Upthrust in Fluids, Archimedes' Principle and Floatation

an upward force

Reason — The upward force exerted on the empty can by water opposing the push is called upthrust or buoyant force.

less than its actual weight

Reason — Due to the upthrust or the buoyant force acting on a body immersed in water, the weight of a body appears to be less than its actual weight.

(i) & (ii)

Reason — Characteristic properties of upthrust :

1. Larger the volume of a body submerged in a fluid, greater is the upthrust.
2. For the same volume of a body inside a fluid, more the density of fluid, greater is the upthrust.
3. The upthrust acts on the body in upward direction at the centre of buoyancy i.e., the centre of gravity of the displaced fluid.

F_B = $\text {Vρg}$

Reason —

Resultant upward thrust on the body

F_B = F₂ – F₂

= h₂ρgA - h₂ρgA

= A(h₂ - h₁)ρg

But, A(h₂ - h₁) = V, the volume of the body submerged in the liquid.

∴ Upthrust (F_B) = Vρg

liquids and gases

Reason — Archimedes' principle applies on both, liquids & gases.

equal to, maximum

Reason — If a body is completely immersed in a liquid, the volume of the liquid displaced will be equal to its own volume and the upthrust will be maximum.

Archimedes' principle

Reason — Archimedes' principle states that when a body is immersed partially or completely in a liquid, it experiences an upthrust, which is equal to the weight of the liquid displaced by it.

slightly more

Reason — A body weighed 'W' in air by a sensitive spring balance, will weigh slightly more in vacuum, because in vacuum no upthrust acts.

the density of water is more than the density of cork.

Reason — Bodies of density greater than that of the liquid sink in it, while bodies of average density equal to or smaller than that of the liquid float on it.

the density of nail is more than the density of water

Reason — Bodies of density greater than that of the liquid sink in it, while bodies of average density equal to or smaller than that of the liquid float on it.

greater, smaller

Reason — Bodies of density greater than that of the liquid sink in it, while bodies of average density equal to or smaller than that of the liquid float on it.

turpentine

Reason — One of the factors affecting the upthrust is the density of the liquid in which the body is submerged. Lesser the density of the liquid, lesser is the upthrust experienced by a body. As amongst the given options turpentine has the least density, hence minimum upthrust will be offered by turpentine.

ρ > ρ_L

Reason — Bodies of density greater than that of liquid, sink in it.

Hence, ρ > ρ_L

The buoyant force on a body due to a liquid acts in the upward direction at the center of buoyancy i.e. the centre of gravity of the displaced liquid.

The upward force exerted on a body by the fluid in which it is submerged, is called the upthrust.

The S.I. unit of upthrust is Newton (N).

A force is needed to keep a block of wood inside water because upthrust due to water on block when fully submerged is more than it's weight.

The upthrust in sea water (F₂) is more than the upthrust in river water (F₁). Hence, F₂ > F₁.

As we know that the sea water is denser than river water, and if the density of the liquid is more it exerts more buoyant force. Hence, the body in sea water will experience more upthrust than in river water.

Given,

• Volume = V
• Density = ρ
• Density of liquid = ρ_L
• Acceleration due to gravity = g

(i) The weight of the body, acting downwards = Vρg

(ii) The upthrust on the body, acting upwards = Vρ_Lg

(iii) The apparent weight of the body in liquid = V(ρ - ρ_L)g

(iv) The loss in weight of the body = Vρ_Lg

(a) Two balls, one of iron and the other of aluminium experience the same upthrust when dipped completely in water if both have equal volumes.

(b) An empty tin container with it's mouth closed has an average density equal to that of a liquid. The container is taken 2m below the surface of that liquid and is left there. Then the container will remain at the same position.

(c) A piece of wood is held under water. The upthrust on it will be more than the weight of the wood piece.

Let density of water be ρ_w

Since radius of both spheres are same hence their volume will be equal.

Let volume of iron sphere = volume of wood sphere = V

Upthrust on Iron sphere = Upthrust_iron = Vρ_wg

Upthrust on Wood sphere = Upthrust_wood = Vρ_wg

∵ Volume of iron sphere and volume of wood sphere is same hence, upthrust by water acting on both the spheres is same.

Comparing the upthrust on the two spheres:

$$\begin{aligned}\frac{\text{Upthrust}_{\text{iron}}}{\text{Upthrust}_{\text{wood}}} = \frac{\text V\text ρ_\text w\text g}{\text V\text ρ_\text w\text g} \\[1em] \frac{\text{Upthrust}_{\text{iron}}}{\text{Upthrust}_{\text{wood}}} = \frac{1}{1}\end{aligned}$$

∴ Upthrust_iron : Upthrust_wood = 1 : 1

(i) When a body is immersed in a liquid such that density of the body is lower than or equal to the density of the liquid then it will float, (i.e.,when ρ ≤ ρ_L)

(ii) When a body is immersed in a liquid such that density of the body is greater than the density of the liquid then it will sink, (i.e., when ρ > ρL)

Archimedes' principle states that when a body is immersed partially or completely in a liquid, it experiences an upthrust, which is equal to the weight of the liquid displaced by it.

The property of liquid to exert an upward force on a body immersed in it, is called buoyancy.

Upthrust or buoyant force arises due to the differences in pressure acting on the submerged object. The fluid exerts pressure on all sides of the object, and the pressure is greater at greater depths. The pressure difference between the top and bottom of an object creates an upward force (i.e., upthrust) that opposes the weight of the object.

Upthrust acts on the body in upward direction at the centre of buoyancy (i.e., the centre of gravity of the displaced liquid).

The upthrust has the following three characteristic properties:

1. Larger the volume of a body submerged in a fluid, greater is the upthrust.
2. For the same volume of a body inside a fluid, more the density of fluid, greater is the upthrust.
3. The upthrust acts on the body in upward direction at the centre of buoyancy i.e., the centre of gravity of the displaced fluid.

When a piece of wood is left under water, two forces act on it, (i) it's weight (i.e., the force due to gravity) W which pulls it downwards, and (ii) the upthrust F_B due to water which pushes the wood upwards. The piece of wood comes to the surface because upthrust on the wood is more than the weight of the wood (i.e.,F_B > W ).

A body weighs more in vacuum than in air when weighed with a spring balance because an upthrust acts on the body in air due to which it appears lighter than it's actual weight.

When a metal solid cylinder tied to a thread, hanging from the hook of a spring balance is gradually immersed into water contained in a jar, then it is observed that the readings on the spring balance decrease.

This loss of weight is due to the upthrust (or buoyant force) of water on the cylinder, hence the reading decreases and the cylinder appears to be lighter.

The factors on which upthrust on the body depends are —

1. Volume of the body immersed in the liquid (fluid), and
2. Density of the liquid (fluid) in which the body is submerged.

Larger the volume of a body submerged in a fluid, greater is the upthrust.

When a bunch of feathers and a stone of same mass are allowed to fall in air, the stone falls faster than the bunch of feathers.

The reason is that upthrust due to air on stone is less than that on the bunch of feathers because the volume of stone is less than that of the bunch of feathers of same mass.

If they both are released simultaneously in vacuum, they will fall together because there will be no upthrust.

When we place a small block of wood in water and another identical block of wood in glycerine, we notice that both the blocks float but the volume of block immersed in glycerine is smaller as compared to the volume of block immersed in water.

The reason is that the density of glycerine is more than that of water and so it exerts more buoyant force and hence, less part of block is immersed.

When a sphere of iron and wood are placed in water then the sphere of iron will sink.

The reason is that ρ_iron > ρ_water, so the weight of iron sphere will be more than upthrust due to water on it. But ρ_wood < ρ_water, so sphere of wood will float will it's that much volume submerged inside water by which upthrust due to water on it balances it's weight.

The bodies of density greater than that of the liquid sink in it, while bodies of average density equal to or smaller than that of the liquid float on it.

It is easier to lift a heavy stone under water than in air because due to the effect of upthrust, the stone experiences an upward buoyant force that balances the true weight of the stone which is acting in the opposite direction.

Hence, the weight of the stone immersed in water appears less than the actual weight.

The upward force exerted on a body by the fluid in which it is submerged is called upthrust or buoyant force.

Experiment —

If a piece of cork is placed on a surface of water in a tub, it floats with nearly $\frac{2}{5}$ of it's volume inside the water. If the cork is pushed into the water and then released, it again comes to the surface of water and floats.

If the cork is kept immersed, our fingers experience some upward force. This upward force is the buoyant force.

The weight of body immersed in a liquid appears to be less than it's actual weight due to the effect of upthrust.

Experiment

Lifting of a bucket full of water from a well —

Take an empty bucket and tie a long rope to it. If the bucket is immersed in the water of a well keeping one end of rope in hand and the bucket is pulled when it is deep inside water, we notice that it is easy to pull the bucket as long as it is inside water.

But as soon it starts coming out of the water surface, it appears to become heavier and now more force is needed to lift it.

