Question

A block of ice of area $A$ and thickness $0.5m$ is floating in the fresh water. In order to just support a man of $100kg$, find the area $A$. (The specific gravity of ice is $0.917$ and density of water $=1000kg/m^3$ ).

• A. $1.40m^2$
• B. $0.75m^2$
• C. $2.41m^2$
• D. None

Answer 1

The correct option is C $2.41m^2$

Draw the situation of the given problem.
Area of ice block $A$
Thickness $=0.5m$
Mass of man $=100kg$
$p_i$ and $p_w$ are the densities of ice and water respectively.

Find the area of the ice block.
For equilibrium, $(m_1+m_2)g=pVg$

Here,
$m_1$ mass of man $=100kg$
$m_2$ mass of ice $=0.917\times 1000V=971V$

$p=1000kg/m^3$

$h-0.5m$

$\therefore 100g+917Vg=pVg=1000Vg$

$\therefore V=\frac{100g}{1000g-917g}$

$\therefore Ah=\frac{100}{83}$

$\Rightarrow A=\frac{100}{83\times 0.5}=2.41m^2$