The relative density of ice is 0.9 and that of sea water is 1.125. What fraction of the whole volume of an iceberg appears above the surface of the sea ?

Options

A. `1//5`
B. `2//5`
C. `3//5`
D. `4//5`

Explanation

To find the fraction of the volume of an iceberg that appears above the surface of the sea, we can use the principle of buoyancy. Here’s a step-by-step solution:

### Step 1: Understand the concept of buoyancy
An iceberg floats in water due to the buoyant force acting on it, which is equal to the weight of the water displaced by the submerged part of the iceberg.

### Step 2: Define the variables
- Let \( V \) be the total volume of the iceberg.
- Let \( V_{\text{displaced}} \) be the volume of the iceberg submerged in water.
- The relative density (specific gravity) of ice is given as \( 0.9 \).
- The relative density of sea water is given as \( 1.125 \).

### Step 3: Calculate the weight of the iceberg
The weight of the iceberg can be expressed as:
\[ \text{Weight of iceberg} = \text{Density of ice} \times V \times g \]
Where \( g \) is the acceleration due to gravity.

Using the relative density, we can express the density of ice as:
\[ \text{Density of ice} = 0.9 \times 1000 \, \text{kg/m}^3 = 900 \, \text{kg/m}^3 \]
Thus, the weight of the iceberg becomes:
\[ \text{Weight of iceberg} = 900 \, V \, g \]

### Step 4: Calculate the buoyant force
The buoyant force acting on the iceberg is equal to the weight of the sea water displaced by the submerged volume:
\[ \text{Buoyant force} = \text{Density of sea water} \times V_{\text{displaced}} \times g \]
Using the relative density of sea water:
\[ \text{Density of sea water} = 1.125 \times 1000 \, \text{kg/m}^3 = 1125 \, \text{kg/m}^3 \]
Thus, the buoyant force becomes:
\[ \text{Buoyant force} = 1125 \, V_{\text{displaced}} \, g \]

### Step 5: Set up the equilibrium condition
At equilibrium, the weight of the iceberg is equal to the buoyant force:
\[ 900 \, V \, g = 1125 \, V_{\text{displaced}} \, g \]
We can cancel \( g \) from both sides:
\[ 900 \, V = 1125 \, V_{\text{displaced}} \]

### Step 6: Solve for the displaced volume
Rearranging the equation gives:
\[ V_{\text{displaced}} = \frac{900}{1125} \, V \]
Calculating the fraction:
\[ V_{\text{displaced}} = \frac{0.9}{1.125} \, V \]

### Step 7: Calculate the fraction of the volume above the water
The fraction of the iceberg that is submerged is:
\[ \text{Fraction submerged} = \frac{0.9}{1.125} \approx 0.8 \]
Thus, the fraction of the iceberg that is above the water is:
\[ \text{Fraction above water} = 1 - \text{Fraction submerged} = 1 - 0.8 = 0.2 \]

### Step 8: Express the fraction in simplest form
The fraction \( 0.2 \) can be expressed as:
\[ \text{Fraction above water} = \frac{1}{5} \]

### Final Answer
The fraction of the volume of the iceberg that appears above the surface of the sea is \( \frac{1}{5} \).

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