This experiment shows that the weight of body appears lighter when it is immersed in water than it's actual weight (in air).

Below diagram shows the direction of forces acting on a body completely immersed inside a liquid:

Whether the body will float or sink will depend on the relative values of F₁ and F₂.

(i) If F₁ ≤ F₂ then the body will float.

(ii) If F₁ > F₂ then the body will sink.

Consider a cylindrical body PQRS of cross-sectional area A immersed in a liquid of density ρ as shown in the figure. Let the upper surface PQ of body be at a depth h₁ while it's lower surface RS be at a depth h₂ below the free surface of liquid.

At depth h₁, the pressure on the upper surface PQ

P₁ = h₁ ρ g

∴ Downward thrust on the upper surface PQ

F₁ = pressure x area = h₁ ρ g A    [Equation 1]

At depth h₂, the pressure on the lower surface RS

P₂ = h₂ ρ g

∴ Upward thrust on the lower surface RS

F₂ = pressure x area = h₂ ρ g A    [Equation 2]

The horizontal thrust at various points on the vertical sides of the body get balanced because liquid pressure is same at all points at the same depth.

From above equations (1) and (2), it is clear that F₂ > F₁ as h₂ > h₁ and hence the body will experience a net upward force.

Resultant upward thrust on the body

F_B = F₂ – F₂

= h₂ρgA - h₂ρgA

= A(h₂ - h₁)ρg

But, A(h₂ - h₁) = V, the volume of the body submerged in the liquid.

∴ Upthrust F_B = Vρg

Vρg = Volume of solid immersed x density of liquid x acceleration due to gravity

Since a solid when immersed in a liquid, displaces liquid equal to the volume of its submerged part, therefore

Vρg = Volume of liquid displaced x density of liquid x acceleration due to gravity

= mass of liquid displaced x acceleration due to gravity

= weight of the liquid displaced by the submerged part of the body.

Hence,

Upthrust = weight of the liquid displaced by the submerged part of the body.

Now in another experiment, we suspend a solid by a thin thread attached to the hook of a spring balance to check its weight. Once the weight is noted, we fill a eureka can with water till its spout and place a measuring cylinder below the spout of the eureka can. The solid is then immersed in water and the displaced water is collected in the cylinder.

When the water stops dripping, the weight of the solid and the volume of water in the cylinder are noted.

In the above figure, the solid weighs 300 gf in air and 200 gf when it is completely immersed in water. The volume of water collected in the measuring cylinder is 100 ml i.e., 100 cm³

∴ Loss in weight = 300 gf - 200 gf = 100 gf   [Equation 1]

Volume of water displaced = Volume of solid = 100 cm³

∵ Density of water = 1 g cm^-3

∴ Weight of water displaced = 100 gf    [Equation 2]

From equations 1 and 2,

Weight of water displaced = Upthrust or loss in weight.

Thus, the weight of water displaced by a solid is equal to the loss in weight of solid.

Experiment —

Take a solid (say a metallic piece). Suspend a solid by a thin thread from the hook of a spring balance. Note it's weight. Now take a eureka can and fill it with water to it's spout. Arrange a measuring cylinder below the spout of the eureka can.

Now, immerse the solid gently into water of the eureka can. The water displaced by it gets collected in the measuring cylinder, as shown in figure. When water stops dripping through the spout, note the weight of the solid and the volume of water collected in the measuring cylinder.

As shown in figure, the solid weighs 300 gf in air and 200 gf when it is completely immersed in water. The volume of water collected in the ,measuring cylinder is 100 ml i.e., 100 cm³.

Therefore, loss in weight = 300 gf - 200 gf = 100 gf     [Equation 1]

Volume of water displaced = Volume of solid = 100 cm³

Since, density of water = 1 g cm^-3

Therefore, weight of water displaced = 100 gf    [Equation 2]

From eqns. (i) and (ii)

Weight of water displaced = Upthrust or loss in weight.

Thus, the weight of water displaced by a solid is equal to the loss in weight of the solid. This verifies Archimedes' principle.

(i) As we know,

Upthrust due to liquid = volume of the solid x density of fluid x acceleration due to gravity

Given,

• Volume of the body = 100 cm³
• Weight of the body in air = 5 kgf
• Density of the liquid = 1.8 x 10³ kg m^-3

Converting cm³ into m³

100 cm = 1 m

And, 100 cm x 100 cm x 100 cm = 1 m³

Therefore, 100 cm³ = $\frac{1 \times 100}{10,00,000}$

Hence, V = 10^-4 m³

Substituting the values in the formula above we get,

Upthrust = 10^-4 x 1.8 x 10³ x g = 0.18 kgf

Hence, the upthrust due to liquid = 0.18 kgf

(ii) weight of the body in liquid = weight of the body in air – upthrust

Substituting the values in the formula above, we get,

weight of the body in liquid = 5 kgf - 0.18 kgf = 4.82 kgf

Hence, weight of the body in liquid = 4.82 kgf

(i) Volume of the body = Density of water x Loss in weight

Given,

• Weight in air = 450 kgf
• Weight in water = 310 kgf
• Loss in weight = 450 – 310 = 140 gf
• Density of water = 1 g cm^-3

Substituting the values in the formula above, we get,

Volume of the body = 1 x 140 = 140 cm³

(ii) Loss in weight = 140 gf

(iii) Upthrust = loss in weight = 140 gf

Assumption in part (i) — density of water = 1.0 g cm^-3

(a) Upthrust due to liquid = volume of the solid x density of fluid x acceleration due to gravity   [Equation 1]

Volume = $\frac{\text {Mass}}{\text {density}}$   [Equation 2]

Given,

• Mass of ball A = 12 g
• Mass of ball B = 12 g
• Volume of the hollow iron ball A = 15 cm³
• Density of iron = 8 g cm³

Substituting the values in equation 2, to find volume of solid iron ball B,

Volume of solid iron ball B = $\frac{12}{8}$ = 1.5 cm³

Substituting the values in equation 1 to get upthrust on hollow ball A

Upthrust on hollow ball A = 15 x 1 x g = 15 gf

Hence, upthrust on hollow ball A = 15 gf

Substituting the values in equation 1 to get upthrust on solid iron ball B

Upthrust on solid iron ball B = 1.5 x 1 x g = 1.5 gf

Hence, upthrust on solid iron ball B = 1.5 gf

(b) Solid iron ball B will sink.

Ball B experiences an upthrust of 1.5 gf in water that is less than it's weight of 12 gf. Hence ball B will sink.

Ball A experiences an upthrust of 15 gf that is greater than it's weight of 12 gf. Hence it will float with its that much part submerged for which upthrust becomes equal to its weight of 12 gf.

Upthrust = Volume of the solid x Density of fluid x acceleration due to gravity    [Equation 1]

and

Volume = $\frac{\text {Mass}}{\text {density}}$   [Equation 2]

Given,

• Density of the solid = 5000 kg m^-3
• Weight of the solid = 0.5 kgf
• Density of water = 1000 kg m^-3

Substituting the values in equation 2 to get volume

Volume = $\frac{\text{0.5}}{\text{5000}}$ = 0.1 x 10^-3

Substituting the values in the formula above we get,

Upthrust = 0.1 x 10^-3 x 1000 x g = 0.1 kgf

apparent weight = true weight – upthrust

Substituting the values we get,

apparent weight = 0.5 - 0.1 = 0.4 kgf

(a)

(i) Weight of sphere = volume of sphere x density of iron x g    [Equation 1]

Upthrust = volume of water displaced x density of water x g    [Equation 2]

Given,

• Density of water = 1 g cm^-3
• Density of sphere A = 0.3 g cm^-3
• Density of sphere B = 8.9 g cm^-3
• Volume of sphere A & B = 100 cm³

Substituting the values in the formula 1, we get weight of sphere A and B,

Weight of sphere A = 100 x 0.3 x g = 30 gf

Weight of sphere B = 100 x 8.9 x g = 890 gf

(ii) Substituting the values in the formula 2, we get upthrust on A and B,

Upthrust on sphere A = 100 x 1 x g = 100 gf

Upthrust on sphere B = 100 x 1 x g = 100 gf

(b) Sphere A will float.

The upthrust on sphere A is 100 gf which is more than its weight of 30 gf hence sphere A will float with its that much part submerged for which upthrust becomes equal to its weight of 30 gf.

The upthrust on sphere B is also 100 gf but it is less than its weight of 890 gf hence sphere B will sink. In general, if the density of the body is less than the density of the liquid then it will float, and as the density of wood is lesser than the density of water hence sphere A floats.

(a) Upthrust = Volume of block x Density of water x g

Given,

• Mass of block = 13.5 kg
• Volume of block = 15 x 10^-3 m³
• Density of water = 1000 kg m^-3

Substituting the values in the formula above, we get, upthrust on block,

Upthrust on the block = 15 x 10^-3 x 1000 x g = 15 kgf

(b) The upthrust on the block when fully immersed is 15 kgf which is more than its weight of 13.5 kgf hence the block will float in water when released.

