Question

Consider a solid sphere of density $\rho \left(r\right) = {\rho_0} \left(1- \dfrac {r^2}{R^2}\right)$ 0< r ≤ R. The minimum density of a liquid in which it float is just
a. $\dfrac {2}{5} {\rho_0}$
b. $\dfrac {2}{3} {\rho_0}$
c. $\dfrac {\rho_0}{5}$
d. $\dfrac {\rho_0}{3}$

Answer 1

Hint: We should know about the meaning of “mass density” before starting the question. The mass per unit volume of an object is referred to as its mass density. The units that can be used to express this parameter are pounds per square foot $\left(\dfrac {lb}{ft^{2}}\right)$ and kilograms per square metre $\left(\dfrac {kg}{m^3}\right)$.

Complete answer:
A substance, material, or object's mass density is a measure of how much mass (or how many particles) it has in relation to the volume it takes up. However, this measurement is not always accurate because mass density is affected by a number of variables, such as temperature and pressure.

For example, when gases are heated, they expand. In other words, their volume rises as the temperature rises. According to the formula above, a gas's density drops as its volume increases.

Any point inside the sphere has density equal to
${{\rho }^{'}}=\rho (r)={{\rho }_{0}}(1-\dfrac{{{r}^{2}}}{{{R}^{2}}})$,
Where r denotes the point's separation from the centre.

As a result, the density will be the same for all spots that are equally spaced from the centre.
Mass of such an elemental shell of thickness dr is
$dM={{\rho }^{'}}dVg={{\rho }^{'}}4\pi {{r}^{2}}dr$
$dM={{\rho }_{0}}(1-\dfrac{{{r}^{2}}}{{{R}^{2}}})4\pi {{r}^{2}}dr.g$
$=4\pi {{\rho }_{0}}(1-\dfrac{{{r}^{2}}}{{{R}^{2}}}){{r}^{2}}dr$

Total mass
$M=\int{dM}=4\pi {{\rho }_{0}}\int\limits_{0}^{k}{[{{r}^{2}}-\dfrac{{{r}^{4}}}{{{R}^{2}}}]dr}$
$=4\pi {{\rho }_{0}}[\dfrac{{{r}^{3}}}{3}-\dfrac{{{r}^{5}}}{5{{R}^{2}}}]\mathop{{}}_{0}^{R}$
$=4\pi {{\rho }_{0}}[\dfrac{{{R}^{3}}}{3}-\dfrac{{{R}^{3}}}{5}]\underset{0}{\overset{R}{\mathop{{}}}}\,$
$=\dfrac{8\pi \rho {{R}^{3}}}{15}$

The sphere is simply submerged and suspended when the liquid reaches the required density. Therefore,
Weight = buoyant force
$Mg=\sigma Vg$
Where,
$\sigma$= density of the liquid
V= volume of the sphere.
$\dfrac{8\pi {{\rho }_{0}}{{R}^{3}}g}{15}=\sigma \dfrac{4}{3}\pi {{R}^{3}}g$
$\dfrac{8{{\rho }_{0}}}{15}=\dfrac{\sigma 4}{3}$
$\sigma =\dfrac{2{{\rho }_{0}}}{5}$

Hence, the option (a) is correct.

Note: Physics and engineering both depend on the concept of density. Density is crucial in deciding whether an object will float when placed on the surface of a fluid since it is closely related to an object's mass. Density is one of the most crucial properties you may learn about a substance, even though it may not be as vital as the fundamental forces.