Question

A cube of ice floats partly in water and partly in kerosene oil. The ratio of volume immersed in water to that in kerosene oil is: 
 

• A. 8:9
• B. 5:4
• C. 9:10
• D. 1:1

Hint:

Floating objects distribute their volume based on density ratios of the fluids they are immersed in.

Answer 1

The Correct Option is D

Step 1: {Define variables} 
Let \( V_1 \) be the volume immersed in water and \( V_2 \) be the volume immersed in oil. 
Step 2: {Equilibrium condition} 
\[ V_1 \rho_w g + V_2 \rho_o g = (V_1 + V_2) \rho_{{ice}} g \] 
Step 3: {Solve for ratio} 
\[ V_1 + 0.8 V_2 = 0.9 (V_1 + V_2) \] \[ 0.1 V_1 = 0.1 V_2 \Rightarrow V_1 : V_2 = 1:1 \] Thus, the correct answer is 1:1. 
 

Answer 2

Step 1: When a body floats in a fluid, the total upward buoyant force equals the weight of the body.
In this case, the cube of ice floats partly in water and partly in kerosene oil, so both contribute to the buoyant force.

Step 2: Let:
- \( V_1 \) = volume of cube immersed in water
- \( V_2 \) = volume of cube immersed in kerosene oil
- \( \rho_w \) = density of water
- \( \rho_k \) = density of kerosene
- \( \rho_i \) = density of ice

Step 3: The total buoyant force equals the weight of ice:
\( \rho_w g V_1 + \rho_k g V_2 = \rho_i g V \)
Since \( g \) is common, it cancels out.

Step 4: Assume the total volume \( V = V_1 + V_2 \), and for ice \( \rho_i = 0.9 \rho_w \), \( \rho_k = 0.8 \rho_w \).
So the equation becomes:
\( \rho_w V_1 + 0.8\rho_w V_2 = 0.9\rho_w (V_1 + V_2) \)
Dividing both sides by \( \rho_w \):
\( V_1 + 0.8V_2 = 0.9(V_1 + V_2) \)

Step 5: Expand and rearrange:
\( V_1 + 0.8V_2 = 0.9V_1 + 0.9V_2 \)
\( V_1 - 0.9V_1 = 0.9V_2 - 0.8V_2 \)
\( 0.1V_1 = 0.1V_2 \Rightarrow V_1 = V_2 \)

Final Answer: The ratio of volume immersed in water to that in kerosene oil is 1:1.