An open-ended U-tube of uniform cross-sectional area contains water (density $10^{3}\,\text{kg m}^{-3}$). Initially the water level stands at $0.29\,\text{m}$ from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density $800\,\text{kg m}^{-3}$ is added to the left arm until its length is $0.1\,\text{m}$, as shown in the schematic figure below. The ratio $\left(\dfrac{h_1}{h_2}\right)$ of the heights of the liquid in the two arms is

Step 1: Identify the given parameters.

Let the uniform cross-sectional area of the U-tube be $A$.

$$\rho_w = 10^{3}\,\text{kg m}^{-3}$$ $$\rho_k = 800\,\text{kg m}^{-3}$$ $$L = 0.1\,\text{m}$$ $$H_0 = 0.29\,\text{m}$$

Step 2: Relate the displacement of water in both arms.

Since the cross-sectional area is uniform, the volume of water is conserved, meaning the level drop in the left arm equals the level rise in the right arm.

$$x_{\text{drop}} = x_{\text{rise}} = x$$

Step 3: Apply the pressure equilibrium condition.

Equate the pressure at the kerosene-water interface in the left arm to the pressure at the same horizontal level in the right arm.

$$P_{\text{left}} = P_{\text{right}}$$ $$P_{\text{atm}} + \rho_k g L = P_{\text{atm}} + \rho_w g (2x)$$ $$\rho_k L = 2\rho_w x$$

Step 4: Calculate the displacement $x$.

Solve the equilibrium equation for $x$.

$$x = \frac{\rho_k L}{2\rho_w}$$ $$x = \frac{800 \times 0.1}{2 \times 10^{3}}$$ $$x = 0.04\,\text{m}$$

Step 5: Determine the total liquid heights in both arms.

Calculate the total height of the liquid in the left arm $h_1$ and the right arm $h_2$.

$$h_1 = (H_0 - x) + L$$ $$h_1 = (0.29 - 0.04) + 0.1$$ $$h_1 = 0.35\,\text{m}$$ $$h_2 = H_0 + x$$ $$h_2 = 0.29 + 0.04$$ $$h_2 = 0.33\,\text{m}$$

Step 6: Calculate the ratio of the heights.

Find the ratio of $h_1$ to $h_2$.

$$\frac{h_1}{h_2} = \frac{0.35}{0.33}$$ $$\frac{h_1}{h_2} = \frac{35}{33}$$

Answer: The ratio of the heights of the liquid in the two arms is $\dfrac{35}{33}$.