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
3533
c
76
d
54
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+ρwgh1−0.1−ρwgh2=P0⇒ρkg0.1+ρwgh1−ρwg×0.1=ρwgh2⇒800×10×0.1+1000×10×h1−1000×10×0.1=1000×10×h2⇒10000h1−h2=200⇒h1−h2=0.02 …….(2)⇒h1=0.35⇒h2=0.33 So, h1h2=3533