Physics - Fluid Question with Solution | TestHub

Solution:

A U-tube of uniform cross-section is partially filled with liquid I. Another immiscible liquid II is poured into one arm. After equilibrium, the free surfaces of both liquids are at the same level, and the level of liquid I rises by $$\begin{gathered} 2 \\ ,cm \end{gathered}$$.

Since the cross-section of the tube is uniform, the rise of liquid I by $$\begin{gathered} 2 \\ ,cm \end{gathered}$$ in one arm implies a fall of $$\begin{gathered} 2 \\ ,cm \end{gathered}$$ in the other arm.

Hence, the difference in heights of liquid I columns in the two arms is:

$$\begin{gathered} h=2+2=4 \\ ,cm \end{gathered}$$

At equilibrium, pressures at the same horizontal level in liquid I must be equal.

Pressure due to liquid II:

$$P_2={\rho }_{2}gh$$

Pressure due to liquid I:

$$P_1={\rho }_{1}gh$$

Since the free surfaces are at the same level,

$${\rho }_{2}gh={\rho }_{1}gh$$

Cancelling common terms,

$${\rho }_2={\rho }_1$$

Given that the specific gravity of liquid I is

$$s_1=1.1$$

the specific gravity of liquid II is

$$s_2=1.1$$

Final Answer:

$$\boxed{1.1}$$