A hemispherical bowl just floats without sinking in a liquid of density 1.2 × 10³ kg/m³. If outer diameter and the density of the bowl are 1 m and 2 × 10⁴ kg/m³ respectively, then the inner diameter of the bowl will be 

Weight of the bowl = mg 

$=\mathrm{V\rho g}=\frac{4}{3}\mathrm{\pi }\left[{\left(\frac{\mathrm{D}}{2}\right)}^3-{\left(\frac{\mathrm{d}}{2}\right)}^3\right]\mathrm{\rho g}$

where D = Outer diameter, 
d = Inner diameter 
$\rho$ = Density of bowl 
Weight of the liquid displaced by the bowl 
$=\mathrm{V\sigma g}=\frac{4}{3}\mathrm{\pi }{\left(\frac{\mathrm{D}}{2}\right)}^3\mathrm{\sigma }\mathrm{g}$

where  $\sigma$ is the density of the liquid.
For the flotation  $\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3\sigma g=\frac{4}{3}\pi \left[{\left(\frac{D}{2}\right)}^3-{\left(\frac{d}{2}\right)}^3\right]\rho g$
$\Rightarrow {\left(\frac{1}{2}\right)}^3\times 1.2\times {10}^3=\left[{\left(\frac{1}{2}\right)}^3-{\left(\frac{\mathrm{d}}{2}\right)}^3\right]2\times {10}^4$
By solving we get d = 0.98 m.