Question

The figure represent a U-tube of uniform cross-section filled with two immiscible liquids. One is water with density $\rho_w$ and other liquid is of density $\rho$. The liquid interface lies 2 cm above base. The relation between $\rho$ and $\rho_w$ is (assume that the level of liquid in two arms is same)

• A. $\rho = \rho_w$
• B. $\rho = 1.02 \rho_w$
• C. $\rho = 1.2 \rho_w$
• D. Insufficient information

Answer 1

Answer: (A)

$\rho = \rho_w$

Answer 2

Let the free surface level be at height $h_0$ in both arms.

• Left arm (only water):
  Pressure at free surface = $\rho_w g h_0$.

• Right arm:
  The U-tube contains water from the base to 2 cm and liquid of density $\rho$ above that up to $h_0$.
  Pressure at free surface = $2\rho_w g + (h_0-2)\rho g$.

Since the free surfaces are at the same level, equate pressures:

$$\rho_w g\,h_0 = 2\rho_w g + (h_0-2)\rho g.$$

Cancel $g$ and rearrange:

$$\rho_w h_0 = 2\rho_w + \rho (h_0-2).$$ $$\rho_w h_0 - \rho h_0 = 2\rho_w - 2\rho.$$ $$h_0 (\rho_w-\rho) = 2(\rho_w-\rho).$$ $$(\rho_w-\rho)(h_0-2) = 0.$$

Since $h_0 \ne 2$ (otherwise, there would be no layer of liquid $\rho$), we must have

$$\rho_w - \rho = 0 \quad \Rightarrow \quad \rho=\rho_w.$$