Question

A cubical block of ice floating in water has to support a metal piece of mass $0.5\text{kg}$. If specific gravity of ice is $0.9$, then minimum edge length (in $\text{m}$) of the block so that it does not sink in water is

• A. ${\left(\frac{1}{200}\right)}^{\frac{1}{3}}\text{m}$
• B. ${\left(\frac{1}{150}\right)}^{\frac{1}{3}}\text{m}$
• C. ${\left(\frac{1}{100}\right)}^{\frac{1}{3}}\text{m}$
• D. ${\left(\frac{1}{250}\right)}^{\frac{1}{3}}\text{m}$

Answer 1

The correct option is A ${\left(\frac{1}{200}\right)}^{\frac{1}{3}}\text{m}$

To support the metal piece, cubical block (of mass $m$) will float fully immersed, considering the case of minimum edge length.


For equilibrium, $\sum F=0$ along vertical direction,
$\Rightarrow F_b-(0.5+m)g=0$
$\Rightarrow {\rho }_{l}gV=(0.5g+{\rho }^{\prime }gV)...\left(i\right)$
where $mg={\rho }^{\prime }gV$
$\left[\begin{array}{c}{\rho }_l\to \text{density of water}\\ {\rho }^{\prime }\to \text{density of ice}\\ V\to \text{volume of ice cube}\end{array}\right]$
From Eq.$\left(i\right)$,
${\rho }_{l}V=0.5+0.9{\rho }_{l}V$
$[\text{Specific gravity}=\frac{{\rho }^{\prime }}{{\rho }_l}=0.9]$
$\Rightarrow 0.1\times {\rho }_{l}V=0.5$
$\Rightarrow V=\frac{0.5}{0.1\times {\rho }_l}$
$\Rightarrow V=\frac{0.5}{0.1\times 1000}$
Volume of cube is $V=x^3$
$\Rightarrow x^3=\frac{1}{200}$
$\therefore x=\sqrt[3]{3\frac{1}{200}}\text{m}$
Therefore, minimum length of the edge of the cube is ${\left(\frac{1}{200}\right)}^{\frac{1}{3}}\text{m}$.