
: An iceberg is floating in the ocean. What fraction of its volume is above the water? (Given: density of ice=\[900\,{\text{kg/}}{{\text{m}}^3}\] and density of ocean water=\[1030\,{\text{kg/}}{{\text{m}}^3}\])
A. \[90/103\]
B. \[13/103\]
C. \[10/103\]
D. \[1/103\]
Answer
561.6k+ views
Hint:Use the concept of Archimedes’ principle. Use the equation for the buoyant force exerted on the object in the liquid. Also use the equation for density of an object. As the iceberg is floating on the water, the buoyant force on the iceberg is equal to the weight of the iceberg submerged in the water.
Formulae used:
The buoyant force \[F\] on an object floating on the liquid is
\[F = \rho Vg\] …… (1)
Here, \[\rho \] is the density of the liquid, \[V\] is the volume of the object and \[g\] is acceleration due to gravity.
The density \[\rho \] of an object is
\[\rho = \dfrac{M}{V}\] …… (2)
Here, \[M\] is the mass of the object and \[V\] is the volume of the object.
Complete step by step answer:
We have given the density of the iceberg is \[900\,{\text{kg/}}{{\text{m}}^3}\] and the density of the ocean water is \[1030\,{\text{kg/}}{{\text{m}}^3}\].
\[{\rho _i} = 900\,{\text{kg/}}{{\text{m}}^3}\]
\[{\rho _w} = 1030\,{\text{kg/}}{{\text{m}}^3}\]
Since the iceberg is floating on the water, the upward buoyant force on the iceberg is equal to the weight of the iceberg submerged in the water in the downward direction.
\[F = {M_i}g\] …… (3)
Rewrite equation (2) for the mass of the iceberg.
\[{M_i} = {\rho _i}{V_i}\]
Here, \[{\rho _i}\] and \[{V_i}\] are the density and total volume of the iceberg respectively.
The volume \[{V_i}\] of the iceberg submerged in the water is equal to the volume \[{V_w}\] of the water displaced by the iceberg.
Substitute \[{V_w}\] for \[{V_i}\] in the above equation.
\[{M_i} = {\rho _i}{V_w}\]
Substitute for \[F\] and \[{\rho _i}{V_w}\] for \[{M_i}\] in equation (3).
\[{\rho _w}{V_i}g = {\rho _i}{V_w}g\]
\[ \Rightarrow {\rho _w}{V_i} = {\rho _i}{V_w}\]
Rearrange the above equation for \[\dfrac{{{V_i}}}{{{V_w}}}\].
\[ \Rightarrow \dfrac{{{V_i}}}{{{V_w}}} = \dfrac{{{\rho _i}}}{{{\rho _w}}}\]
Substitute \[900\,{\text{kg/}}{{\text{m}}^3}\] for \[{\rho _i}\] and \[1030\,{\text{kg/}}{{\text{m}}^3}\] for \[{\rho _w}\] in the above equation.
\[ \Rightarrow \dfrac{{{V_i}}}{{{V_w}}} = \dfrac{{900\,{\text{kg/}}{{\text{m}}^3}}}{{1030\,{\text{kg/}}{{\text{m}}^3}}}\]
\[ \therefore \dfrac{{{V_i}}}{{{V_w}}} = \dfrac{{90}}{{103}}\]
Hence, the fraction of the volume of the iceberg submerged in water is \[\dfrac{{90}}{{103}}\].
The fraction of iceberg which is above the ocean water is \[1 - \dfrac{{90}}{{103}} = \dfrac{{13}}{{103}}\].
Therefore, the fraction of the volume of iceberg above the water is \[\dfrac{{13}}{{103}}\].Hence, the correct option is B.
Note:The students may forget to subtract the volume fraction obtained after applying Archimedes’ principle to the floating iceberg. The fraction of the volume obtained will be the fraction of volume of the iceberg submerged in the ocean water. This obtained volume fraction must be subtracted from 1 to get the correct answer.
Formulae used:
The buoyant force \[F\] on an object floating on the liquid is
\[F = \rho Vg\] …… (1)
Here, \[\rho \] is the density of the liquid, \[V\] is the volume of the object and \[g\] is acceleration due to gravity.
The density \[\rho \] of an object is
\[\rho = \dfrac{M}{V}\] …… (2)
Here, \[M\] is the mass of the object and \[V\] is the volume of the object.
Complete step by step answer:
We have given the density of the iceberg is \[900\,{\text{kg/}}{{\text{m}}^3}\] and the density of the ocean water is \[1030\,{\text{kg/}}{{\text{m}}^3}\].
\[{\rho _i} = 900\,{\text{kg/}}{{\text{m}}^3}\]
\[{\rho _w} = 1030\,{\text{kg/}}{{\text{m}}^3}\]
Since the iceberg is floating on the water, the upward buoyant force on the iceberg is equal to the weight of the iceberg submerged in the water in the downward direction.
\[F = {M_i}g\] …… (3)
Rewrite equation (2) for the mass of the iceberg.
\[{M_i} = {\rho _i}{V_i}\]
Here, \[{\rho _i}\] and \[{V_i}\] are the density and total volume of the iceberg respectively.
The volume \[{V_i}\] of the iceberg submerged in the water is equal to the volume \[{V_w}\] of the water displaced by the iceberg.
Substitute \[{V_w}\] for \[{V_i}\] in the above equation.
\[{M_i} = {\rho _i}{V_w}\]
Substitute for \[F\] and \[{\rho _i}{V_w}\] for \[{M_i}\] in equation (3).
\[{\rho _w}{V_i}g = {\rho _i}{V_w}g\]
\[ \Rightarrow {\rho _w}{V_i} = {\rho _i}{V_w}\]
Rearrange the above equation for \[\dfrac{{{V_i}}}{{{V_w}}}\].
\[ \Rightarrow \dfrac{{{V_i}}}{{{V_w}}} = \dfrac{{{\rho _i}}}{{{\rho _w}}}\]
Substitute \[900\,{\text{kg/}}{{\text{m}}^3}\] for \[{\rho _i}\] and \[1030\,{\text{kg/}}{{\text{m}}^3}\] for \[{\rho _w}\] in the above equation.
\[ \Rightarrow \dfrac{{{V_i}}}{{{V_w}}} = \dfrac{{900\,{\text{kg/}}{{\text{m}}^3}}}{{1030\,{\text{kg/}}{{\text{m}}^3}}}\]
\[ \therefore \dfrac{{{V_i}}}{{{V_w}}} = \dfrac{{90}}{{103}}\]
Hence, the fraction of the volume of the iceberg submerged in water is \[\dfrac{{90}}{{103}}\].
The fraction of iceberg which is above the ocean water is \[1 - \dfrac{{90}}{{103}} = \dfrac{{13}}{{103}}\].
Therefore, the fraction of the volume of iceberg above the water is \[\dfrac{{13}}{{103}}\].Hence, the correct option is B.
Note:The students may forget to subtract the volume fraction obtained after applying Archimedes’ principle to the floating iceberg. The fraction of the volume obtained will be the fraction of volume of the iceberg submerged in the ocean water. This obtained volume fraction must be subtracted from 1 to get the correct answer.
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