Question

A hemispherical bowl just floats without sinking in a liquid of density 1.2 $\times$ ${10}^{3}kg/m^3$. If outer diameter and the density of the bowl are 1 m and 2 $\times$ 10${}^4$ kg/m${}^3$ respectively, then the inner diameter of the bowl will be

• A. 0.94 m
• B. 0.97 m
• C. 0.98 m
• D. 0.99 m

Answer 1

The correct option is C

0.98 m

Weight of the bowl = mg

=$V\rho g=\frac{4}{3}\pi [⟮\frac{D}{2}{⟯}^3-⟮\frac{d}{2}{⟯}^3]\rho g$

where D = Outer diameter,

d = Inner diameter

$\rho$= Density of bowl

Weight of the liquid displaced by the bowl

=$V\sigma g=\frac{4}{3}\pi ⟮\frac{D}{2}{⟯}^3\sigma g$

where $\sigma$ is the density of the liquid.

For the flotation $\frac{4}{3}\pi ⟮\frac{D}{2}{⟯}^3\sigma g=\frac{4}{3}\pi [⟮\frac{D}{2}{⟯}^3-⟮\frac{d}{2}{⟯}^3]\rho g$

$\Rightarrow ⟮\frac{1}{2}{⟯}^3\times 1.2\times {10}^3=[⟮\frac{1}{2}{⟯}^3-⟮\frac{d}{2}{⟯}^3]2\times {10}^4$

By solving we get d = 0.98 m.