Explanation

(c) Let $D_{1}$ be the inner diameter of the hemispherical bowl and $D_{2}$ be the outer diameter of the bowl. As bowl is just floating so
$\frac{4}{3}\pi\left( \frac{1}{2} \right)^{3} \times 1.2 \times 10^{3}$
$= \frac{4}{3}\pi\left\lbrack \left( \frac{1}{2} \right)^{3} - \left( \frac{D_{1}}{2} \right)^{3} \right\rbrack \times (2 \times 10^{4})$
Or $\frac{1.2 \times 10^{3}}{2 \times 10^{4}} = 1 - D_{1}^{3}$
Or $D_{1} = \left( 1 - \frac{1.2}{20} \right)^{1/3} = \left( \frac{18.8}{20} \right)^{1/3}$
On solving, $D_{1} = 0.98$ m