(c) The block will float with its that much part submerged for which upthrust becomes equal to its weight of 13.5 kgf. Hence while floating, upthrust on the block will be 13.5 kgf.

Given,

• Weight of brass piece in air = 175 gf
• Weight of the brass piece in water = 150 gf
• Density of water = 1.0 g cm^-3

(i) Upthrust on brass piece = volume of brass piece x density of water x g

and

Upthrust on brass piece = Loss in weight = 175 gf - 150 gf = 25 gf

∴ 25 x g = volume of brass piece x density of water x g

⇒ 25 x g = volume of brass piece x 1 x g

⇒ volume of brass piece = 25 cm³

(ii) The brass piece weighs lesser in water due to the upthrust.

The brass piece experiences an upward buoyant force that balances the true weight of the piece which is acting in the opposite direction.

Hence, the weight of the brass piece immersed in water appears less than the actual weight.

Given,

• Edges of cube = 5 cm
• Volume = 5 x 5 x 5 = 125 cm³
• Density of metal = 9.0 g cm^-3
• Density of liquid = 1.2 g cm^-3

Mass of the cube = volume of the cube x density of the cube
= 125 x 9
= 1125 g

Upthrust on cube = volume of the cube x density of the liquid x g
= 125 x 1.2 x g
= 150 gf

Tension in thread = Apparent weight of the cube in liquid
= true weight – upthrust
= 1125 - 150
= 975 kgf
= 9.75 N    [∵ g = 10 ms^-2]

Hence, tension in thread = 9.75 N

Given,

• g = 9.8 N kg^-1
• Density of water = 1000 kg m³
• Dimensions of wooden block = 50 cm x 50 cm x 50 cm = 0.5 m x 0.5 m x 0.5 m    [converting to metres]
• Volume of wooden block = 0.5 m x 0.5 m x 0.5 m = 0.125 m³

Buoyant force = volume of block x density of water x g
= 0.125 x 1000 x 9.8
= 1225 N

Given,

• Mass = 3.5 kg

Therefore, weight = 3.5 kgf

Volume of displaced water = 1000 cm³

(i) Volume of body = volume of the water displaced = 1000 cm³

Hence, the volume of body = 1000 cm³

(ii) Upthrust on body = volume of the body x density of water x g
= 1000 x 0.001 x g
= 1 kgf

(iii) The apparent weight of body in water = true weight – upthrust
= 3.5 - 1
= 2.5 kgf

decreases with an increase in temperature

Reason — Most of the substances expand on heating and contract on cooling, but their mass remains unchanged. Therefore, density of most of the substances decreases with the increase in temperature and increases with the decrease in temperature.

7800 kg/m³

Reason —

Density = $\frac{\text{mass}}{\text{volume}}$ = $\frac{7.8}{1}$ g/cm³

Converting g cm^-3 into kg m^-3,

We know, 1 g = 10^-3 kg

1 cm = 10^-2 m

1 cm³ = 10^-6 m³

Therefore,

7.8 $\frac{\text { g}}{\text { cm}^3}$

= 7.8 $\frac{10^{-3}\text { kg}}{10^{-6} \text{ m}^3}$

= 7.8 $\frac{10^{3}\text { kg}}{1 \text{ m}^3}$

= 7.8 x 1000 kg m^-3

Hence, 7.8 g/cm³ = 7800 kg/m³

4, 1000

Reason — Due to anomalous expansion of water, when cooled it first contracts in volume but below 4°C, it starts expanding and continues to do so till the temperature reaches 0°C, the point at which it freezes into ice. Thus, The density of water is maximum at 4°C and is 1000 kg/m³.

water

Reason — Relative density of a substance is also defined as the ratio of the mass of a certain volume of a substance to the mass of an equal volume of water at 4°C.

no unit

Reason — Since relative density is a pure ratio, it has no unit.

R.D. = $\frac{\text{Density of substance in g cm}^{-3}}{1.0{\text{ g cm}^{-3}}}$

Reason — In C.G.S. system, density of water at 4°C is 1 g cm^-3, so the relative density of a substance is equal to the numerical value of density of that substance.

Thus, R.D. = $\frac{\text{Density of substance in g cm}^{-3}}{1.0{\text{ g cm}^{-3}}}$

R.D. = $\frac{\text{W}_1}{\text{W}_1 - \text{W}_2}$

Reason —

$$\begin{aligned}\text {R.D.} = \frac{\text{Weight of body}}{\text{Weight of water displaced by the same body}} \\[1em] = \frac{\text{Weight of body}}{\text{Loss of weight of body in water (or upthrust)}} \\[1em] = \frac{\text{Weight of body in air }(\text W_1)}{\text{Weight of body in air} (\text W_1) - \text{Weight of body in water}(\text W_2)}\end{aligned}$$

Thus, R.D. = $\frac{\textbf{W}_1}{\textbf{W}_1 - \textbf{W}_2}$

R.D. of liquid = $\frac{\text{W}_1 - \text{W}_2}{\text{W}_1 - \text{W}_3}$

Reason —

$$\begin{aligned}\text {R.D. of liquid} = \frac{\text{Weight of liquid displaced by a body}}{\text{Weight of water displaced by the same body}} \\[1em] = \frac{\text{Weight of body in air}\ (\text W_1) - \text{Weight of body in liquid}\ (\text W_2)}{\text{Weight of body in air}\ (\text W_1) - \text{Weight of body in water}\ (\text W_3)}\end{aligned}$$

Thus, R.D. of liquid = $\frac{\textbf{W}_1 - \textbf{W}_2}{\textbf{W}_1 - \textbf{W}_3}$

The density of a substance is it's mass per unit volume. i.e.,

$\text{Density} =\frac{\text {Mass of the substance}}{\text {Volume of the substance}}$

It is a scalar quantity and is represented by the letter ρ (rho) or d.

(i) In C.G.S system, unit of density is g cm^-3.

(ii) In S.I. system, unit of density is kg m^-3.

Relation between S.I. and C.G.S. units —

1 kg m^-3 = $\frac{1 \text {kg}}{1 \text {m}^{3}}$ = $\frac{1000 \text{g}}{(100 \text {cm})^{3}}$ = $\frac{{1}}{1000}$ g cm^-3

Thus,

1 kg m^-3 = 10^-3 g cm^-3 or

1 g cm^-3 = 1000 kg m^-3

The statement tells that the mass of 1 m³ of iron is 7800 kg.

The density of water at 4°C in S.I. unit is 1000 kg m^-3.

The parameters are affected in the following ways —

(i) Mass — It remains unchanged with increase in temperature.

(ii) Volume — It increases with an increase in the temperature.

(iii) Density — Most of the substances expand on heating and contract on cooling, but their mass remains unchanged. Therefore, density of most of the substances decreases with the increase in temperature.

(i) Mass = volume x density

(ii) S.I. unit of density is kg m^-3

(iii) Density of water is 1000 kg m^-3

(iv) Density in kg m^-3 = 1000 x density in g cm^-3

Relative density has no unit.

Given,

Weight of body in air = W

Weight of body in water = W₁

(i) Let V be the volume of the body.

Upthrust = loss in weight when immersed in water = (W - W₁) gf    ...[Eq 1]

Weight of water displaced = Volume of water displaced x density of water x g

Volume of water displaced = Volume of the body = V

Density of water = 1 g cm^-3

∴ Weight of water displaced = V x 1 x g = V gf     ...[Eq 2]

But weight of water displaced is equal to upthrust

∴ From eqns 1 & 2,

V = (W - W₁) cm³

∴ Volume of the body = V = (W - W₁) cm³

(ii) Upthrust = loss in weight when immersed in water = (W - W₁) gf

(iii) Relative density of the material of the body

= $\frac{\text {Weight of solid in air}}{\text {Weight in air - Weight in water}}$

= $\frac{\text W_1}{\text W_1 - \text W_2}$

On heating from 0°C, the density of water increases up to 4°C and then decreases beyond 4°C.

Relative density of a substance is defined as the ratio of the mass of a certain volume of a substance to the mass of an equal volume of water at 4°C.

Relative density is also defined as a ratio of mass of a certain volume of a substance to the mass of an equal volume of water at 4°C.

Relative density of a solid denser than water and insoluble in it

Procedure —

(i) Suspend a piece of the given solid with a thread from hook of the left pan of a physical balance and find it's weight W₁.

(ii) Now balance a wooden bridge over the left pan of balance and place a beaker nearly two-third filled with water on the bridge. Take care that the bridge and beaker do not touch the pan of balance,

(iii) Immerse the solid completely in water such that it does not touch the walls and bottom of beaker and find the weight W₂ of solid in water.

