the object is

given by

Weightobject =m·g,

where mis the mass of the object and gis the acceleration due to gravity

(9.8 m/s2). Since density is mass per unit volume, we can rearrange the formula

for density to find the mass of the object:

Density = m

Vobject

⇒m= Density ×Vobject.

Substitute the values: Density = 800 kg/m3and Vobject = 24 m3.

m= 800 kg/m3×24 m3= 19200 kg.

Therefore, the weight of the object is

Weightobject = 19200 kg ×9.8 m/s2= 188160 N.

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

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

calculated using Archimedes’ principle:

Buoyant force = Weight of water displaced

= Density of water ×g×Vobject.

Given that the density of water is 1000 kg/m3, substitute the values and solve:

Buoyant force = 1000 kg/m3×9.8 m/s2×24 m3= 235200 N.

Therefore, the buoyant force acting on the object is 235200 N.

Question 32

Question

A cube of side length Land density ρ1is floating in a liquid of density ρ2with

a fraction fof its volume submerged. Determine the relationship between the

densities ρ1and ρ2, the side length L, and the fraction f.

Solution

Step 1: Start by considering the forces acting on the cube. The weight of the

cube is acting downwards and is given by W=ρ1gV , where Vis the volume of

the cube. The buoyant force acting upwards is given by FB=ρ2gV , where V

is the volume of the submerged part of the cube. Step 2: At equilibrium, the

weight of the cube is balanced by the buoyant force. Therefore, W=FB. Step

3: Substitute the expressions for Wand FBinto the equilibrium condition:

ρ1gV =ρ2gV

Step 4: Using the relationship between density ρ, volume V, and mass m(ρ=

V), we can rewrite the equilibrium condition in terms of masses:

Vg=m2

Step 5: Since m1=ρ1Vand m2=ρ2V, we can simplify the equation to:

ρ1=ρ2

Step 6: Therefore, the relationship between the densities ρ1and ρ2is that they

must be equal for the cube to float in the liquid with a fraction fof its volume

submerged.

Question 33

Question

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

water. The density of water is 1000 kg/m3. Calculate the buoyant force acting

on the object and determine if it will sink or float in water. (Hint: Assume the

object is made of aluminum with a density of 2700 kg/m3.)

Solution

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

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

the given values: V=π(0.05 m)2(0.10 m). Calculating, V= 0.000785 m3.

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

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

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

formula m=ρV , where ρis the density of the object. Substitute the values:

m= (2700 kg/m3)×0.000785 m3. Calculating, m= 2.1145 kg. Now calculate

the weight: W= (2.1145 kg) ×(9.81 m/s2). Thus, W≈20.77 N.

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

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

volume of water displaced by the cylinder is the same as the volume of the

cylinder, so Vwater = 0.000785 m3. The weight of this volume of water is given

by Wwater =ρwaterVwaterg. Substitute the values: Wwater = (1000 kg/m3)×

0.000785 m3×9.81 m/s2. Calculating, Wwater ≈7.72 N. Therefore, the buoyant

force is Fbuoyant =Wwater = 7.72 N.

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

buoyant force must be greater than or equal to the weight of the object. In this

case, the buoyant force is 7.72 N and the weight of the object is 20.77 N. Since

the weight of the object is greater than the buoyant force, the object will sink

in water.

Question 34

Question

A cubical box with a density of 800 kg/m3is floating in water. The box has a

side length of 0.5 m. Calculate the depth to which the box floats in the water.

Assume the density of water is 1000 kg/m3and the acceleration due to gravity

is 9.81 m/s2.

Solution

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

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

the weight of the fluid displaced by the object.

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

gravity is 9.81 m/s2, the weight of the water displaced by the box is equal to

the weight of the box itself. Therefore, the buoyant force on the box is equal to

the weight of the box, which is mg.

The mass of the box can be calculated as its volume multiplied by its density:

Volume of box = (side length)3= (0.5 m)3= 0.125 m3

Mass of box = Volume ×Density = 0.125 m3×800 kg/m3= 100 kg

Thus, the buoyant force is:

Fbuoyant = Weight of box = mg = 100 kg ×9.81 m/s2= 981 N

Step 2: Calculate the depth to which the box floats in the water. Let dbe

the depth to which the box floats in the water. The volume of water displaced

by the box at depth dis equal to the volume of the box itself:

Volume of water displaced = Volume of box = 0.125 m3

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

on the box:

Weight of water displaced = Buoyant force = 981 N

The weight of the water displaced can be calculated using its density and volume:

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

981 N = 1000 kg/m3×0.125 m3×9.81 m/s2

d=Weight of water displaced

Density of water ×g

Solving for d, we find:

d=981

1000 ×9.81 ≈0.1 m

Therefore, the box floats to a depth of approximately 0.1 meters in the water.

Question 35

Question

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

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

water.

Solution

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

kg/m3, we can use this value in further calculations.

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

by Vcube = (side length)3= (0.1 m)3.

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

Wcube = densitywood ×Vcube ×g, where g= 9.81 m/s2is the acceleration due to

gravity.

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

Fbuoyant is equal to the weight of the water displaced by the cube. We can

calculate this using Archimedes’ principle: Fbuoyant = densitywater ×Vsubmerged ×

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

Step 5: Set up the equilibrium condition. At equilibrium, the weight of the

cube is equal to the buoyant force acting on it: Wcube =Fbuoyant.

Step 6: Substitute the expressions for Wcube and Fbuoyant, and solve for

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

Question 2

Question

A spherical balloon with a radius of 5 m is filled with helium at a density of 0.18

kg/m3. The balloon is tethered to the ground with a string. Find the tension

in the string when the balloon is released and begins to rise.

Solution

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

the weight of the fluid displaced by the object. In this case, the buoyant force

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

Step 2: The volume of the balloon can be calculated using the formula for

the volume of a sphere: V=4

3πr3, where ris the radius of the sphere.

Step 3: Substitute the given radius r= 5 m into the formula to find the

volume of the balloon.

Step 4: The weight of the air displaced by the balloon is equal to the mass

of the displaced air times the acceleration due to gravity. The mass of the

displaced air can be calculated using the density of air (ρair = 1.2 kg/m3) and

the volume of the balloon.

Step 5: The buoyant force is then given by Fb=ρair ·V·g, where gis the

acceleration due to gravity.

Step 6: The tension in the string can be found using Newton’s second law:

T−Fb=m·a, where Tis the tension in the string, Fbis the buoyant force, m

is the mass of the balloon (which is equal to the mass of the helium filling it),

and ais the acceleration of the balloon.

Step 7: Since the balloon is in equilibrium just as it is released, the acceler-

ation ais equal to zero. Thus, T=Fb, and we can substitute in the calculated

value for the buoyant force to find the tension in the string.

Question 3

Question

A cube of side length 10 cm and mass 2 kg is completely submerged in water. If

the cube is released from rest, what is the acceleration of the cube? (Assume the

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

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 of the cube is 10

cm, we convert it to meters: s= 0.1 m. So, V= (0.1 m)3= 0.001 m3.