Question

A sphere of radius $R$ made up of material of relative density $\sigma$ has a concentric spherical cavity of radius $r$. It just floats when placed in a tank full of water. The value of the ratio $\frac{R}{r}$ will be

• A. ${\left(\frac{\sigma }{\sigma -1}\right)}^{1/3}$
• B. ${\left(\frac{\sigma -1}{\sigma }\right)}^{1/3}$
• C. ${\left(\frac{\sigma +1}{\sigma -1}\right)}^{1/3}$
• D. ${\left(\frac{\sigma -1}{\sigma +1}\right)}^{1/3}$

Answer 1

The correct option is A ${\left(\frac{\sigma }{\sigma -1}\right)}^{1/3}$

Sphere will float if its weight is balanced by buoyant force acting on it.
$\Rightarrow F_B=mg$ .............$\left(i\right)$


Since, the sphere is just floating $i.e.$ it is floating with fully submerged.
$F_B={\rho }_{w}gV$
Here, $V=\frac{4}{3}\pi R^3$
(volume of liquid displaced)
Mass of sphere,
$m={\rho }_s{V}_s=\frac{4}{3}\pi (R^3-r^3){\rho }_s$
Substituting in eq. $\left(i\right)$,
${\rho }_{w}g\times \frac{4}{3}\pi R^3=\frac{4}{3}\pi (R^3-r^3){\rho }_{s}g$
$\Rightarrow \frac{{\rho }_w}{{\rho }_s}=1-\frac{r^3}{R^3}$
$\Rightarrow \frac{1}{\sigma }=1-\frac{r^3}{R^3}$
$\Rightarrow \frac{r^3}{R^3}=1-\frac{1}{\sigma }$
$\Rightarrow \frac{r^3}{R^3}=\frac{\sigma -1}{\sigma }$
$\Rightarrow \frac{R^3}{r^3}=\frac{\sigma }{\sigma -1}$
​​​​​​​$\Rightarrow \frac{R}{r}={\left(\frac{\sigma }{\sigma -1}\right)}^{1/3}$

Why this question? Caution: In this case the volume of liquid displaced by the sphere will be equal to the volume of the entire sphere (of radius $R$) because the sphere is fully submerged. Bottom line: Mass of the sphere will correspond to its volume excluding cavity.