A body is floating on mercury keeping its `1/4`th part inside mercury. Now, water is poured into the container such that the body remains just immersed totally in water. Now what part of the body will remain immersed in mercury? [Density of mercury = `13.6g*cm^(-3)`]
Text Solution
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Let the volume of the body be V, density of its material be d. In the case of floatation in mercury, `Vxxd=V/4xx13.6or,d=13.6/4=3.4g*cm^(-3)`. When water is added, suppose x part of the volume of the body remains immersed in mercury and (1 - x) part of its volume in water. `therefore" "Vxx3.4=Vxx x xx13.6+V(1-x)xx1` or, 3.4 = 13.6x + 1 - x or, 12.6x = 2.4 or, `x=2.4/12.6=4/21` So, `4/21` th part of the given body will remain immersed in mercury.
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