Q.

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 h1h2of the heights of the liquid in the two arms is-

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a

1514

b

76

c

54

d

3533

answer is B.

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Detailed Solution

h1+h2=0.29×2+0.1h1+h2=0.68       1ρk= density of kerosene &ρw= density of water P0+ρkg0.1+ρwgh10.1ρwgh2=P0ρkg0.1+ρwgh1ρwg×0.1=ρwgh2800×10×0.1+1000×10×h11000×10×0.1=1000×10×h210000h1h2=200

h1h2=0.02 .(2)h1=0.35h2=0.33 So, h1h2=3533

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