Question

100 cm cube of iron of density 7.8 g/cm 3 floats on mercury of density 13.6.What volume of Iron is immersed in mercury.

Answer 1

Explanation:

let x be the volume of iron which is submerged or the volume which is displaced.

100 * 7.8*g = x*13.6*g

x= 100*7.8/13.6 = 57.35 cm^3

Answer 2

Answer:

Explanation:

Correct Question :-

If 100 cc iron of specific gravity 7.8 floats on mercury of specific gravity 13.6, what volume of iron is immersed ?

Answer :-

Solution :-

Let V be the volume of iron inside mercury.

Volume of mercury displaced = 13.6 × V × g

Weight of iron

= Volume of iron × specific gravity of iron × g

= 100 × 7.8 × 6

= 780 g

Using the principle of floatation, weight of mercury displaced = weight of iron

⇒ V × 13.6 × g = 780 g

⇒ V = 780/13.6

⇒ V = 57.4 cc

Hence, the volume of Iron is immersed is 57.4 cc.