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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 mercury level.
`(rho_(Hg)=13.6g//cm^(3),rho_(Fe)=7.2g//cm^(3))`

Text Solution

Verified by Experts

From the law of floatation, `V_(b)rho_(b)g=V_("in")rho_(L)g`
`rArr (5)^(3)xx(7.2)=(5^(2)x)xx(13.6), therefore x=2.65 cm`
Then, the height of the block above mercury level `=5cm -x=2.35cm`
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