#### To determine

(a)

**To show:**

The percentage of volume of the object above the surface of the liquid is

100ρf-ρ0ρf%

#### Answer

The percentage of volume of the object above the surface of the liquid is

100ρf-ρ0ρf%

#### Explanation

**1) Concept:**

Archimedes’ Principle

**2) Theorem:**

Archimedes’ Principle states that the buoyant force on an object partially or fully submerged in a fluid is equal to the weight of the fluid that the object displaces, that is, F=W.

**3) Given:**

The buoyant force is

F=ρfg∫-h0Aydy

Weight of the object is

W=ρ0g∫-hL-hAydy

**4) Calculations:**

Volume above the surface = whole volume – volume under water.

Therefore, the volume above the surface is

∫0L-hAydy=∫-hL-hAydy-∫-h0Aydy

So, to find the proportion of volume above the surface, divide above equation by the whole volume.

∫0L-hAydy∫-hL-hAydy=∫-hL-hAydy-∫-h0Aydy∫-hL-hAydy⋯(*)

Now, by Archimedes’ Principle,

F=W

Therefore,

ρfg∫-h0Aydy=ρ0g∫-hL-hAydy

Dividing by ρfg,

∫-h0Aydy=ρ0ρf∫-hL-hAydy

Therefore, equation (*) becomes

∫0L-hAydy∫-hL-hAydy=∫-hL-hAydy-ρ0ρf∫-hL-hAydy∫-hL-hAydy

Factoring out ∫-hL-hAydy from the right side of the above equation,

∫0L-hAydy∫-hL-hAydy=1-ρ0ρf∫-hL-hAydy∫-hL-hAydy

After simplification,

∫0L-hAydy∫-hL-hAydy=1-ρ0ρf=ρf-ρ0ρf

To convert this to percentage multiply by 100.

Therefore, the percentage volume above the surface is

100ρf-ρ0ρf %

**Conclusion:**

The percentage of volume of the object above the surface of the liquid is

100ρf-ρ0ρf%

#### To determine

(b)

**To find:**

Percentage of the volume of an iceberg above water.

#### Answer

Percentage of the volume of an iceberg above water is 11

#### Explanation

**1) Given:**

ρf=1030 kg/m3

ρ0=917 kg/m3

**2) Calculation:**

From part (a), the percentage of volume of the object above the surface of the liquid is

100ρf-ρ0ρf%

Therefore, the percentage of volume of the iceberg above water is

1001030-9171030%

After simplifying,

1001030-9171030%≈11%

The percentage of volume of the iceberg above water is 11%.

**Conclusion:**

The percentage of volume of iceberg above water is 11%.

#### To determine

**(c)**

Whether the water overflows when the ice melts.

#### Answer

The water does not overflow when the ice melts.

#### Explanation

Let Vi be the volume of ice cube and Vw be the volume of water, which results from melting.

Then by the formula derived in part (a),

The volume of ice cube above the surface of water is

ρf-ρ0ρfVi,

So the volume below the surface is

Vi-ρf-ρ0ρfVi=ρ0ρfVi………….(1)

Now, the mass of the ice cube and mass of the water formed when it melts is same.

That is,m=ρ0·Vi=ρf·Vw

Therefore,

Vw=ρ0ρfVi……….(2)

So from (1) and (2), when the ice cube melts, the volume of the resulting water is the same as the underwater volume of the ice cube, and so the water does not overflow.

**Conclusion:**

The water does not overflow when the ice melt.

#### To determine

(d)

The work required to submerge the sphere completely.

#### Answer

The work required to submerge the sphere completely is 1.05×103 J.

#### Explanation

**1) Given:**

Radius of sphere is 0.4 m.

Density of water is 1000kg/m3.

**2) Calculations:**

The figure shows the instant when height of the exposed part of sphere is y.

The formula for the volume of a segment of height h of a sphere of radius r is

V=13πh2(3r-h).

Therefore, the volume of the submerged part of the sphere is 13πh2(3r-h)

Substitute the values r=0.4 and h=0.8-y in the above formula.

Therefore,

13πh23r-h

=13π0.8-y230.4-0.8-y

Let s=0.8-y.

Then when y=0, s=0.8 and when y=0.8,s=0.

Therefore, the volume of the submerged part of the sphere is

13πs21.2-s

Then the weight of water displaced when sphere is submerged to height h is

1000·g·13πs21.2-s

Then the work done to fully submerge the sphere is

W=∫00.8fds

Where f is the force (weight) so

W=∫00.810003·g·πs21.2-sds

=10003gπ∫00.81.2s2-s3ds

After integrating,

W=10003gπ1.2s33-s440.80

Applying the Fundamental Theorem of Calculus,

W=10003gπ1.20.833-0.844-0

W=100039.8π0.2048-0.1024

=100039.8π0.1024

=1051.30

Therefore,

W=1051.30=1.05×103J

**Conclusion:**

The work required to submerge the sphere completely is 1.05×103 J.