Liquid Height Ratio in U Tube with Two Immiscible Fluids

To find the ratio of the fluid heights in the two limbs $\left(\frac{h_1}{h_2}\right)$, we apply the principles of hydrostatic equilibrium for immiscible liquids in a U-tube.

1. Initial State:

Initially, water of density $\rho_w = 10^3 \text{ kg m}^{-3}$ stands at a height of $H_0 = 0.29 \text{ m}$ in both arms of the U-tube from the bottom.

2. Final State after adding Kerosene:

Kerosene of density $\rho_k = 800 \text{ kg m}^{-3}$ and column length $l_k = 0.1 \text{ m}$ is poured into the left limb.

Because the U-tube has a uniform cross-sectional area:

• The water level in the left limb moves down by a distance $x$.
• The water level in the right limb moves up by the same distance $x$.

Thus:

• The height of the water column in the left limb from the bottom becomes: $H_{w,1} = H_0 - x = 0.29 - x$
• The height of the water column in the right limb from the bottom becomes: $H_{w,2} = H_0 + x = 0.29 + x$

3. Hydrostatic Pressure Balance:

Equating the hydrostatic pressure at the bottom of both limbs: $P_0 + \rho_k g l_k + \rho_w g (0.29 - x) = P_0 + \rho_w g (0.29 + x)$

Canceling $P_0$ and $g$ from both sides: $\rho_k l_k + \rho_w (0.29 - x) = \rho_w (0.29 + x)$

$\rho_k l_k = 2 \rho_w x$

$x = \frac{\rho_k l_k}{2 \rho_w}$

Substitute the given numerical values ($\rho_k = 800 \text{ kg m}^{-3}$, $\rho_w = 1000 \text{ kg m}^{-3}$, and $l_k = 0.1 \text{ m}$): $x = \frac{800 \times 0.1}{2 \times 1000} = \frac{80}{2000} = 0.04 \text{ m}$

4. Calculating Heights $h_1$ and $h_2$:

• Total height of liquid in the left limb ($h_1$): $h_1 = H_{w,1} + l_k = (0.29 - x) + 0.1 = 0.39 - 0.04 = 0.35 \text{ m}$

• Total height of liquid in the right limb ($h_2$): $h_2 = H_{w,2} = 0.29 + x = 0.29 + 0.04 = 0.33 \text{ m}$

5. Ratio of Heights:

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

Thus, the correct option is (B).