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

Answer: (D)

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:
$V_w : V_k = 0.5 : 0.5 = 1 : 1.$