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An open-ended U-tube of uniform cross-sectional area contains water. Initially the water level stands at 0.29 m from the bottom in each arm. Kerosene oil of density $400 k g / m^{3}$ is added to the left arm until its length is 0.1 m. The ratio $\frac{h_{1}}{h_{2}}$ of the heights of the liquid in two arms is

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$\frac{35}{33}$

$\frac{37}{31}$

$\frac{38}{32}$

$\frac{39}{42}$

answer is B.

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Detailed Solution

$h_{1} + h_{2} = (0.29 \times 2) + 0.1 = 0.68$

$\Rightarrow ρ_{k} g (0.1) + ρ_{w} g h_{1} - ρ_{w} g (0.1) = ρ_{w} g h_{2}$

$400 \times 10 \times 0.1 + (1000 \times 10 \times h_{1}) - (1000 \times 10 \times 0.1) = 1000 \times 10 \times h_{2}$

$400 + (10^{4} h_{1}) - 10^{3} = 10^{4} h_{2}$

$(h_{1} - h_{2}) 10^{4} = 600$

$h_{1} - h_{2} = \frac{600}{10^{4}} = 0.06$

$h_{1} + h_{2} = 0.68$

$2 h_{1} = 0.74 \Rightarrow h_{1} = 0.37$

$h_{2} = 0.31$

$∴ \frac{h_{1}}{h_{2}} = \frac{37}{31}$

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