A U-tube of uniform cross section (see Fig) is partially filled with liquid I. Another liquid II which does not mix with liquid I is poured into one side. It is found that the liquid levels of the two sides of the tube are the same, while the level of liquid I has risen by $2\ \text{cm}$. If the specific gravity of liquid I is $1.1$, the specific gravity of liquid II must be

Step 1: Analyze the change in liquid levels.

Let the initial level of liquid I in both limbs be the reference level.

$$\Delta h = 2\text{ cm}$$

Step 2: Determine the heights of the liquid columns above the interface.

The height of liquid I above the interface in the right limb is $h_1$.

$$h_1 = 2\Delta h$$ $$h_1 = 4\text{ cm}$$

Since the free surfaces of both liquids are at the same level, the height of liquid II is $h_2$.

$$h_2 = h_1$$ $$h_2 = 4\text{ cm}$$

Step 3: Apply the pressure balance at the interface.

Equate the pressures at the interface level in both limbs.

$$P_0 + \rho_2 g h_2 = P_0 + \rho_1 g h_1$$ $$\rho_2 h_2 = \rho_1 h_1$$

Step 4: Calculate the specific gravity of liquid II.

Substitute $h_2 = h_1$ into the relation.

$$\rho_2 = \rho_1$$ $$s_2 = s_1$$ $$s_2 = 1.1$$

Answer: The specific gravity of liquid II is $1.1$.