An open-ended U-tube of uniform cross-sectional area contains water (density $$10^{3}kg m^{−3}$$). Initially the water level stands at 0.29 m from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density 800 kg$$ m^{−3}$$ is added to the left arm until its length is 0.1 m, as shown in the schematic figure below. The ratio $$\left(\frac{ℎ1}{ℎ2}\right)$$ of the heights of the liquid in the two arms is 

Initially both arms contain only water of density $$\rho_w = 1000 \, \text{kg m}^{-3}$$ up to a height $$h_0 = 0.29 \, \text{m}$$ from the bottom.

When kerosene (density $$\rho_k = 800 \, \text{kg m}^{-3}$$) of length $$0.10 \, \text{m}$$ is gently poured into the left arm, it floats on water. Let the water-kerosene interface in the left arm fall by $$x \, \text{m}$$. Because the tube has uniform cross-section, the water level in the right arm rises by the same $$x \, \text{m}$$.

Therefore, after pouring kerosene:
Left arm: height of water column = $$h_0 - x = 0.29 - x$$, height of kerosene column = $$0.10 \, \text{m}$$.
Right arm: height of water column = $$h_0 + x = 0.29 + x$$.

Pressures at the common bottom level in the two arms must be equal (both tops are open to atmosphere). Equating hydrostatic pressures:

$$\rho_w g(0.29 - x) + \rho_k g(0.10) = \rho_w g(0.29 + x)$$

Cancel $$g$$ on both sides and substitute the densities:

$$1000(0.29 - x) + 800(0.10) = 1000(0.29 + x)$$

$$290 - 1000x + 80 = 290 + 1000x$$

$$370 - 1000x = 290 + 1000x$$

$$80 = 2000x \; \Longrightarrow \; x = 0.04 \, \text{m}$$

Total height of liquid in each arm:
Left arm: $$h_1 = (0.29 - x) + 0.10 = 0.39 - 0.04 = 0.35 \, \text{m}$$.
Right arm: $$h_2 = 0.29 + x = 0.33 \, \text{m}$$.

Required ratio:

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

Option B which is: $$\dfrac{35}{33}$$