The density of ice is `0.92gcm^(-3)` and that of sea water is `1.025gcm^(-3)` Find the total volume of an iceberg which floats with its volume `800cm^(3)` above water.

Answer

To solve the problem of finding the total volume of an iceberg that floats with a volume of 800 cm³ above water, we can follow these steps:

### Step 1: Understand the relationship between the volumes
According to Archimedes' principle, the volume of the iceberg submerged in water (let's denote it as \( V_b \)) and the total volume of the iceberg (denote it as \( V \)) are related by the densities of ice and seawater. The relationship can be expressed as:

\[ \frac{V_b}{V} = \frac{\text{Density of Ice}}{\text{Density of Sea Water}} \]

### Step 2: Identify the known values
From the problem, we know:
- Density of ice, \( \rho_{ice} = 0.92 \, \text{g/cm}^3 \)
- Density of seawater, \( \rho_{water} = 1.025 \, \text{g/cm}^3 \)
- Volume of the iceberg above water, \( V_{above} = 800 \, \text{cm}^3 \)

### Step 3: Calculate the ratio of densities
Now, we can find the ratio of the densities:

\[ \frac{\rho_{ice}}{\rho_{water}} = \frac{0.92}{1.025} \approx 0.8976 \]

### Step 4: Relate the submerged volume and total volume
Let \( V_b \) be the volume of the iceberg submerged in water. Then, we can express \( V_b \) in terms of \( V \):

\[ \frac{V_b}{V} = 0.8976 \implies V_b = 0.8976 V \]

### Step 5: Relate the volumes above and below water
Since the volume of the iceberg above water is given, we can express it as:

\[ V_{above} = V - V_b \]

Substituting \( V_b \):

\[ 800 = V - 0.8976 V \]

### Step 6: Solve for the total volume \( V \)
This simplifies to:

\[ 800 = V(1 - 0.8976) \implies 800 = V(0.1024) \]

Now, we can solve for \( V \):

\[ V = \frac{800}{0.1024} \approx 7812.5 \, \text{cm}^3 \]

### Final Answer
The total volume of the iceberg is approximately:

\[ \boxed{7812.5 \, \text{cm}^3} \]

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