Question

A hemispherical blown just float without sinking in a liquid of density 1.2×1000kg/m3.If outer diameter and the density of the bowl are 1m and 2× 10000 kg/m3 respectively ,then the inner diameter of the bowl will be ;

Answer 1

Answer on Question #74136, Chemistry / Other

A hemispherical blown just float without sinking in a liquid of density $1.2 \times 1000 \, \mathrm{kg/m^3}$. If outer diameter and the density of the bowl are $1 \, \mathrm{m}$ and $2 \times 10000 \, \mathrm{kg/m^3}$ respectively, then the inner diameter of the bowl will be.

Solution:

Weight of the bowl

$$W = mg = V \rho_b g = \frac{4}{3} \pi \left[ \frac{D}{2} \right] - \frac{d}{2} \left[ \frac{d}{2} \right]$$

Where D is outer diameter, d is Inner diameter, $\rho_b$ is density of the bowl

Weight of the liquid displaced by the bowl

$$W = V \rho_l g = \frac{4}{3} \pi (D / 2)^3 \rho_l g$$

Where $\rho_l$ is the density of the liquid.

$$\frac{4}{3} \pi \times \left(\frac{D}{2}\right)^3 \times \rho_l \times g = \frac{4}{3} \pi \times \left[ \left(\frac{D}{2}\right)^3 - \left(\frac{d}{2}\right)^3 \right] \times \rho_b \times g$$ $$d = D \sqrt[3]{1 - \frac{\rho_l}{\rho_b}}$$ $$d = 1 \times \sqrt[3]{1 - \frac{1.2 \times 10^3}{2 \times 10^4}} = 0.98 \, m$$

Answer: 0.98 m

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