Step-by-step Solution

1. Understand the Initial and Final States

   • Initially, the U-tube contains water of density $\rho_w = 10^3$ kg/m³ up to a height of $H_0 = 0.29$ m from the bottom in both arms.
   • Then, kerosene of density $\rho_k = 800$ kg/m³ and column length $L_k = 0.1$ m is added to the left arm.
   • The addition of kerosene pushes the water level in the left arm down by a distance $x$ and, due to the uniform cross-section, raises the water level in the right arm by the same distance $x$.
   • The new height of the water level in the left arm (the interface between water and kerosene) is $H_0 - x$ from the bottom.
   • The new height of the water level in the right arm is $H_0 + x$ from the bottom.

2. Apply Pressure Balance

   • We can balance the pressure at a common horizontal level in the continuous fluid (water). The lowest such level is the interface between kerosene and water in the left arm.
   • Let point A be at the kerosene-water interface in the left arm, and point B be at the same horizontal level in the right arm.
   • The pressure at these two points must be equal: $P_A = P_B$.
   • The pressure at point A is due to the atmospheric pressure ($P_{atm}$) plus the pressure from the kerosene column of length $L_k$. $P_A = P_{atm} + \rho_k g L_k$
   • The pressure at point B is due to the atmospheric pressure ($P_{atm}$) plus the pressure from the water column above it. The height of this water column is the difference in water levels between the two arms. Height of water column above B = $(H_0 + x) - (H_0 - x) = 2x$. $P_B = P_{atm} + \rho_w g (2x)$

3. Solve for the displacement, x

   • Equating the expressions for $P_A$ and $P_B$: $P_{atm} + \rho_k g L_k = P_{atm} + \rho_w g (2x)$ $\rho_k L_k = 2 \rho_w x$
   • Now, we can solve for $x$ by substituting the given values: $800 \times 0.1 = 2 \times 1000 \times x$ $80 = 2000x$ $x = {{80} \over {2000}} = {{4} \over {100}} = 0.04\ \text{m}$

4. Calculate the final heights, h₁ and h₂

   • The height $h_1$ is the height of the free surface of the liquid (kerosene) in the left arm from the bottom. It is the sum of the height of the water column in the left arm and the length of the kerosene column. $h_1 = (H_0 - x) + L_k$ $h_1 = (0.29 - 0.04) + 0.1 = 0.25 + 0.1 = 0.35\ \text{m}$
   • The height $h_2$ is the height of the free surface of the liquid (water) in the right arm from the bottom. $h_2 = H_0 + x$ $h_2 = 0.29 + 0.04 = 0.33\ \text{m}$

5. Determine the Ratio

   • The required ratio is $\left( {{{{h_1}} \over {{h_2}}}} \right)$. $\frac{h_1}{h_2} = \frac{0.35}{0.33} = \frac{35/100}{33/100} = \frac{35}{33}$

   • This matches option B.