Observation —

Weight of solid in air = W₁ gf

Weight of solid in water= W₂ gf

Calculation —

Loss in weight of solid when immersed in water = (W₁ - W₂) gf

$\text{R.D.} = \frac{\text{ Weight of solid in air}}{\text{Loss in weight of solid in water}}$

or

$\text{R.D.} = \frac{\text W_{1}}{\text W_1 - \text W_2}$

Relative density of a solid denser than water and soluble in it —

Procedure —

If solid is soluble in water, instead of water, we take a liquid of known relative density in which solid in insoluble and it sinks in that liquid. Then the process described above is repeated. Now

$\text{R.D.} = \frac{\text{Weight of solid in air}}{\text{Loss in weight of solid in liquid}} \times \text{R.D. of liquid}$

By definition, relative density of a liquid is given as —

$\text{R.D.} = \frac{\text{Weight of given volume of liquid}}{\text{Weight of same volume of water}}$

By Archimedes' Principle if a solid is immersed in a liquid or water, it displaces the liquid or water equal to it's volume. Therefore, above equation takes the form —

$$\begin{aligned}\text{R.D.} = \frac{\text{Weight of liq displaced by body}}{\text{Weight of water displaced by body}} \\[1em] = \frac{(\text{Weight of body})_\text{air}-(\text{Weight of body})_\text{liquid}}{(\text{Weight of body})_\text{air}-(\text{Weight of body})_\text{water}}\end{aligned}$$

Thus, to find the relative density of a liquid using Archimedes' Principle, we take a body which is heavier than both the given liquid and water and also insoluble in both. The body is first weighed in air, then in liquid and then after washing it with water and drying, it is weighed in water. If the weight of the body in air is W₁ gf, in liquid is W₂ gf and in water is W₃ gf, then from above we get,

$\text{R.D. of liquid} = \frac{\text W_1 - \text W_2}{\text W_1 - \text W_3}$

Given,

• Weight of the body in air = W₁ gf
• Weight of the body in liquid = W₂ gf
• Weight of the body in water = W₃ gf

(i) Let V be the volume of the body.

Upthrust due to water = loss in weight when immersed in water = (W₁ - W₃) gf    ...[Eq 1]

Weight of water displaced = Volume of water displaced x density of water x g

Volume of water displaced = Volume of the body = V

Density of water = 1 g cm^-3

∴ Weight of water displaced = V x 1 x g = V gf     ...[Eq 2]

But weight of water displaced is equal to upthrust due to water

∴ From eqns 1 & 2,

V = (W₁ - W₃) cm³

∴ Volume of the body = V = (W₁ - W₃) cm³

(ii) Upthrust due to liquid = loss in weight when immersed in liquid = (W₁ - W₂) gf

(iii) Relative density of the solid

$$\begin{aligned}\text{R.D. of solid} = \frac{(\text{Weight of body})_\text{air}}{(\text{Weight of body})_\text{air}-(\text{Weight of body})_\text{water}} \\[1em] = \frac{\text W_1}{\text W_1 - \text W_3}\end{aligned}$$

(iv) Relative density of the liquid

$$\begin{aligned}\text{R.D. of liquid} = \frac{(\text{Weight of body})_\text{air}-(\text{Weight of body})_\text{liquid}}{(\text{Weight of body})_\text{air}-(\text{Weight of body})_\text{water}} \\[0.5em] = \frac{\text W_1 - \text W_2}{\text W_1 - \text W_3}\end{aligned}$$

Given,

• Density of copper = 8.83 g cm^-3

Converting g cm^-3 into kg m^-3,

We know, 1 g = 10^-3 kg

1 cm = 10^-2 m

1 cm³ = 10^-6 m³

Therefore,

8.83 $\frac{\text {g}}{\text {cm}^3}$

= 8.83 $\frac{10^{-3}\text {kg}}{10^{-6} \text{m}^3}$

= 8.83 $\frac{10^{3}\text {kg}}{1 \text{m}^3}$

= 8.83 x 1000 kg m^-3

Hence, 8.83 g cm^-3 = 8830 kg m^-3

Given,

• Relative density of mercury = 13.6

(i) In C.G.S. unit,

R.D. = $\frac{\text {Density of substance in g cm}^{-3}}{\text{1.0 g cm}^{-3}}$

Substituting the values we get,

$$\begin{aligned}13.6 = \frac{\text {Density of substance in g cm}^{-3}}{\text{1.0 g cm}^{-3}} \\[1em] \Rightarrow \text {Density of substance in g cm}^{-3} = 13.6 \times \text{1.0 g cm}^{-3}\\[1em] \Rightarrow \text {Density of substance in g cm}^{-3} = 13.6 \text{ g cm}^{-3}\\[1em]\end{aligned}$$

Therefore, density of mercury in C.G.S. unit = 13.6 g cm^-3

(ii) In S.I. unit,

$$\begin{aligned}13.6 = \frac{\text {Density of substance in kg m}^{-3}}{\text{1000 kg m}^{-3}} \\[1em] \Rightarrow \text {Density of substance in kg m}^{-3} = 13.6 \times \text{1000 kg m}^{-3}\\[1em] \Rightarrow \text {Density of substance in kg m}^{-3} = 13.6 \times 10^{-3} \text{ kg m}^{-3}\end{aligned}$$

Therefore, density of mercury in S.I. unit = 13.6 x 10³ kg m^-3

We know,

R.D. = $\frac{\text {Density of substance in kg m}^{-3}}{\text{1000 kg m}^{-3}}$

Given,

• Density of iron = 7.8 x 10³ kg m^-3

Substituting the values in the formula above we get,

R.D. = $\frac{7.8 \times 1000 \text { kg m}^{-3}}{\text{1000 kg m}^{-3}}$

Therefore, relative Density of iron = 7.8

We know,

R.D. = $\frac{\text {Density of substance in kg m}^{-3}}{\text{1000 kg m}^{-3}}$

Given,

• Relative density of silver = 10.8

Substituting the values in the formula above we get,

$$\begin{aligned}10.8 = \frac{\text {Density of silver}}{\text{1000 kg m}^{-3}} \\[1em] \therefore \text {Density of silver} = (10.8 \times 1000) \text { kg m}^{-3}\end{aligned}$$

As we know,

Mass = Density x Volume

Given,

• Volume = 2 m³
• R.D. = 0.52

Now,

R.D. = $\frac{\text {Density of substance in kg m}^{-3}}{\text{1000 kg m}^{-3}}$

Substituting the values in the formula for R.D. we get,

0.52 = $\frac{\text {Density of body}}{\text{1000 kg m}^{-3}}$

Therefore,

Density of body = 0.52 x 10³ kg m^-3

Substituting the values in the formula for mass, we get,

$$\begin{aligned}\text {Mass} = 0.52 \times 1000 \times 2 \\[1em] \Rightarrow \text {Mass} = 1040 \text { kg}\end{aligned}$$

Therefore, mass of body = 1040 kg

Given,

• Density of air = 1.3 kg m^-3
• Dimensions of room = 4.5 m x 3.5 m x 2.5 m

∴ Volume of room = 4.5 x 3.5 x 2.5 = 39.375 m³

Mass = density x volume
= 1.3 x 39.375
= 51.1875 kg

Therefore, mass of body = 51.1875 kg ≈ 51.19 kg

Given,

• Mass of stone = 113 g

Rise in the level of water is equivalent to the volume occupied by the stone

Rise in water level = 40 ml – 30 ml = 10 ml

∴ Volume occupied by the stone = 10 cm³

$$\begin{aligned}\text{Mass} = \text{Density} \times \text{Volume} \\[1em] \Rightarrow \text{Density} = \frac{\text{Mass}}{\text{Volume}} \\[1em] = \frac{113}{10} \\[1em] = 11.3 \text { g cm}^{-3}\end{aligned}$$

Therefore, density of stone = 11.3 g cm^-3

$$\begin{aligned}\text{R.D. of stone} = \frac{\text {Density of stone in g cm}^{-3}}{\text{1.0 g cm}^{-3}} \\[0.5em] = \frac{11.3}{1} \\[0.5em] = 11.3\end{aligned}$$

Hence, relative density of stone = 11.3

Given,

• Volume of body = 100 cm³
• Weight of the body in air (W₁) = 1 kgf = 1000 gf

Let weight of the body in water be W₂

$$\begin{aligned}\text{Density} = \frac{\text{Mass}}{\text{Volume}} \\[1em] = \frac{1000}{100} \\[1em] = 10 \text { g cm}^{-3}\end{aligned}$$

Hence,

Density of body = 10 g cm^-3

$$\begin{aligned}\text{R.D. of body} = \frac{\text {Density of body in g cm}^{-3}}{\text{1.0 g cm}^{-3}} \\[0.5em] = \frac{10}{1} \\[0.5em] = 10\end{aligned}$$

