Question

an iceberg is floating in water. the density of ice in iceberg is 917 kg m-3.the density of water is 1024 kg m-3. what percentage fraction of iceberg would be visible?

Answer 1

The buoyancing force due to water = Weight of iceberg
V₁ρ₁g = mg = V₂ρ₂g
(V₁ × 1024) = (V₂ × 917)
V₁ = 0.89V₂

Iceberg visible = V₂ - V₁
= V₂ - 0.89V₂
= 0.11V₂

Fraction of iceberg visible = 0.11V₂/V₂ = 0.11
% Fraction of iceberg visible = 11 %

Answer 2

Answer:

The percentage fraction of the visible iceberg measured is $11\%$.

Explanation:

Given data,

The iceberg's density, $\rho_{iceberg}$ = $917kg/m^{3}$

The water's density, $\rho_{water}$ = $1024kg/m^{3}$

The percentage fraction of the visible iceberg =?

Let the volume of water = V₁

Let the volume of the iceberg = V₂

As given,

• Weight of water displaced = weight of iceberg

As we know,

• Weight = mass × gravity

Also,

• Mass = volume × density

Therefore,

• Weight = $V\rho g$

Thus,

• Weight of water displaced = weight of iceberg
• $V_{1} \rho_{water} g$ = $V_{2} \rho_{iceberg} g$
• $V_{1} \rho_{water}$ = $V_{2} \rho_{iceberg}$
• V₁ = $\frac{V_{2} \rho_{iceberg}}{\rho_{water}}$
• V₁ = $\frac{V_{2}\times 917 }{1024}$
• V₁ = $0.89V_{2}$

Now, the volume of the visible iceberg, V₃ = $V_{2} - V_{1}$ = $V_{2} - 0.89V_{2}$ = $0.11V_{2}$.

Thus,

• The fraction of the visible iceberg = $\frac{V_{3} }{V_{2} }\times 100$ = $\frac{0.11V_{2}}{V_{2}}\times 100$ = $11\%$.