A block of wood (specific gravity = 0.96) is floating at the interface between a layer of gasoline and a layer of water as shown in figure. What fraction of the wood block is below the interface if specific gravity of gasoline is 0.74 and that of water is 1?

Concept:

For an object floating at the interface of two fluids, the buoyant force acting on the object equals its weight. The buoyant force is contributed by both the gasoline and water layers.

The fraction of the wood block submerged in gasoline is denoted by \( x \) and the fraction submerged in water is \( (1 - x) \).

The equilibrium condition can be written as:

\( \text{Weight of the wood block} = \text{Buoyant force due to gasoline} + \text{Buoyant force due to water} \)

Calculation:

Given data:

• Specific gravity of wood: \( S_w = 0.96 \)
• Specific gravity of gasoline: \( S_g = 0.74 \)
• Specific gravity of water: \( S_w = 1 \)

Using the equilibrium equation:

\( 0.96 = 0.74x + 1(1 - x) \)

Solving for \( x \):

\( 0.96 = 0.74x + 1 - x \)

\( 0.96 - 1 = -0.26x \)

\( -0.04 = -0.26x \)

\( x = \frac{0.04}{0.26} = \frac{2}{13} \)

Fraction of the wood block below the interface (submerged in water):

\( 1 - x = 1 - \frac{2}{13} = \frac{11}{13} \)

Closest to this value among options is \(\frac{5}{6} \)