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 $800 k g m^{- 3}$ is added to the left arm until its length is 0.1 m, as shown in the schematic figure below. The ratio $(\frac{h_{2}}{h_{1}})$ of the heights of the liquid in the two arms is $\frac{33}{5 k}$ . Then the value of $k$ is

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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,
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$800 \times g \times 0.1 = 1000 \times g \times 2 x$
$\Rightarrow x = \frac{8}{200} m = \frac{1}{25} m = 0.04 m$
Hence, $h_{2} = 0.29 + 0.04 = 0.33 m$
And $h_{1} = 0.29 - 0.04 + 0.1 = 0.35 m$
So, $\frac{h_{1}}{h_{2}} = \frac{35}{33}$