A uniform solid cylinder of density $0.8\,\text{g/cm}^{3}$ floats in equilibrium in a combination of two non-mixing liquids $A$ and $B$ with its axis vertical. The densities of the liquids $A$ and $B$ are $0.7\,\text{g/cm}^{3}$ and $1.2\,\text{g/cm}^{3}$, respectively. The height of liquid $A$ is $h_{A} = 1.2\,\text{cm}$. The length of the part of the cylinder immersed in liquid $B$ is $h_{B} = 0.8\,\text{cm}$. (a) Find the total force exerted by liquid $A$ on the cylinder. (b) Find $h$, the length of the part of the cylinder in air. (c) The cylinder is depressed in such a way that its top surface is just below the upper surface of liquid $A$ and is then released. Find the acceleration of the cylinder immediately after it is released.

Step 1: Set up the equilibrium condition.

Let $S$ be the cross-sectional area of the cylinder and $L$ be its total length.

$$\rho_A h_A S g + \rho_B h_B S g = \rho_{\text{cyl}} L S g$$

Dividing by $S g$ gives the simplified equilibrium relation:

$$\rho_A h_A + \rho_B h_B = \rho_{\text{cyl}} L$$

Step 2: Calculate the total length of the cylinder.

Substitute the given values into the equilibrium relation.

$$(0.7 \times 1.2) + (1.2 \times 0.8) = 0.8 L$$ $$0.84 + 0.96 = 0.8 L$$ $$1.80 = 0.8 L$$ $$L = 2.25\,\text{cm}$$

Step 3: Find the length of the cylinder in air (Part b).

The length of the cylinder in air, $h$, is the total length minus the submerged lengths.

$$h = L - h_A - h_B$$ $$h = 2.25 - 1.2 - 0.8 = 0.25\,\text{cm}$$

Step 4: Find the total force exerted by liquid A on the cylinder (Part a).

The force exerted by liquid A is the buoyant force due to the volume of liquid A displaced.

$$F_A = \rho_A h_A S g$$ $$F_A = 0.7 \times 1.2 \times S g = 0.84 S g\,\text{upward}$$

Step 5: Find the acceleration of the cylinder immediately after release (Part c).

When the cylinder is depressed so that its top is just below the surface of liquid A, the extra length $h = 0.25\,\text{cm}$ is pushed into liquid B.

$$\Delta F = \rho_B S h g$$ $$\Delta F = 1.2 \times 0.25 \times S g = 0.30 S g$$

The mass of the cylinder is given by its density and total volume.

$$m = \rho_{\text{cyl}} L S = 0.8 \times 2.25 \times S = 1.80 S$$

Calculate the upward acceleration immediately after release.

$$a = \frac{\Delta F}{m}$$ $$a = \frac{0.30 S g}{1.80 S} = \frac{g}{6} \approx 1.63\,\text{m/s}^2$$

Answer: (a) $0.84 S g$ upward (where $S$ is the cross-sectional area of the cylinder) (b) $h = 0.25\,\text{cm}$ (c) $a = \frac{g}{6} \approx 1.63\,\text{m/s}^2$ upward