An ice-cube of density 900 text kg/m3 is floating in water of density 1000 text kg/m3 . The percentage of volume of ice-cube outside the water is

Question

An ice-cube of density 900 text kg/m3 is floating in water of density 1000 text kg/m3 . The percentage of volume of ice-cube outside the water is (A) 20% (B) 35% (C) 10% (D) 25%

Tardigrade Admin · Accepted Answer

The percentage of volume of ice cube outside the water is =(ρ water-ρ ice/ρ water)× 100 =(1000-900/1000)× 100=10%

Q. An ice-cube of density $900 k g / m^{3}$ is floating in water of density $1000 k g / m^{3}$ . The percentage of volume of ice-cube outside the water is

20%

35%

10%

25%

The percentage of volume of ice cube outside the water is $= \frac{ρ _{w a t er} - ρ _{i ce}}{ρ _{w a t er}} \times 100$ $= \frac{1000 - 900}{1000} \times 100 = 10$

Q. An ice-cube of density 900kg/m3 is floating in water of density 1000kg/m3 . The percentage of volume of ice-cube outside the water is

20%

35%

10%

25%

The percentage of volume of ice cube outside the water is =ρwater​ρwater​−ρice​​×100 =10001000−900​×100=10

Q. An ice-cube of density $900 k g / m^{3}$ is floating in water of density $1000 k g / m^{3}$ . The percentage of volume of ice-cube outside the water is

The percentage of volume of ice cube outside the water is $= \frac{ρ _{w a t er} - ρ _{i ce}}{ρ _{w a t er}} \times 100$ $= \frac{1000 - 900}{1000} \times 100 = 10$