Question

A cubical block of ice floating in water has to support a metal piece of 0.8 kg. What can be the minimum length of the edge of the block so that it does not sink in the water? Assume specific gravity of ice as 0.9.

• A. 10 cm
• B. 20 cm
• C. 18 cm
• D. 27 cm

Answer 1

The correct option is B 20 cm

Mass of metal piece, m = 0.8 kg
Relative density of ice = 0.9
Relative density = $\frac{\text{Density of given substance}}{\text{Density of water}}$


Density of ice, ${\rho }_1=\text{Relative density of ice}\times \text{Density of water}\Rightarrow {\rho }_1=0.9\times 1000(\because {\rho }_w={10}^{3}kg/m^3\Rightarrow {\rho }_1=900kg/m^3$

To keep the metal piece floating:

$W_m+W_i=U\dots ..\left(i\right)$

Here, $W_m$ is weight of metal piece, $W_i$ is weight of ice cube and U is upthrust applied by water

$\Rightarrow W_m=mg=0.8g{W}_i=Mg\text{(M = mass of ice block)}{W}_i=\text{density of ice}\times \text{volume of ice}\times g$

Let edge of the cuboid ice block be a.

$\Rightarrow W_i={\rho }_1\times a^3\times g=900a^{3}g(\because \text{Volume of the cube}=a^3)$

$\text{Upthrust, U = mass of water displaced}\times g\Rightarrow U={\rho }_W\times a^3\times g=1000a^{3}g(\because =1000kg/m^3)$

Putting the values of $W_m,W_{i}andU$ in (i), we get:

$0.8g+900\times a^3\times g=1000\times a^3\times g\Rightarrow 0.8+900a^3=1000a^3\Rightarrow 1000a^3–900a^3=0.8\Rightarrow 100a^3=0.8\Rightarrow a^3=\frac{08}{100}\Rightarrow a=0.2m=0.2\times 100cm=20cm$

Thus, the required length of the edge of the cube is 20 cm.

Hence, the correct answer is option (b).