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A cubical block of iron 5 cm on each sid...

A cubical block of iron 5 cm on each side is floating on mercury in a vessel. (i) what is the height of the block above mercury level ?
(ii) Water is poured in the vessel until it just covers the iron block. What is the height of water column . density of mercury `13.6 gm//cm^(3)`. Density of iron 7.2 `gm//cm^(3)`

Text Solution

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Case-I : Suppose h be the height of cubical block of iron above mercury.
Volume of iron block = 5 ×5 × 5 `= 125" cm"^(3 )`
Mass of iron block `= 125 × 7.2 = 900" gm"`
Volume of mercury displaced by the block
`= 5 × 5 × (5 – h)" cm"^(3)`
Mass of mercury displaced
`= 5 × 5 (5 – h) × 13.6` gm
By the law floatation,
weight of mercury displaced = weight of iron block
`5 × 5 (5 – h) × 13.6 = 900`
or `" "(5-h)=(900)/(25xx13.6)2.65`
`rArr " "h=5-2.65=2.35" cm"`
Case - II :Suppose in this case height of iron block in water be x. The height of iron block in mercury will be (5 – x) cm.

Mass of the water displaced `=5xx5xx(X)xx1`
Mass of mercury displaced `=5xx5xx(5-x)xx13.6`
So, weight of water displaced +weight of mercury displaced=weight of iron block
or `" "5xx5xx x xx1+5xx5xx(5-x)xx13.6=900`
or `" "x=(5-x)xx13.6=36`
`x=2.54`cm
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