Question

A cubical iron block of side 10 cm is floating on Mercury in a vessel what is the height of the block above Mercury level given that density of iron is 7.2 gm per cc and density of mercury is 13.6 gm/cc

Answer 1

Answer: 4 .71 cm

Step-by-step explanation:

The block will float in mercury if its weight is equal to the weight of mercury displaced by it

Density of iron pi = 7.2 g/ cm³

Density of mercury of mercury Pm = 13.6 g /cm³

Length of each side of the iron block = 10 cm

let x be the height of the block above the mercury level in the vessel.

Then , the volume of the block inside the mercury

Vm = 10 ×10 × (10 - x)

This is also equal to the volume of mercury displaced by the block

Weight of this volume of mercury

W = VmPmg

W= Vm × 13.6 ×981= upthrust on block

Total volume of the iron block is

V = 10× 10 × 10 = 1000 cm³

Hence , weight  is

W = V Pi g = V × 7.2 × 981

The iron float ,

Weight of mercury displaced = Weight of the iron block

100 × (10 -x )× 13.6 ×981 = 1000 ×7.2 ×981

(10 -x) ×1360 = 7200

10 - x  = 7200 / 3600

10 - x = 5.29

- x = 5.29 -10

x = 4.71

So, the height of the block abovr mercury level = 4.71 cm