Floatation Class 9 Numericals

In this post, you will find a set of Numericals on Floatation for class 9. In the solutions, you may note the formulas that are used to solve these Floatation class 9 Numericals.

Floatation Class 9 Numericals (with solution)

Question 1] A body experiences an upthrust F₁ in river water and F₂ in seawater when dipped up to the same level. Which is more F₁ or F₂? Give reason.

Solution:

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

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

Question 2] A body of volume V and density ρ is kept completely immersed in a liquid of density ρ_L. If g is the acceleration due to gravity, write expressions for the following —

(i) The weight of the body,
(ii) The upthrust on the body,
(iii) The apparent weight of the body in liquid,
(iv) The loss in weight of the body.

Solution:

Given,

volume = V

density = ρ

density of liquid = ρ_L

acceleration due to gravity = g

(i) The weight of the body, acting downwards = V ρ g

(ii) The upthrust on the body, acting upwards = V ρ_L g

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

(iv) The loss in weight of the body = V ρ_L g

Question 3] A sphere of iron and another of wood of the same radius are held under water. Compare the upthrust on the two spheres. [Hint — Both have equal volume inside water.]

Solution:

Let the density of water be ρ_w

Since the radius of both spheres is the same hence their volume will be equal.
Let the volume of the iron sphere = volume of the wood sphere = V

Upthrust on Iron sphere = Upthrust_iron = Vρ_wg

Upthrust on Wood sphere = Upthrust_wood = Vρ_wg

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

Comparing the upthrust on the two spheres:

Upthrust_iron/Upthrust_wood=Vρ_wg / Vρ_wg
=>Upthrust_iron/Upthrust_wood = 1/1
Upthrust_iron : Upthrust_wood = 1:1

∴ Upthrust_iron : Upthrust_wood = 1 : 1

Question 4] A body of density ρ is immersed in a liquid of density ρ_L. State condition when the body will (i) float, (ii) sink, in liquid.

Solution:

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

Question 5] A sphere of iron and another of wood, both of the same radius are placed on the surface of water. State which of the two will sink. Give a reason for your answer.

Solution:

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

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

Question 6] A body of volume 100 cm³ weighs 5 kgf in air. It is completely immersed in a liquid of density 1.8 x 10³ kg m^-3. Find (i) the upthrust due to liquid and (ii) the weight of the body in liquid.

Solution:

(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

Converting cm³ into m³

100 cm = 1 m

So, 100 cm x 100 cm x 100 cm = 1 m³
=>10⁶ cm³ = 1 m³
=> 1 cm³ = (1/10⁶) m³

Hence, 100 cm³ = (1/10⁶)×100​ m³

Therefore, V = 10^-4 m³

Weight of the body in air = 5 kgf

Density of the liquid = 1.8 x 10³ kg m^-3

Substituting the values in the formula above we get,

Upthrust = volume x density x g =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, the weight of the body in liquid = 4.82 kgf

Question 7] A piece of brass weighs 175 gf in air and 150 gf when fully immersed in water. The density of water is 1.0 g cm^-3. (i) What is the volume of the brass piece? (ii) Why does the brass piece weigh less in water?

Solution:

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.

Question 8] A body weighs 450 gf in air and 310 gf when completely immersed in water. Find (i) the volume of the body, (ii) the loss in weight of the body, and (iii) the upthrust on the body. State the assumption made in part (i).

Solution:

(i) Volume of the body = density of water x loss in weight

density of water = 1 g cm^-3

Given,

weight in air = 450 kgf

weight in water = 310 kgf

loss in weight = 450 – 310 = 140 gf

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

Question 9] A solid of density 5000 kg m^-3 weighs 0.5 kgf in air. It is completely immersed in water of density 1000 kg m^-3. Calculate the apparent weight of the solid in water.

Solution:

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

and

Volume = Mass/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 = 0.5/5000​ = 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

Question 10] Two spheres A and B, each of volume 100 cm³ are placed on water (density = 1.0 g cm^-3). Sphere A is made of wood of density 0.3 g cm^-3 and sphere B is made of iron of density 8.9 g cm^-3. (a) Find (i) the weight of each sphere, and (ii) the upthrust on each sphere. (b) Which sphere will float? Give reason.

(a) 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 the weight of spheres 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. 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.