An open-ended U-tube of uniform cross-sectional area contains water (density = 10³ 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 schematic figure below. The ratio $(h_{1} / h_{2})$ of the heights of the liquid in the two arms is

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$\frac{7}{6}$

$\frac{15}{4}$

$\frac{5}{4}$

$\frac{35}{33}$

answer is B.

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Detailed Solution

Let water level falls by x meter due to kerosene oil. Equating pressure on both sides of the tube,

$P_{0} + 800 \times g \times 0.1 = P_{0} + 1000 \times g \times 2 x$

$\Rightarrow x = \frac{800}{2000} = 0.04 m$

$\Rightarrow h_{1} = 0.29 + 0.1 - 0.04 = 0.35 m$ and $h_{2} = 0.29 + 0.04 = 0.33 m$

$\Rightarrow \frac{h_{1}}{h_{2}} = \frac{0.35}{0.33} = \frac{35}{33}$

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