Therefore, relative Density of body = 10

$$\begin{aligned}\text{R.D. of body} = \frac{\text W_{1}}{\text W_{1} − \text W_{2}} \\[1 em] \Rightarrow 10 = \frac{1000}{1000 − \text W_{2}} \\[1 em] \Rightarrow 10 \times (1000 − \text W_{2}) = 1000 \\[1 em] \Rightarrow 10000 - 10\text W_{2} = 1000 \\[1 em] \Rightarrow 10\text W_{2} = 9000 \\[1 em] \Rightarrow \text W_{2} = \frac{9000}{10} \\[1 em] \Rightarrow \text W_{2} = 900 \text { gf}\end{aligned}$$

Hence,

(i) Weight in water = 900 gf

(ii) Relative density of body = 10

Given,

• Mass of body = 70 kg
• Volume of water displaced = 20,000 cm³

Converting cm³ to m³, we get,

100 cm = 1m

100 cm x 100 cm x 100 cm = 1 m³

Therefore, 20,000 cm³ = $\frac{1}{10,00,000}$ x 20,000 = 0.02 m³

Hence, volume of water displaced = 0.02 m³

(i) Mass of body immersed in water = Mass of the water displaced = Volume of water displaced x Density of water
= 0.02 x 1000
= 20 kg

• Weight of the body = mg = 70 x g = 70 kgf
• Weight of water displaced = mg = 20 x g = 20 kgf

Weight of body in water = weight of body in air - upthrust due to liquid
= 70 kgf - 20 kgf    [∵ upthrust is equal to weight of water displaced]
= 50 kgf

(ii) Formula for density is:

$$\begin{aligned}\text{Density} = \frac{\text{Mass}}{\text{Volume}} \\[1 em] = \frac{70}{0.02} \\[1 em] = 3500 \text { Kg m}^{-3}\end{aligned}$$

$$\begin{aligned}\text{R.D. of body} = \frac{\text {Density of body in kg m}^{-3}}{\text{1000 kg cm}^{-3}} \\[1em] = \frac{3500}{1000} \\[1em] = 3.5\end{aligned}$$

Therefore, relative density of body = 3.5

Given,

Weight of the solid in air (W₁) = 120 gf

Weight of the solid when completely immersed in water (W₂) = 105 gf

$$\begin{aligned}\text{Relative density of solid} = \big(\frac{\text W_{1}}{\text W_{1} − \text W_{2}}\big) \\[0.5em] = \big(\frac{120}{120 - 105}\big) \\[0.5em] = 8\end{aligned}$$

∴ Relative density of solid = 8

Given,

• Weight of the solid in air (W₁) = 32 gf
• Weight of the solid immersed completely in water (W₂) = 28.8 gf

$$\begin{aligned}\text{Relative density of solid} = \big(\frac{\text W_{1}}{\text W_{1} − \text W_{2}}\big) \\[1 em] = \big(\frac{32}{32 - 28.8}\big) \\[1 em] = \big(\frac{32}{3.2}\big) \\[1 em] = 10 \\[1.5em] \text{But, R.D. of solid} = \frac{\text {Density of solid in g m}^{-3}}{\text{1 g cm}^{-3}} \\[1 em] \Rightarrow 10 = \frac{\text {Density of solid in g m}^{-3}}{\text{1 g cm}^{-3}} \\[1 em] \Rightarrow \text {Density of solid} = 10 \text{ g m}^{-3} \\[1 em] \text{Density} = \frac{\text{Mass}}{\text{Volume}} \\[1em] \therefore \text{Volume} = \frac{\text{Mass}}{\text{Density}} \\[1 em] = \frac{32}{10} \\[1 em] = 3.2 \text { cm}^{3}\end{aligned}$$

Hence, volume of solid = 3.2 cm³

Let weight of solid in liquid of density 0.9 g cm^-3 be W.

$$\begin{aligned}\text{R.D. of solid} = \Big(\frac{\text W_{1}}{\text W_{1} − \text W_{2}}\Big) \times \text{R.D. of liquid} \\[1em]\end{aligned}$$

And

$$\begin{aligned}\text{R.D. of liquid} = \frac{\text {Density of liquid in g cm}^{-3}}{\text{1.0 g cm}^{-3}} \\[1 em] = \frac{0.9}{1} \\[1 em] = 0.9\end{aligned}$$

Substituting the values in the formula for relative density of solid :

$$\begin{aligned}10 = \Big(\frac{32}{32 − \text W}\Big) \times 0.9 \\[1 em] 10 = \frac{28.8}{32 − \text W} \\[1 em] 10 \times (32 - \text W) = 28.8 \\[1 em] 320 - 10\text W = 28.8 \\[1 em] 10\text W = 320 - 28.8 \\[1 em] \text W = \frac{291.2}{10} \\[1 em] \text W = 29.12 \text{ gf}\end{aligned}$$

Summarizing the answers:

(i) Volume of solid = 3.2 cm³

(ii) Relative density of solid = 10

(iii) Weight of solid in a liquid of density 0.9 g cm^-3 = 29.12 gf

Given,

• Weight of the solid in air (W₁) = 20 gf
• Weight of the solid in water (W₂) = 18 gf

$$\begin{aligned}\text{Relative density of body} = \big(\frac{\text W_{1}}{\text W_{1} − \text W_{2}}\big) \\[1em] = \big(\frac{20}{20 - 18}\big) \\[1em] = \big(\frac{20}{2}\big) \\[1em] = 10\end{aligned}$$

Given,

• Weight of the solid in air (W₁) = 1.5 kgf
• Weight of the solid in liquid (W₂) = 0.9 kgf
• Density of the liquid = 1.2 x 10³ kg m^-3

$$\begin{aligned}\text{R.D. of liquid} = \frac{\text {Density of liquid in kg m}^{-3}}{\text{1000 kg cm}^{-3}} \\[1em] = \frac{1.2 \times 10^3}{1000} \\[1em] = 1.2\end{aligned}$$

And

$$\begin{aligned}\text{R.D. of solid} = \Big(\frac{\text W_{1}}{\text W_{1} − \text W_{2}}\Big) \times \text{R.D. of liquid} \\[1em] = \Big(\frac{1.5}{1.5 - 0.9}\Big) \times 1.2 \\[1em] = \Big(\frac{1.5}{0.6}\Big) \times 1.2 \\[1em] = 3.0\end{aligned}$$

Given,

• Relative density of pure gold = 19.3
• Weight of the bangle in air (W₁) = 25.25 gf
• Weight of the bangle in water (W₂) = 23.075 gf

$$\begin{aligned}\text{Relative density of bangle} = \big(\frac{\text W_{1}}{\text W_{1} − \text W_{2}}\big) \\[1 em] = \big(\frac{25.25}{25.25 - 23.075}\big) \\[1 em] = \big(\frac{25.25}{2.175}\big) \\[1 em] = 11.6\end{aligned}$$

It is given that relative density of gold is 19.3 where as relative density of bangle is 11.6.

Hence, we can deduce that the bangle is not made of pure gold.

Given,

• Weight of the solid in air (W₁) = 44.5 gf
• Density of the iron = 8.9 x 10³ kg m^-3

$$\begin{aligned}\text{R.D. of iron} = \frac{\text {Density of iron in kg m}^{-3}}{\text{1000 kg cm}^{-3}} \\[1 em] = \frac{8.9 \times 10^3}{1000} \\[1 em] = 8.9 \\[1.5em]\end{aligned}$$

From relation,

$$\begin{aligned}\text{R.D.} = \frac{\text W_{1}}{\text W_{1} − \text W_{2}} \\[1 em] \Rightarrow 8.9 = \frac{44.5}{44.5 - \text W_{2}} \\[1 em] \Rightarrow 8.9 (44.5 - \text W_{2}) = 44.5 \\[1 em] \Rightarrow (8.9 \times 44.5) - (8.9\text W_{2} ) = 44.5 \\[1 em] \Rightarrow 396.05 - 8.9\text W_{2} = 44.5 \\[1 em] \Rightarrow \text W_{2} = \frac{396.05 - 44.5}{8.9} \\[1 em] \Rightarrow \text W_{2} = \frac{351.55}{8.9} \\[1 em] \Rightarrow \text W_{2} = 39.5 \text{ gm} \\[1 em]\end{aligned}$$

Hence, weight of iron piece when immersed in water = 39.5 gm

Given,

• Mass of the stone = 15.1 g
• Weight of the stone in liquid = 10.9 gf
• Weight of the stone in water = 9.7 gf

(a) Weight = mass x acceleration due to gravity
= 15.1 x g
= 15.1 gf

Hence, weight of the piece of stone in air (W₁) = 15.1 gf

(b) Upthrust on the stone

= loss in weight when immersed in water

= Weight in air (W₁) - Weight in water (W₃)

= 15.1 – 9.7

= 5.4 gf

Let volume of the piece of stone be V.

