Question

A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is $\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$, and the density of iron is $\mathbf{7}\mathbf{.}\mathbf{87}\mathbf{}\raisebox{1ex}{$\mathbf{g}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{cm}}^{\mathbf{3}}$}\right.$ . Find the inner diameter.

Short answer

Inner diameter of sphere is $57.34\mathrm{cm}$

Step-by-step solution

The given data

i) Density of water,${\mathrm{\rho }}_{\text{water}}=1.0\mathrm{g}/{\mathrm{cm}}^3$

ii) Density of iron,${\mathrm{\rho }}_{\text{ivon}}=7.87\mathrm{g}/{\mathrm{cm}}^3$

iii) Outer radius,${\mathrm{r}}_0=60\mathrm{cm}$

Understanding the concept of buoyancy

Using the concept of buoyancy, we can say that the force of buoyancy is equal to the weight of water displaced. We can find the inner diameter with the help of both densities.

Formula:

Volume of a hollow sphere, $\mathrm{V}=\frac{4}{3}\mathrm{\pi }\left({\mathrm{r}}_{\text{out}}^3-{\mathrm{r}}_{\mathrm{i\pi }}^3\right)$ (i)

Force applied on a body, $\mathrm{F}=\mathrm{m}\times \mathrm{g}$(ii)

Buoyant force exerted by fluid on a body, $\mathrm{F}=\mathrm{\rho gV}$(iii)

Calculation of inner diameter of hollow sphere

Buoyant force is equal to the weight of water displaced. The mass of sphere equals the mass of water displaced.

Mass of hollow sphere = Mass of water displaced

${\mathrm{M}}_{\mathrm{s}}=\frac{4}{3}{\mathrm{\pi r}}^3\times {\mathrm{p}}_{\text{water}}$

${\mathrm{M}}_{\mathrm{s}}=\frac{4}{3}(3.14)(30{)}^3\times (1)$

$M_s=113040\mathrm{gm}$

Using the formula of volume of hollow sphere in mass equation (1), we get

$\frac{4}{3}\mathrm{\pi }\left({\mathrm{r}}_{\text{out}}^3-{\mathrm{r}}_{\text{in}}^3\right){\mathrm{\rho }}_{\text{iron}}=113040\mathrm{gm}$

role="math" localid="1657548174348" $\frac{4}{3}(3.14)\left({30}^3-{\mathrm{r}}_{\text{in}}^3\right)\times 7.87=113040\mathrm{gm}$

$\left({\mathrm{r}}_{\mathrm{m}}^3\right)=23569.3$

$r_{\text{in}}=28.67\mathrm{cm}$

So, the inner diameter is$2\times r_{in}=57.34\mathrm{cm}$