Question

A cube of ice floats partly in water and partly in kerosene oil. The radio of volume of ice immersed in water to that in kerosene oil (specific gravity of Kerosene oil = 0.8, specific gravity of ice = 0.9)

• A. 8 : 9
• B. 5 : 4
• C. 9 : 10
• D. 1 : 1

Answer 1

The Correct Option is D

To solve this problem, we need to understand the floating condition of the ice cube in both water and kerosene oil. The principle behind this is Archimedes' principle, which states that the weight of the fluid displaced by a submerged object is equal to the weight of the object.

The specific gravity gives us the density relative to water. For our problem:

• Specific gravity of ice (\(\text{S}_\text{ice}\)) = 0.9
• Specific gravity of kerosene oil (\(\text{S}_\text{ker}\)) = 0.8
• Specific gravity of water = 1.0 (by definition)

The density of ice is 0.9 times that of water, and the density of kerosene oil is 0.8 times that of water.

The buoyant force on ice from both water and kerosene must balance the weight of the ice. Therefore, we have:

\(\text{Weight of Ice} = \text{Buoyant Force in Water} + \text{Buoyant Force in Kerosene}\)

Let \( V_w \) be the volume of ice immersed in water, and \( V_k \) be the volume of ice immersed in kerosene. Then:

• Weight of ice = \(\rho_\text{ice} \cdot V_\text{total} \cdot g\)
• Buoyant force in water = \(\rho_\text{water} \cdot V_w \cdot g\)
• Buoyant force in kerosene = \(\rho_\text{ker} \cdot V_k \cdot g\)

Equating the weight of the ice to the total buoyant force:

\(\rho_\text{ice} \cdot V_\text{total} = \rho_\text{water} \cdot V_w + \rho_\text{ker} \cdot V_k\)

Substituting the specific gravities:

\(0.9 \cdot V_\text{total} = 1 \cdot V_w + 0.8 \cdot V_k\)

Assuming the entire volume is submerged, \( V_\text{total} = V_w + V_k \).

Let’s simplify:

\(V_w + 0.8 \cdot V_k = 0.9 \cdot (V_w + V_k)\)

This simplifies to:

\(V_w + 0.8 \cdot V_k = 0.9 \cdot V_w + 0.9 \cdot V_k\)

\(0.1 \cdot V_w = 0.1 \cdot V_k\)

Thus, the ratio \(w : V_k\) is \(1 :\).

Therefore, the correct answer is 1 : 1.

Answer 2

Let the volume of the cube immersed in water be \(V_w\) and the volume immersed in kerosene oil be \(V_k\). Using the principle of floatation, the weight of the ice cube is equal to the total upthrust provided by water and kerosene oil.
The weight of the ice cube is:
\[W_{\text{ice}} = \rho_{\text{ice}} Vg,\]
where \(\rho_{\text{ice}} = 0.9 \rho_{\text{water}}\).
The total upthrust is:
\[U = \rho_{\text{water}} V_w g + \rho_{\text{kerosene}} V_k g.\]
Equating the weight of the ice cube to the total upthrust:
\[0.9 \rho_{\text{water}} V g = \rho_{\text{water}} V_w g + 0.8 \rho_{\text{water}} V_k g.\]
Cancel \(\rho_{\text{water}} g\) from all terms:
\[0.9 V = V_w + 0.8 V_k.\]
Divide through by \(V\):
\[0.9 = \frac{V_w}{V} + 0.8 \frac{V_k}{V}.\]
Since the total volume of the ice cube is \(V\), \(\frac{V_w}{V} + \frac{V_k}{V} = 1\). Let \(x = \frac{V_w}{V}\) and \(y = \frac{V_k}{V}\), then \(x + y = 1\).
Substituting \(y = 1 - x\) into the equation:
\[0.9 = x + 0.8(1 - x).\]
Simplify:
\[0.9 = x + 0.8 - 0.8x.\]
\[0.9 - 0.8 = 0.2x \implies x = 0.5.\]
Thus, \(y = 1 - 0.5 = 0.5\). The ratio of \(V_w\) to \(V_k\) is:
\[V_w : V_k = 0.5 : 0.5 = 1 : 1.\]