From the relation,

Upthrust on stone = Volume of stone x Density of water x Acceleration due to gravity

⇒ 5.4 x g = V x 1 x g
⇒ V = 5.4 cm³

∴ Volume of piece of stone = 5.4 cm³

(c) From the relation,

$$\begin{aligned}\text{Relative density of stone} = \big(\frac{\text W_{1}}{\text W_{1} − \text W_{3}}\big) \\[1 em] = \big(\frac{15.1}{15.1 - 9.7}\big) \\[1 em] = \big(\frac{15.1}{5.4}\big) \\[1 em] = 2.79 \approx 2.8\end{aligned}$$

Hence, relative density of stone = 2.8

(d) From the relation,

$\text{Relative density of liquid} = \frac{\text W_{1} - \text W_{2}}{\text W_{1} − \text W_{3}}$

where,

• W₁ is the weight of piece of stone in air,
• W₂ is the weight of piece of stone in liquid,
• W₃ is the weight of piece of stone in water

Substituting the values in the formula above we get,

$$\begin{aligned}\text{Relative density of liquid} = \frac{15.1 - 10.9}{15.1 - 9.7} \\[1em] = \frac{4.2}{5.4} \\[1em] = 0.777 \approx 0.78\end{aligned}$$

Hence, the relative density of the liquid = 0.777 ≈ 0.78

completely immersed inside the liquid

Reason — As F_B = volume of submerged part of body x density of liquid x g

Hence, F_B will be maximum when the body is completely immersed inside the liquid.

W < F^'_B

Reason — When a body weighing W floats partially above and partially below the liquid, it implies that the weight of the body is less than the maximum upthrust experienced by it when fully immersed inside the liquid.

Hence, W < F^'_B.

zero

Reason — When W = F_B i.e., the weight of the body is equal to the weight of the displaced liquid then the body will float and the apparent weight is equal to zero.

$\frac{\text v}{\text V}$ = $\frac{\text ρ_\text s}{\text ρ_\text L}$

Reason —

Given,

• Volume of body = V
• Volume of body submerged in liquid = v
• Density of body = ρ_s
• Density of liquid = ρ_L

Let weight of the body be W.

W = volume of the body x density of the body x g = V ρ_s g

Weight of liquid displaced by the body will be equal to upthrust. Let it be F_B.

F_B = volume of the liquid displaced x density of the liquid x g = v ρ_L g

From principle of floatation,

W = F_B

⇒ V ρ_s g = v ρ_L g

⇒ $\frac{\text v}{\text V}$ = $\frac{\text ρ_\text s}{\text ρ_\text L}$

Hence proved.

ρ₁ > ρ₂

Reason — Density of liquid A (ρ₁) is greater than the density of liquid B (ρ₂) as the body is partially immersed in liquid A whereas it is fully immersed in liquid B.

Plimsoll line

Reason — Plimsoll line indicates the safe limit for loading the ship in water of density 10³ kg m^-3. A ship is not allowed to load further if its plimsoll line starts touching the water level, so that when it sails in sea water of density more than 10³ kg m^-3, only the part of it below the plimsoll line remains submerged in water.

only (i)

Reason — A loaded ship is submerged more while an unloaded ship is submerged less because loaded ship has more weight and exerts more force vertically downwards.

A ship begins to be submerged more as it sails from sea water to river water. The water of a river is of low density than that of a sea. Therefore, a ship sails from sea water to river water, it sinks further. The reason is that according to the law of floatation, to balance the weight of ship, a greater volume of water is required to be displaced in water of lower density in river.

lower, stable

Reason — An unloaded ship floats with very small volume inside water. As a result, its centre of gravity is higher and its equilibrium is unstable. There is a danger that it may get blown over on its side by strong winds. Therefore, an unloaded ship is filled with sand (or stones), called ballast, at its bottom. This lowers its centre of gravity to make its equilibrium stable.

the density of sea water is more than that of river water

Reason — It is easier to swim in sea water because the density of sea water is more than that of the river water, so our weight is balanced in sea water with less part of the body submerged inside it.

filled with water, greater

Reason — If a submarine is to dive, its ballast tanks are, filled with water so that the average density of the submarine becomes greater than the density of sea water

larger, smaller

Reason — As ice bergs are lighter than water, they float on the surface of water with their larger portion inside the water surface and smaller portion above the water surface.

the density of air decreases with altitude

Reason — A balloon does not rise indefinitely because the density of air decreases with altitude. Therefore, as the balloon gradually goes up, the weight of the displaced air decreases. It keeps on rising as long as the upthrust on it exceeds its weight. When upthrust becomes equal to its weight, it stops rising further.

increases, decreases

Reason — When a fish has to rise up in water, it diffuses oxygen gas from its blood into the bladder, so its volume increases and its average density decreases

This increases the volume of water displaced by the fish and so the upthrust on the fish increases due to which it rises up.

Reason —

Given,

• Density of oil : ρ_oil
• Density of ball : ρ
• Density of water : ρ_water

And the relation : ρ_oil < ρ < ρ_water

This means :

(a) The ball is denser than oil, so it will sink in oil.

(b) The ball is less dense than water, so it will float in water.

Since oil and water are immiscible, they form two distinct layers — oil on top and water below. So, the ball will sink through the oil (because it is denser than oil) and partially float in the water (since it is less dense than water).

Therefore, it will come to rest at the oil-water interface, partially submerged in both fluids.

ρ₃ > ρ₂ > ρ₁

Reason

Given,

An object is floating in three liquids A, B, and C and the fraction of volume outside the liquid is :

• A : $$\begin{aligned}\frac{1}{9} \text { outside} \Rightarrow \frac{8}{9} \text { submerged} \\[1em]\end{aligned}$$
• B : $$\begin{aligned}\frac{2}{11} \text { outside} \Rightarrow\frac{9}{11} \text { submerged} \\[1em]\end{aligned}$$
• C : $$\begin{aligned}\frac{3}{7} \text { outside} \Rightarrow\frac{4}{7} \text { submerged} \\[1em]\end{aligned}$$

Making the denominator common for all the submerged fractions:

LCM of 9, 11, 7 = 693

• A : $$\begin{aligned}\frac{8}{9} = \frac{8 \times 77}{9 \times 77} = \frac{616}{693} \\[1em]\end{aligned}$$
• B : $$\begin{aligned}\frac{9}{11} = \frac{9 \times 63}{11 \times 63} = \frac{567}{693}\\[1em]\end{aligned}$$
• C : $$\begin{aligned}\frac{4}{7} = \frac{4 \times 99}{7 \times 99} = \frac{396}{693}\\[1em]\end{aligned}$$

For a floating object :

$$\begin{aligned}\text {Fraction submerged} =\frac {\text {Volume submerged}}{\text {Total volume}} \\[1em] =\frac{\text {Density of object}}{\text {Density of liquid}}\end{aligned}$$ ​ Since the object is the same in all three cases then submerged fraction is inversely proportional to density of liquid i.e., less submerged volume → greater liquid density and vice versa.

On comparing submerged fractions:

C : $\frac{396}{693}$ < B : $\frac{567}{693}$ < A : $\frac{616}{693}$

So, the order of density of liquids is : ρ₃ > ρ₂ > ρ₁

The principle of floatation states that the weight of a floating body is equal to the weight of the liquid displaced by it's submerged part.

The piece of the wood will float on water and while floating, apparent weight is zero. Hence, the reading on the scale of spring balance will be zero.

The body will float if ρ ≤ ρ_L and it will sink if ρ_S > ρ_L.

When a body floats in a liquid then the weight of the floating body is equal to the upthrust.

According to law of flotation, a body will float in a fluid if it has less density than the fluid. So, in order for the ship to float, its average density must be less than the density of water.

(a) Ice will have more volume submerged inside water, because density of ice is more than the density of wood and we know that the body with higher density will have more volume submerged in water.

(b) Ice will have more volume submerged inside water and hence it will experience greater upthrust than the piece of wood. This is so because the upthrust is always equal to the volume of water displaced by the submerged part of the body and ice will displace more water than the piece of wood.

When a floating piece of ice melts into water, it contracts by the volume equal to the volume of ice piece above the water surface while floating on it. Hence, the level of water does not change when the ice floating on it melts.

The three forces that keep the buoy in equilibrium are —

1. Weight of buoy acting in vertically downward direction.
2. Upthrust of water on buoy acting in vertically upward direction.
3. Tension in thread acting in vertically downward direction.

(a) When a solid iron ball of mass 500 g is dropped in mercury contained in a beaker, the ball will float because the density of iron ball is less than the density of mercury.

(b) The apparent weight of the ball is zero because while floating upthrust is equal to the weight of the body.

Iron nail floats on mercury because the density of iron is less than that of mercury but it sinks in water because the density of iron is more than the density of water.

(a) When a homogenous block floats on water with it's body partly immersed then centre of buoyancy B will lie vertically below centre of gravity, G.

(b) When a homogenous block floats on water with it's body completely immersed then centre of buoyancy B will coincide with centre of gravity, G.

The liquid C has highest density.

The upthrust on the body by each liquid is the same and it is equal to the weight of body.

But. upthrust = volume submerged x ρ_L x g.

