Explanation

(b) Let total volume of iceberg $= V$
Volume of visible part of iceberg$= V_{1}$
$\therefore\ \ \ \ \ volume\ of\ iceberg\ inside\ water = \left( V - V_{1} \right)$
Now, volume of water displaced by iceberg $= (V - V_{1})$
Let $density\ of\ iceberg = d$
and $density\ of\ water = D$
Then, applying law of floatation, at equilibrium,
weight of iceberg = weight of displaced water
$\ \ Vdg = \left( V - V_{1} \right)Dg$
$\Longrightarrow \ \ \ \frac{V - V_{1}}{V} = \frac{d}{D}$
$\Longrightarrow \ \ \ 1 - \frac{V_{1}}{V} = 1 - \frac{d}{D}$
$\ \ \ \ \ \frac{V_{1}}{V} = 1 - \frac{d}{D}$
$\therefore$ Percentage fraction of visible iceberg
$\ \ \frac{V_{1}}{V} \times 100\% = \left( 1 - \frac{d}{D} \right) \times 100\%$
Here, $d = 917\ kgm^{- 3}$ , $D = 1024\ kgm^{- 3}$
$\therefore\ \frac{V_{1}}{V} \times 100\% = \left( 1 - \frac{917}{1024} \right) \times 100\%$
$\ \ = \frac{10700}{1024}\% \approx 10\%$