Question
The relative density of ice is 0.92. What fraction of the volume of a piece of ice will be submerged when floating on:
Water
A liquid of density 1.11. Take the density of water as 1gcm^-3
Water
A liquid of density 1.11. Take the density of water as 1gcm^-3
Answers
GPT-4o mini
To determine the fraction of the volume of a piece of ice that will be submerged when it is floating, we can use the principle of buoyancy, which states that the weight of the fluid displaced by the floating object is equal to the weight of the floating object itself.
### Given:
- Relative density (RD) of ice = 0.92
- Density of water (\( \rho_{water} \)) = 1 g/cm³
- Density of the liquid = 1.11 g/cm³
### Step 1: Calculate the density of ice.
The density of ice can be calculated using its relative density:
\[
\rho_{ice} = RD \times \rho_{water} = 0.92 \times 1 \text{ g/cm}^3 = 0.92 \text{ g/cm}^3
\]
### Step 2: Calculate the fraction submerged in water.
Let \( V \) be the volume of ice and \( V_{sub} \) be the submerged volume.
Using the buoyancy principle:
\[
\text{Weight of ice} = \text{Weight of displaced water}
\]
\[
\rho_{ice} \times V = \rho_{water} \times V_{sub}
\]
The fraction of the volume of ice that is submerged when floating on water is:
\[
\frac{V_{sub}}{V} = \frac{\rho_{ice}}{\rho_{water}} = \frac{0.92 \text{ g/cm}^3}{1 \text{ g/cm}^3} = 0.92
\]
### Step 3: Calculate the fraction submerged in the liquid of density 1.11 g/cm³.
Using the same principle of buoyancy:
\[
\rho_{ice} \times V = \rho_{liquid} \times V_{sub}
\]
The fraction of the volume of ice that is submerged when floating on a liquid of density 1.11 g/cm³ is:
\[
\frac{V_{sub}}{V} = \frac{\rho_{ice}}{\rho_{liquid}} = \frac{0.92 \text{ g/cm}^3}{1.11 \text{ g/cm}^3}
\]
Calculating this value:
\[
\frac{V_{sub}}{V} \approx 0.8293
\]
### Summary of Results:
- When floating on **water**, about **92%** of the volume of ice will be submerged.
- When floating on a **liquid of density 1.11 g/cm³**, about **82.93%** of the volume of ice will be submerged.
### Given:
- Relative density (RD) of ice = 0.92
- Density of water (\( \rho_{water} \)) = 1 g/cm³
- Density of the liquid = 1.11 g/cm³
### Step 1: Calculate the density of ice.
The density of ice can be calculated using its relative density:
\[
\rho_{ice} = RD \times \rho_{water} = 0.92 \times 1 \text{ g/cm}^3 = 0.92 \text{ g/cm}^3
\]
### Step 2: Calculate the fraction submerged in water.
Let \( V \) be the volume of ice and \( V_{sub} \) be the submerged volume.
Using the buoyancy principle:
\[
\text{Weight of ice} = \text{Weight of displaced water}
\]
\[
\rho_{ice} \times V = \rho_{water} \times V_{sub}
\]
The fraction of the volume of ice that is submerged when floating on water is:
\[
\frac{V_{sub}}{V} = \frac{\rho_{ice}}{\rho_{water}} = \frac{0.92 \text{ g/cm}^3}{1 \text{ g/cm}^3} = 0.92
\]
### Step 3: Calculate the fraction submerged in the liquid of density 1.11 g/cm³.
Using the same principle of buoyancy:
\[
\rho_{ice} \times V = \rho_{liquid} \times V_{sub}
\]
The fraction of the volume of ice that is submerged when floating on a liquid of density 1.11 g/cm³ is:
\[
\frac{V_{sub}}{V} = \frac{\rho_{ice}}{\rho_{liquid}} = \frac{0.92 \text{ g/cm}^3}{1.11 \text{ g/cm}^3}
\]
Calculating this value:
\[
\frac{V_{sub}}{V} \approx 0.8293
\]
### Summary of Results:
- When floating on **water**, about **92%** of the volume of ice will be submerged.
- When floating on a **liquid of density 1.11 g/cm³**, about **82.93%** of the volume of ice will be submerged.