Question

A cube of ice floats partly in water and partly in kerosene oil. Specific gravity of kerosene oil and that of ice is $0.8$ and $0.9$ respectively. Ratio of the volume of ice immersed in water to that in kerosene oil is

• A. $2:1$
• B. $3:1$
• C. $1:2$
• D. $1:1$

Answer 1

The correct option is D $1:1$

Let the volume of ice immersed in water be $V_1$ and that in kerosere oil be $V_2$.
Buouant Force always acts in upward direction, as shown in the FBD of ice cube.


Applying the equilibrium condition in vertical direction as the body is floating,
$\sum F=0$
$\Rightarrow F_{bk}+F_{bw}-Mg=0...\left(i\right)$
$F_{bk}$ is the buoyant force on ice cube due to kerosene oil
$F_{bw}$ is the buoyant force on ice cube due to water

Using $F_b=V{\rho }_{l}g$, for buoyant force
Total volume of ice cube, $V=V_1+V_2$
From Eq.$\left(i\right)$,
${\rho }_{k}g{V}_2+{\rho }_{w}g{V}_1-\rho g(V_1+V_2)=0$
$\Rightarrow \frac{{\rho }_k}{{\rho }_w}{V}_2+V_1-\frac{\rho }{{\rho }_w}(V_1+V_2)=0$
$\Rightarrow 0.8V_2+V_1-0.9(V_1+V_2)=0$
$\Rightarrow -0.1V_2+0.1V_1=0\Rightarrow \frac{V_1}{V_2}=\frac{1}{1}\therefore V_1:V_2=1:1$