A cubical block of wood 10 cm on a side floats at the interface between oil and water with its lower surface horizontal and 4 cm below the interface. The density of oil is 0.6 gcm⁻¹.The mass of block is

The correct option is 760 g.

Step-by-Step Explanation:

According to the principle of floatation, when a body floats in a fluid, the upward buoyant force acting on it must equal the total downward gravitational force (weight) of the body.

Let:
- Side length of the cubical block, L = 10 cm
- Cross-sectional area of the block, A = L × L = 10 cm × 10 cm = 100 cm²
- Height of the block submerged in water, h_w = 4 cm
- Height of the block submerged in oil, h_o = 10 cm - 4 cm = 6 cm
- Density of water, ρ_w = 1.0 g cm^-3
- Density of oil, ρ_o = 0.6 g cm^-3

First, we calculate the volume of the block submerged in water (V_w) and the volume submerged in oil (V_o):

$V_w=A\times h_w=100{\text{ cm}}^2\times 4\text{ cm}=400{\text{ cm}}^3$

$V_o=A\times h_o=100{\text{ cm}}^2\times 6\text{ cm}=600{\text{ cm}}^3$

The total buoyant force (F_b) is the sum of the buoyant forces exerted by water and oil:

$F_b=({\rho }_w{V}_w+{\rho }_o{V}_o)g$

The weight of the block (W) is given by:

$W=Mg$

Where M is the mass of the block, and g is the acceleration due to gravity. Setting the weight equal to the buoyant force:

$Mg=({\rho }_w{V}_w+{\rho }_o{V}_o)g$

Simplifying by cancelling g from both sides:

$M={\rho }_w{V}_w+{\rho }_o{V}_o$

Substituting the values into the equation:

$M=(1.0{\text{ g cm}}^{-3}\times 400{\text{ cm}}^3)+(0.6{\text{ g cm}}^{-3}\times 600{\text{ cm}}^3)$

$M=400\text{ g}+360\text{ g}=760\text{ g}$

Thus, the mass of the block is 760 g.