An ice-cube of density 900 kg / m 3 is floating in water of density 1000 kg / m 3. The percentage of volume of ice-cube outside the water is

Question

An ice-cube of density 900 kg / m 3 is floating in water of density 1000 kg / m 3. 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 text water -ρ textice/ρ text 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
$= \frac{ρ _{water} - ρ _{ice}}{ρ _{water}} \times 100$
$= \frac{1000 - 900}{1000} \times 100$
$= 10%$

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
$= \frac{ρ _{water} - ρ _{ice}}{ρ _{water}} \times 100$
$= \frac{1000 - 900}{1000} \times 100$
$= 10%$