An open-ended U-tube of uniform cross-sectional area contains water (density 10³kg 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 $(\frac{h_{1}}{h_{2}})$ of the heights of the liquid in the two arms is-

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$\frac{15}{14}$

$\frac{7}{6}$

$\frac{5}{4}$

$\frac{35}{33}$

answer is B.

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Detailed Solution

$\begin{array}{l} h_{1} + h_{2} = 0.29 \times 2 + 0.1 \\ \begin{array}{l} h_{1} + h_{2} = 0.68 \dots \dots (1) \\ [ρ_{k} = density of kerosene & ρ_{w} = density of water] \end{array} \\ \Rightarrow P_{0} + ρ_{k} g (0.1) + ρ_{w} g (h_{1} - 0.1) - ρ_{w} g h_{2} = P_{0} \\ \Rightarrow ρ_{k} g (0.1) + ρ_{w} g h_{1} - ρ_{w} g \times (0.1) = ρ_{w} g h_{2} \\ \Rightarrow 800 \times 10 \times 0.1 + 1000 \times 10 \times h_{1} - 1000 \times 10 \times 0.1 = 1000 \times 10 \times h_{2} \\ \Rightarrow 10000 (h_{1} - h_{2}) = 200 \end{array}$

$\begin{array}{l} \Rightarrow h_{1} - h_{2} = 0.02 \dots \dots . (2) \\ \Rightarrow h_{1} = 0.35 \\ \Rightarrow h_{2} = 0.33 \\ So, \frac{h_{1}}{h_{2}} = \frac{35}{33} \end{array}$