For liquid C, since volume submerged is least, so density ρ₃ must be maximum.

Centre of buoyancy B, is the centre of gravity of the displaced liquid.

The centre of buoyancy B, and centre of gravity G, will coincide if the body is fully immersed in the liquid.

In the case when the body is partially above and partially below the surface of liquid, the centre of buoyancy B is vertically below the centre of gravity G.

When air from jar is pumped out the balloon will sink.

As air from jar is pumped out, the density of air in jar decreases, so upthrust on balloon decreases, hence the weight of balloon exceeds the upthrust on it and the balloon sinks.

(a) The block floats with some part outside water when salt is added to the water.

On adding some salt to water, the density of water increases, so upthrust on block of wood increases and hence the block rises up till the weight of salty water displaced by the submerged part of the block becomes equal to the weight of block.

(b) When the water is heated, the block sinks.

On heating the water, the density of water decreases, so upthrust on block decreases and weight of block exceeds the upthrust due to which it sinks.

Floating ice is less submerged in brine than in water because the density of brine is more than the density of water.

(i) In each case the weight of water displaced will be equal to the weight of the man.

Hence, the ratio weights of sea water and river water displaced by him will be 1:1.

(ii) The man finds it easier to swim in sea water because the density of sea water is more than that of the river water, so his weight is balanced in sea water with his less part submerged inside it.

An iron nail sinks in water because the density of iron is greater than the density of water, so the weight of the nail is more than the upthrust of water on it.

On the other hand, the ship is hollow and the empty space in it contains air which makes it volume large and average density less than that of water. Therefore, even with a small portion of ship submerged in water, the weight of water displaced by the submerged part of ship becomes equal to the total weight of ship and therefore, it floats.

The ship begins to submerge more as it sails from sea water to river water.

The water of a river is of low density than that of a sea. Therefore, when a loaded cargo ship sails from sea water to river water, it sinks further. The reason is that according to the law of floatation, to balance the weight of ship, a greater volume of water is required to be displaced in water of lower density in river.

(i) The two forces that act on the body immersed in liquid are —

1. The weight W of the body acting vertically downwards, through the centre of gravity G of the body. This force has a tendency to sink the body.

2. The upthrust F_B of the liquid acting vertically upwards, through the centre of buoyancy B i.e., the centre of gravity of the displaced liquid. This force has a tendency to make the body float.

(ii) Depending upon whether the maximum upthrust F'_B is less than, equal to or greater than the weight W, the body will either sink or float in liquid.

1. When W > F'_B i.e., the weight of the body is greater than the weight of the displaced liquid then the body will sink.
2. When W = F'_B i.e., the weight of the body is equal to the weight of the displaced liquid then the body will float just below the surface of liquid.
3. When W < F'_B i.e., the weight of the body is less than the weight of the displaced liquid the body will float partially submerged in the liquid. Only that much portion of the body gets submerged by which the weight of displaced liquid becomes equal to the weight of the body.

(iii) The net force on the body if it sinks/floats is as follows:

1. In the case when the body sinks, the net force acting on the body is the weight of the body itself.
2. In the case when the body floats, the net force acting on the body is the upthrust due to the liquid.

Below diagram shows the forces acting on a body floating in water with it's some part submerged:

The forces acting on the body are —

(i) The weight W of body acting vertically downwards, through the centre of gravity G of the body. This force has a tendency to sink the body.

(ii) The upthrust F_B of the liquid acting vertically upwards, through the centre of buoyancy B i.e., the centre of gravity of the displaced liquid. This force has a tendency to make the body float.

The weight of the water displaced by the floating body is equal to the weight of the floating body.

Given,

• Volume of body = V
• Volume of body submerged in liquid = v
• Density of body = ρ_s
• Density of liquid = ρ_L

Let weight of the body be W.

W = volume of the body x density of the body x g = V ρ_s g

Weight of liquid displaced by the body will be equal to upthrust. Let it be F_B.

F_B = volume of the liquid displaced x density of the liquid x g = vρ_Lg

From principle of floatation,

W = F_B

⇒ V ρ_s g = v ρ_L g

⇒ $\frac{\text v}{\text V}$ = $\frac{\text ρ_\text s}{\text ρ_\text L}$

Hence proved.

Icebergs being lighter than water, float on water with their major part (nearly 90%) inside water and only a small portion (say 10%) outside water. Since, the portion of iceberg inside water surface depends on the density of sea water, therefore, for the driver of ship, it becomes difficult to estimate the size of iceberg. Thus, an iceberg is very dangerous for a ship as it may collide with the ship and cause damage.

A strong salt solution is denser than fresh water (i.e. density of strong salt solution is more than that of fresh water), hence it will exert more upthrust on the egg that balances the weight of the egg. Therefore, the egg floats in strong salt solution and sinks in fresh water.

When a light gas like hydrogen (density much less than that of air) is filled in a balloon, the weight of air displaced by the inflated balloon (i.e., upthrust) becomes more than weight of the gas filled and it rises up.

However, reverse happens when the balloon is filled with a heavier gas like carbon dioxide (i.e., the weight of air displaced by the inflated balloon (i.e., upthrust) becomes less than weight of the gas filled and hence, the balloon sinks to the bottom.)

As a ship in harbour is being unloaded, it slowly rises higher in water because the weight of the ship decreases and it displaces less water and therefore, the ship rises in water till the weight of water displaced balances the weight of unloaded ship.

A balloon filled with hydrogen rises to a certain height and then stops rising further, because the density of air decreases with altitude. Therefore, as the balloon gradually goes up, the weight of the displaced air (i.e., upthrust) decreases. It keeps on rising as long as the upthrust on it exceeds it's weight. When upthrust becomes equal to its weight, it stops rising further.

The ship begins to submerge more as it sails from sea water to river water. The water of river is of low density than that of a sea. Therefore, when a loaded cargo ship sails from sea water to river water, it sinks further. The reason is that according to the law of floatation, to balance the weight of ship, a greater volume of water is required to be displaced in water of lower density i.e., river.

Let,

• V = total volume of rubber ball
• ρ = density of the rubber ball

Given,

• $\frac{1}{3}$rd of V is outside water

Therefore, volume inside water (V')

= V - $\frac{1}{3}$V

= $\frac{2}{3}$V

By the principle of flotation,

Weight of ball = Weight of water displaced by the immersed part of the ball

And

Density of water = 1000 kg m^-3

i.e., V' x ρ x g = $\frac{2}{3}$ V x 1000 x g

Therefore,

ρ = $\frac{\textbf {2000}}{\textbf 3}$ = 667 kg m^-3

Hence, density of rubber = 667 kg m^-3

(a) Given,

• Mass = 24 kg
• Volume = 0.032 m³

Upthrust = Volume of block below the water x density of liquid x acceleration due to gravity

Hence,

$$\begin{aligned}24 \text { kgf} = \text v \times 1000 \times \text g \\[1em] \text v = \frac{24}{1000} \\[1em] \Rightarrow \text v = 0.024 \text{ m}^{3}\end{aligned}$$

Hence, volume of block below the surface of water = 0.024 m³

(b) By the principle of floatation,

$\frac{\text {Volume of immersed part }}{\text{Total volume}} = \frac{\text {Density of wood}}{\text {Density of water}}$

And

Density of water = 1000 kg m^-3

Substituting the values in the formula we get,

$$\begin{aligned}\frac{0.024}{0.032} = \frac{\text {Density of wood}}{1000} \\[1em] \Rightarrow \text{Density of wood} = \frac{0.024}{0.032} \times 1000 \\[1em] = 7.5 \times 10^{2} \text{ kg m}^{-3}\end{aligned}$$

Given,

• Side of wooden cube = 10 cm

Hence,

• Volume of wooden cube = 10 cm x 10 cm x 10 cm = 1000 cm³
• Mass = 700 g

$$\begin{aligned}\text{Density} = \frac{\text{Mass}}{\text {Volume}} \\[1em] \therefore \text{Density of wooden cube} = \frac{700}{1000} \\[1em] = 0.7 \text{ g cm}^{-3}\end{aligned}$$

By the principle of floatation,

$\frac{\text {Volume of immersed part}}{\text{Total volume}} = \frac{\text {Density of wood}}{\text {Density of water}}$

• Density of water = 1 g cm^-3
• Density of wooden cube = 0.7 g cm^-3

$\therefore \frac{\text {Volume of immersed part}}{\text{Total volume}} = \frac{0.7}{1}$

Thus,

• Fraction submerged = 0.7
• Height of wooden cube = 10 cm
• Part of woden cube which is submerged = 10 x 0.7 = 7 cm

Therefore, part above water = 10 - 7 = 3 cm

Hence, 3 cm of height of wooden cube remains above water while floating.

