A block of mass 20 g sits at rest on a plate that is at the top of the fluid on the right side of a U-tube. The U-tube contains two different fluids with densities ρ1 = 900 kg/m^3 and ρ2 = 630 kg/m^3 and has a cross sectional area A = 5.3x10^-4 m^2. The surfaces are offset by an amount h. If the fluid on the right side of the U-tube has the greater density, what is h?

To find the offset height (h), we need to consider the equilibrium of forces acting on the block.

Let's analyze the forces acting on the block:
1. Weight force (mg): The weight force is acting vertically downwards with a magnitude of mg, where m is the mass of the block (20 g or 0.02 kg) and g is the acceleration due to gravity (9.8 m/s^2).
2. Buoyant force (Fb1): The buoyant force acts in the upward direction and its magnitude depends on the density of the fluid and the volume of the displaced fluid. As the block sits at the top of the fluid, the buoyant force only depends on the first fluid with density ρ1.

The equation for the buoyant force is given by:
Fb1 = ρ1 * g * V
where V is the volume of the fluid displaced by the block.

Now, let's calculate the volume of the displaced fluid:
V = A * h
where A is the cross-sectional area of the U-tube (5.3x10^-4 m^2) and h is the offset height we want to find.

The net force acting on the block is given by:
Net Force = Fb1 - mg
Since the block is at rest, the net force must be zero:
Fb1 - mg = 0

Substituting the expressions for Fb1 and V, we get:
ρ1 * g * A * h - mg = 0

Now we can solve this equation for h:
h = mg / (ρ1 * g * A)

Applying the given values:
m = 0.02 kg
g = 9.8 m/s^2
ρ1 = 900 kg/m^3
A = 5.3x10^-4 m^2

h = (0.02 kg * 9.8 m/s^2) / (900 kg/m^3 * 5.3x10^-4 m^2)

Calculating the result:
h ≈ 0.043 m or 4.3 cm

Therefore, the offset height (h) between the fluid surfaces is approximately 4.3 cm.