Question

A cube made of material having a density of $0.9\times {10}^{3}kg/m^3$ floats between water and a liquid of density $0.7\times {10}^{3}kg/m^3,$ which is immiscible with water. What part of the cube is immersed in water?

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $\frac{3}{7}$

Answer 1

Answer: (B)

Let $l=$ side of the tube
$x=$ side of cube immersed in water
$l-x=$ side of cube immersed in liquid
According to law of floatation,
$l^3\times 0.9\times 10^3\times g=\left(l^2\times x\right)\times 1000g$
$+l^2(l-x)\times 0.7\times 10^{3}g$
$l\times 0.9=x+(l-x)\times 0.7$
or $0.3x=0:2l$
or $\frac{x}{l}=\frac{2}{3}$