Given,

• Density of wax = 0.95 g cm^-3
• Density of brine = 1.1 g cm^-3

By the principle of floatation,

$$\begin{aligned}\frac{\text {Volume of immersed part}}{\text{Total volume}} = \frac{\text {Density of wax}}{\text {Density of brine}} \\[1em] \therefore \frac{\text {Volume of immersed part}}{\text{Total volume}} = \frac{0.95}{1.1} \\[1em] = 0.86\end{aligned}$$

Hence, the piece of wax floats with 0.86^th part of it's volume immersed in brine surface.

Given,

• Density of ice = 0.9 g cm^-3
• Density of sea water = 1.1 g cm^-3

By the principle of floatation,

$$\begin{aligned}\frac{\text {Volume of immersed part}}{\text{Total volume}} = \frac{\text {Density of ice}}{\text {Density of sea water}} \\[1em] \therefore \frac{\text {Volume of immersed part}}{\text{Total volume}} = \frac{0.9}{1.1} \\[1 em] = \frac{9}{11} \\[1 em] = 0.818\end{aligned}$$

Hence, the volume of iceberg that will remain below the surface of water in the sea is $\frac{\textbf 9}{\textbf {11}}$th of total volume (or 0.818th) part.

Given,

• Height of the wood = 15 cm
• Height of the wood in water = 10 cm
• Height of the wood in spirit = 12 cm

As the block of wood is of uniform cross-sectional area, the height of the block is proportional to the volume

By the principle of floatation,

$$\begin{aligned}\frac{\text {Volume of immersed part}}{\text{Total volume}} = \frac{\text {Density of wood}}{\text {Density of water}} \\[1em] \therefore \frac{10}{15} = \frac{\text {Density of wood}}{1} \\[1 em] \Rightarrow \text{Density of wood} = \frac{10}{15} = 0.667 \text {g cm}^{-3}\end{aligned}$$

In the case of spirit, substituting the values in the formula above, we get,

$$\begin{aligned}\frac{\text {Volume of immersed part}}{\text{Total volume}} = \frac{\text {Density of wood}}{\text {Density of spirit}} \\[1em] \frac{12}{15} = \frac{0.667}{\text{Density of spirit}} \\[1em] \text{Density of spirit} = \frac{15}{12} \times 0.667 = 0.833 \text { g cm}^{-3}\end{aligned}$$

(a) Let volume of wooden block be V.

Given,

• Volume of block submerged in water = $\frac{2}{3}\text V$
• Density of water = 1000 kg m^-3

By the principle of floatation,

$$\begin{aligned}\frac{\text {Volume of immersed part}}{\text{Total volume}} = \frac{\text {Density of wood}}{\text {Density of water}} \\[1em] \therefore \frac{\frac{2}{3}\text V}{\text V} = \frac{\text {Density of wood}}{1000} \\[1 em] \Rightarrow \text{Density of wood} = \frac{2}{3} \times 1000 \\[1 em] = 666.6666 \text { kg m}^{-3} \\[1 em] \approx 667 \text { kg m}^{-3}\end{aligned}$$

Hence, the density of wood is approximately 667 kg m^-3.

(b) Volume of block submerged in oil = $\frac{3}{4}\text V$

By the principle of floatation,

$$\begin{aligned}\frac{\text {Volume of immersed part}}{\text{Total volume}} = \frac{\text {Density of wood}}{\text {Density of oil}} \\[1em] \therefore \frac{\frac{3}{4}\text V}{\text V} = \frac{\text {667}}{\text{Density of oil}} \\[1em] \Rightarrow \text{Density of oil} = \frac{4}{3} \times 667 \\[1 em] = 889.33 \text { kg m}^{-3} \\[1 em] \approx 889 \text { kg m}^{-3}\end{aligned}$$

Hence, the density of oil is approximately 889 kg m^-3.

Given,

• Density of ice = 0.92 g cm^-3
• Density of sea water = 1.025 g cm^-3
• Volume of iceberg above sea water = 800 cm³

Let total volume of iceberg be V cm³

• Volume of iceberg immersed in sea water = (V - 800) cm³

By the principle of floatation,

$$\begin{aligned}\frac{\text {Volume of immersed part}}{\text{Total volume}} = \frac{\text {Density of ice}}{\text {Density of water}} \\[1em] \therefore \frac{(\text V - 800)}{\text V} = \frac{0.92}{1.025} \\[0.5em] \Rightarrow 1.025\text V - 820 = 0.92\text V \\[1em] \Rightarrow 1.025\text V - 0.92\text V = 820 \\[1em] \Rightarrow 0.105\text V = 820 \\[1em] \Rightarrow \text V = \frac{820}{0.105} = 7809.52 \text{ cm}^3\end{aligned}$$

Hence, total volume of iceberg = 7809.52 cm³

Given,

• Volume of the plastic balloon = 15 m³
• Density of hydrogen = 0.09 kg m^-3
• Mass of the unfilled balloon = 7.15 kg
• Density of air = 1.3 kg m^-3

(i) As we know,

Mass = volume of balloon x density of hydrogen

Substituting the value in the formula we get,

15 x 0.09 = 1.35 kg

Hence, mass of hydrogen in the balloon = 1.35 kg

(ii) Mass of hydrogen and balloon = mass of hydrogen in the balloon + mass of unfilled balloon = 7.15 + 1.35 = 8.5 kg

Hence, mass of hydrogen and balloon = 8.5 kg

(iii) Given,

mass of equipment = x kg

Hence, total mass of hydrogen, balloon and equipment = (8.5 + x) kg

(iv) Weight of the air displaced = upthrust = Volume of the balloon x density of the balloon x g

Hence, mass of the air displaced = volume of balloon x density of air = 15 x 1.3 = 19.5 kg

(v) By law of floatation,

Mass of the air displaced = total mass of hydrogen, balloon and equipment

Substituting the values in the formula we get,

$$\begin{aligned}19.5 = 8.5 + \text x \\[1em] \text x = 19.5 - 8.5 \\[1em] \text x = 11 \text { kg}\end{aligned}$$

Hence, mass of the equipment = 11 kg

(a) From the observations :

• Object A sinks in L₂ but floats in L₁ and L₃, which means L₂ has the least density i.e., ρ_L2 < ρ_A < ρ_L1, ρ_L3.
• Object B sinks in L₁ but floats (fully submerged) in L₃, indicating that L₃ has the highest density i.e., ρ_L1 < ρ_B < ρ_L3.
• Since L₁ lies between these two cases, its density is intermediate.

Hence, the increasing order of density is ρ_L2 (kerosene) < ρ_L1 (water) < ρ_L3 (concentrated salt solution).

(b) The condition of flotation is:

W = F'_B

where,
W = weight of body acting vertically downwards
F'_B = Maximum upthrust that can act on the body

As object B floats fully submerged in L₃, hence, we can say that density of object B is equal density of liquid L₃

∴ ρ_object = ρ_L3

(c) The statement is correct because although object C floats in all three liquids, the upthrust acting on it is equal to its weight in each case, but the volume of liquid displaced is different depending on the density of the liquid.

In denser liquids, a smaller volume is displaced, while in less dense liquids, a larger volume is displaced so, while upthrust equals weight in all cases, the surrounding conditions like submerged volume differ.

(d) When object A is heated, its volume increases while its mass remains the same, leading to a decrease in its density so as a result, it will displace same amount of liquid to balance its weight but due to its larger volume it will be submerged less and hence will float higher in liquid L₁.

(e) The ship floating in fresh water displaces more volume because fresh water has lower density compared to sea water so to balance its weight, the ship must displace a larger volume of fresh water, whereas in denser sea water, a smaller volume is sufficient to provide the required upthrust.

(a) From the figure :

• Cube P is half submerged in liquid A,
• Cube Q is 3/4 submerged in liquid C, and
• Cube R is fully submerged in liquid B.

For identical objects, a greater submerged portion indicates a lower density of the liquid. Thus, liquid B (fully submerged) has the least density, liquid C (3/4 submerged) has intermediate density and liquid A (half submerged) has the highest density.

Hence, increasing order of density is ρ_B < ρ_C < ρ_A.

(b) The condition of flotation is:

W = F'_B

where,
W = weight of cube R acting vertically downwards
F'_B = Maximum upthrust that can act on cube R

As cube R floats fully submerged in liquid B, hence, we can say that density of cube R is equal density of liquid B.

∴ ρ_{cube R} = ρ_B

(c) All three cubes are identical and floating, so the upthrust acting on each cube is equal to its weight and since their weights are the same, the upthrust on P, Q, and R is equal, even though the submerged volumes differ.

Hence, upthrust on P = Q = R.

(d) We know that Cube R floats fully submerged in liquid B. As all the cubes are identical, so cube P when placed in liquid B will also float fully submerged.

Therefore, cube P will remain fully submerged in liquid B.

(e) Liquid A exerts the maximum buoyant force on a cube when it is fully immersed.

Reason : When a cube is fully immersed, the buoyant force depends on the density of the liquid and the volume displaced and since all cubes have the same volume, the liquid with the highest density will exert the maximum buoyant force, because buoyant force is directly proportional to the density of the liquid.