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A cubical block of iron of side 5 cm is floating in mercury taken in a vessel. What is the height of the block above mercuery level.
`(rho_(Hg)=13.6g//cm^(3),rho_(Fe)=7.2g//cm^(3))`

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Verified by Experts

From the law of floatation `V_(b)rho_(b)g=V_(I n)roh_(L)g`
`implies(5)^(2)xx(7.2)=(5^(2)x)xx(13.6),becausex=2.65cm` Then the height of the block above mercury level `=5cm-x=2.35cm`
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