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

Apparent weight of floating ice-berg is zero.
For ice-cube floating in water only that much portion of the body will be immersed by which the weight of the liquid displaced balances the total weight of the body.
Thus, the apparent weight of the body will be zero.
For floatation

W_{1} = W_{2}

V d g = v d_{L} g

\Rightarrow \frac{v}{V} = \frac{d}{d _{L}}

\frac{Volume of immersed part of body}{Total volume of body}

= \frac{density of body}{density of liquid}

= \frac{900}{1000} = 0.9

%

of volume outside

= (1 - 0.9) \times 100 = 10%

Note : Density of ice is less than density of water for floating.

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 :

$20%$

$35%$

$10%$

$25%$

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%

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 :

$20%$

$35%$

$10%$

$25%$