A cube of ice partly floats in water and partly in kerosene . Find the ratio of volume of ixe immersed in water to that of kerosene oil. (Specefic Gravity of kerosene is 0.8)

To determine the ratio of the volume of ice immersed in water to that of kerosene, we need to consider the principles of buoyancy and the specific gravities of the liquids involved. Let's break this down step by step.

Understanding Buoyancy

When an object is placed in a fluid, it experiences an upward force known as buoyancy. This force is equal to the weight of the fluid displaced by the object. For our scenario, we have a cube of ice that is floating in two different fluids: water and kerosene.

Key Concepts

• Specific Gravity: This is the ratio of the density of a substance to the density of a reference substance, typically water for liquids. The specific gravity of kerosene is given as 0.8, which means it is less dense than water.
• Density of Water: The density of water is approximately 1 g/cm³.
• Density of Kerosene: Since the specific gravity of kerosene is 0.8, its density can be calculated as 0.8 g/cm³.

Calculating the Volumes

Let’s denote the volume of the ice cube as V. The weight of the ice cube can be expressed as:

Weight of Ice = Volume × Density of Ice

Since the density of ice is about 0.9 g/cm³, the weight of the ice cube becomes:

Weight of Ice = V × 0.9 g/cm³

Weight of Displaced Fluids

When the ice cube floats, it displaces a volume of water and a volume of kerosene. Let’s denote:

• Vw: Volume of ice immersed in water
• Vk: Volume of ice immersed in kerosene

The weight of the water displaced by the ice is:

Weight of Water Displaced = Vw × 1 g/cm³

And the weight of the kerosene displaced is:

Weight of Kerosene Displaced = Vk × 0.8 g/cm³

Setting Up the Equation

According to the principle of buoyancy, the total weight of the ice cube must equal the total weight of the fluids displaced:

Weight of Ice = Weight of Water Displaced + Weight of Kerosene Displaced

Substituting the expressions we have:

V × 0.9 = Vw × 1 + Vk × 0.8

Relating the Volumes

Since the ice cube is floating, we can also express the total volume of the ice cube as:

V = Vw + Vk

Solving the Equations

Now we have two equations:

1. V × 0.9 = Vw + 0.8Vk
2. V = Vw + Vk

From the second equation, we can express Vk in terms of V and Vw:

Vk = V - Vw

Substituting this into the first equation gives:

V × 0.9 = Vw + 0.8(V - Vw)

Expanding this, we get:

V × 0.9 = Vw + 0.8V - 0.8Vw

Combining like terms results in:

V × 0.9 = 0.2Vw + 0.8V

Rearranging gives:

0.2Vw = V × 0.9 - 0.8V

0.2Vw = 0.1V

Thus, we find:

Vw = 0.5V

Finding the Ratio

Now that we have Vw, we can find Vk:

Vk = V - Vw = V - 0.5V = 0.5V

Finally, the ratio of the volume of ice immersed in water to that in kerosene is:

Ratio = Vw : Vk = 0.5V : 0.5V = 1 : 1

In conclusion, the ratio of the volume of ice immersed in water to that in kerosene is 1:1. This means that the ice cube displaces equal volumes of both fluids while floating.