An open-ended U-tube of uniform cross-sectional area contains water (density ). Initially the water level stands at 0.29 m from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density is added to the left arm until its length is 0.1 m

Question

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 $800 kg m^{- 3}$ 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 [2020]

Smart Achievers · Accepted Answer

(2)

$Pressure will be same at the same horizontal level.$

$∴$ $P_{A} = P_{B}$

$\Rightarrow P_{0} + ρ_{k} g \times 0.1 + ρ_{w} g \times (h_{1} - 0.1) = ρ_{w} g h_{2} + p_{0}$

$\Rightarrow 80 + 1000 (h_{1} - 0.1) = 1000 h_{2}$

$\Rightarrow 80 + 1000 h_{1} - 100 = 1000 h_{2}$

$\Rightarrow h_{1} - h_{2} = \frac{20}{1000}$

$\Rightarrow h_{1} - h_{2} = 0.02 \dots (i)$

$Also, h_{1} - 0.1 + h_{2} = 2 \times 0.29$

$\Rightarrow h_{1} + h_{2} = 0.68 \dots (i i)$

$Solving (i) & (ii), we get$

$2 h_{1} = 0.70$

$\Rightarrow h_{1} = 0.35 m$

$and h_{2} = 0.33 m$

$So, \frac{h_{1}}{h_{2}} = \frac{35}{33}$

Solution

1 $\frac{15}{14}$

2 $\frac{35}{33}$

3 $\frac{7}{6}$

4 $\frac{5}{4}$

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Solution

1 1514

2 3533

3 76

4 54

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1 $\frac{15}{14}$

2 $\frac{35}{33}$

3 $\frac{7}{6}$

4 $\frac{5}{4}$