A cubical block of density 600 kg/m3 is floating in a liquid of density 900 kg/m3. The height of cube immersed in liquid is (cube side = 10 cm)

The correct option is 6.67 cm.

Step-by-Step Explanation:

To find the height of the cubical block immersed in the liquid, we can apply the principle of floatation. According to this principle, for a floating body, the upward buoyant force acting on the body must be equal to the downward gravitational force (weight) acting on the body.

Let:
- The side length of the cubical block be $L=10\text{cm}=0.1\text{m}$
- The density of the cubical block be ${\rho }_{\text{block}}=600{\text{kg/m}}^3$
- The density of the liquid be ${\rho }_{\text{liquid}}=900{\text{kg/m}}^3$
- The height of the cube immersed in the liquid be $h$

The total volume of the cubical block is:
$V_{\text{total}}=L^3$
The weight of the cubical block is given by:
$W=V_{\text{total}}\times {\rho }_{\text{block}}\times g=L^3\times {\rho }_{\text{block}}\times g$
where $g$ is the acceleration due to gravity.

The volume of the block immersed in the liquid is:
$V_{\text{immersed}}=L^2\times h$
The buoyant force exerted by the liquid on the block is equal to the weight of the liquid displaced:
$F_B=V_{\text{immersed}}\times {\rho }_{\text{liquid}}\times g=L^2\times h\times {\rho }_{\text{liquid}}\times g$

For the block to float in equilibrium, the buoyant force must equal the weight of the block:
$F_B=W$
Substituting the expressions:
$L^2\times h\times {\rho }_{\text{liquid}}\times g=L^3\times {\rho }_{\text{block}}\times g$

We can simplify this equation by dividing both sides by $L^2\times g$:
$h\times {\rho }_{\text{liquid}}=L\times {\rho }_{\text{block}}$
Solving for the immersed height $h$:
$h=L\times \frac{{\rho }_{\text{block}}}{{\rho }_{\text{liquid}}}$

Now, substitute the given values into the equation:
$h=10\text{cm}\times \frac{600}{900}$
$h=10\times \frac{2}{3}\text{cm}$
$h=\frac{20}{3}\text{cm}\approx 6.67\text{cm}$

Therefore, the height of the cube immersed in the liquid is approximately 6.67 cm.