Questions
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 of the heights of the liquid in the two arms is-
detailed solution
Correct option is B
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=3533Talk to our academic expert!
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A homogeneous solid cylinder of length , cross-sectional area is immersed such that it floats with its axis vertical at the liquid interface with length in the denser liquid as shown in Figure. The lower density liquid is open to atmosphere having pressure P0. Then, density D of solid is given by
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