Question

A sphere of solid material of specific gravity 8 has a concentric spherical cavity and just sinks in water. The ratio of radius of cavity to that of outer radius of the sphere must be
a. \[{{4}^{\dfrac{1}{3}}}\]
b. $\dfrac{{{5}^{\dfrac{1}{3}}}}{2}$
c. $\dfrac{{{9}^{\dfrac{1}{3}}}}{2}$
d. $\dfrac{{{7}^{\dfrac{1}{3}}}}{2}$

Answer 1

Hint: In the question it is given that the sphere and the cavity are concentric, that is they have the same centre. We can use this point and solve the question to find the correct answer.

Complete step by step answer:
Let us assume that the radius of the sphere is $R$ and that of the cavity is $r$. And also we know that they are concentric. That is we are to find $\dfrac{r}{R}$. It is also given that the sphere with the cavity “just sinks”. That is the weight of the object must be equal to the force of buoyancy when it is just about to sink.
Therefore, we can write the equation as $mg={{F}_{b}}$ where $m$ is the mass of the object and ${{F}_{b}}$ is the buoyancy. And mass is volume multiplied by density.
$\left( \dfrac{4}{3}\pi {{R}^{3}}-\dfrac{4}{3}\pi {{r}^{3}} \right)\rho g=\dfrac{4}{3}\pi {{R}^{3}}{{\rho }_{w}}g$
By cancelling the terms on both side, we can write the equation as:
$\left( {{R}^{3}}-{{r}^{3}} \right)\rho ={{R}^{3}}{{\rho }_{w}}$
And by shifting the terms, we get the following
$\dfrac{{{R}^{3}}-{{r}^{3}}}{{{R}^{3}}}=\dfrac{{{\rho }_{w}}}{\rho }$
And the specific gravity of water is $1$ and that of the object is $8$ as per the question.
Now If we put the values and simplify the terms, we get
$1-\dfrac{{{r}^{3}}}{{{R}^{3}}}=\dfrac{1}{8}$
Which can be written as
$\dfrac{r}{R}=\dfrac{{{7}^{\dfrac{1}{3}}}}{2}$.

So, the correct answer is “Option d”.

Note: We neglected the mass of the cavity because it has no mass and so there is no question of volume and density. Therefore we solved the question taking only the mass of the sphere into consideration and subtracting the extra mass of the cavity.