Question

A cubical block of iron 5cm on each side is floating on mercury in a vessel:- 

1) What is the height of the block above mercury level? 

2) Water is poured into vessel so that it just covers the iron block. What is the height of the water column?

 Given Density of mercury = 13.6g|cm³ and Density of iron = 7.2g|cm³ 

Answer 1

1) Let h = height of cubical block above mercury level 

Volume of cubical Iron Block = l × b × h 

= 5 × 5 × 5 

= 125 cm³.

Mass of cubical Iron Block = Volume × Density of Iron

= 125 × 7.2

= 900 g → 1) 

Volume of Mercury displaced = Length × Breadth × Decreased height 

= l × b × (5 – h) 

= 5 × 5 × (5 – h) 

Mass of Mercury displaced = Volume of Mercury × Density of Mercury 

= 5 × 5 × (5 – h) × 13.6 → 2) 

2) When water is poured, let x = height of block in water

∴Depth of block in mercury = (5 – x) cm

Mass of water displaced = 5 × 5 × x × 1 = 25 x gm 

1 = Density of water in g|cm³ 

Mass of Mercury displaced = 5 × 5 × (5 – x) × 13.6 = 25 X 13.6 (5 – x) gm 

Acc. to principle of floatation, 

Weight of Iron Block = Weight of water displaced + Weight of mercury displaced