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 1000 \,g $
$+l^{2}(l-x) \times 0.7 \times 10^{3} g $
$l \times 0.9 =x+(l-x) \times 0.7 $
or $ 0.3 x=0: 2 l $
or $ \frac{x}{l}=\frac{2}{3}$