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A concrete sphere of radius \[R\] has a cavity of radius $r$ which is packed with saw-dust. The specific gravities of concrete and saw-dust are $2.4$ and $0.3$ respectively. The sphere floats with its entire volume submerged under water. Calculate the ratio of mass of concrete and saw-dust.
A) $1$
B) $2$
C) $3$
D) $4$

Last updated date: 13th Jun 2024
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Hint:Recall what is called specific gravity. From this definition you should be able to understand the problem. Now we will have to relate the floatation of the sphere along with the saw-dust and we shall be able to reach our final answer.

Formula used:
Specific gravity is simply a ratio between density of a substance to the same of a reference substance, which in this case is water. It is otherwise called as relative density $(R.D)$
Mathematically, $R.D = \dfrac{{{\rho _1}}}{{{\rho _0}}}$
Where ${\rho _1}$ and ${\rho _0}$ are densities of the substance of concern and water respectively. The value of ${\rho _0}$ is $1gc{m^{ - 3}}$ or $1kg{m^{ - 3}}$.

Complete step by step solution:
Let’s start the problem from the beginning. There is a concrete sphere that has a carving in it which makes it a hollow sphere and we are given that the radius of the sphere is $R$ and the radius of the carving is $r$.
This central carving or cavity is filled with saw-dust.
As the question mentions, the specific gravity of concrete and saw dust are respectively $2.4$ and $0.3$.
From the given data and the formula mentioned, we can calculate the densities as follows.
${\rho _1} = 2.4kg{m^{ - 3}}$ and ${\rho _2} = 0.3kg{m^{ - 3}}$
Where ${\rho _1}$ and ${\rho _0}$ are densities of concrete and saw-dust respectively.
From the Archimedes’ Principle, we are familiar with the idea that a body floats if the upward thrust generated equals the weight of the body we consider for floating.
Weight of the body can be figured out if we happen to know its mass which we can calculate from its volume and density
Volume of the concrete sphere is ${V_1} = \dfrac{4}{3}\pi {R^3} - \dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi ({R^3} - {r^3})$ ………. (1)
Volume of the saw-dust sphere is ${V_2} = \dfrac{4}{3}\pi {r^3}$ ………. (2)
Volume of the sphere would be the combined volume of the saw-dust sphere and the concrete sphere. We will need this volume for calculating thrust due to the fluid.
So the combined volume of the sphere is
\[V = {V_1} + {V_2}\]
$ \Rightarrow V = \dfrac{{4\pi }}{3}{R^3}$ ………. (3)
Let’s calculate the mass of each sphere.
For concrete sphere,
${m_1} = {V_1}{\rho _1}$
$ \Rightarrow {m_1} = \dfrac{{4\pi }}{3}({R^3} - {r^3}){\rho _1}$ ……… (4)
Let’s leave density as ${\rho _1}$ for the ease of calculations later.
For the saw dust sphere,
${m_2} = {V_2}{\rho _2}$
$ \Rightarrow {m_2} = \dfrac{{4\pi }}{3}{r^3}{\rho _2}$ ……… (5)
So the ratio of masses for the concrete sphere and the saw-dust sphere would be,
$\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{({R^3} - {r^3}){\rho _1}}}{{{r^3}{\rho _2}}}$ ………. (6)
We have discussed earlier that the upward thrust on a body must be equal to its weight for its floating.
Let’s discuss thrust. It’s the opposition provided by the fluid to the body. This is given by the formula
$U = \rho gV$
Where $U$ is the up-thrust
$\rho $ is the density of the fluid
$g$ is acceleration due to gravity and
$V$ is the volume of the floating object
So in the given question,
$U = \dfrac{{4\pi }}{3}{R^3} \times 1 \times g$ ………. (7)
Weight of the whole sphere is
$W = {m_1}g + {m_2}g$
Where $g$ is the acceleration due to gravity.
$ \Rightarrow W = \dfrac{{4\pi }}{3}({R^3} - {r^3}){\rho _1}g + \dfrac{{4\pi }}{3}{r^3}{\rho _2}g$ ………. (8)
Again as we have discussed the weight of the body must be equal to the thrust provided by the fluid which is water in this case.
So, equating equations (7) and (8) we get
$W = U$
$ \Rightarrow \dfrac{{4\pi }}{3}({R^3} - {r^3}){\rho _1}g + \dfrac{{4\pi }}{3}{r^3}{\rho _2}g = \dfrac{{4\pi }}{3}{R^3} \times 1 \times g$
Diving by $\dfrac{{4\pi }}{3}g$ both sides, we get
$ \Rightarrow ({R^3} - {r^3}){\rho _1} + {r^3}{\rho _2} = {R^3}$
$ \Rightarrow {R^3}{\rho _1} - {R^3} = {r^3}{\rho _1} - {r^3}{\rho _2}$
$ \Rightarrow {R^3}({\rho _1} - 1) = {r^3}({\rho _1} - {\rho _2})$
$ \Rightarrow \dfrac{{{R^3}}}{{{r^3}}} = \dfrac{{{\rho _1} - {\rho _2}}}{{{\rho _1} - 1}}$
Applying properties of double ration we get
$ \Rightarrow \dfrac{{{R^3} - {r^3}}}{{{r^3}}} = \dfrac{{{\rho _1} - {\rho _2} - ({\rho _1} - 1)}}{{{\rho _1} - 1}}$
$ \Rightarrow \dfrac{{{R^3} - {r^3}}}{{{r^3}}} = \dfrac{{1 - {\rho _2}}}{{{\rho _1} - 1}}$
Multiplying each side of the equation by $\dfrac{{{\rho _1}}}{{{\rho _2}}}$ we get,
$ \Rightarrow \dfrac{{({R^3} - {r^3}){\rho _1}}}{{{r^3}{\rho _2}}} = (\dfrac{{1 - {\rho _2}}}{{{\rho _1} - 1}})\dfrac{{{\rho _1}}}{{{\rho _2}}}$
Putting the respective values from equation (6) we get
$ \Rightarrow \dfrac{{{m_1}}}{{{m_2}}} = (\dfrac{{1 - {\rho _2}}}{{{\rho _1} - 1}})\dfrac{{{\rho _1}}}{{{\rho _2}}}$
Substituting the values of ${\rho _1}$ and ${\rho _2}$ we get
$ \Rightarrow \dfrac{{{m_1}}}{{{m_2}}} = (\dfrac{{1 - 0.3}}{{2.4 - 1}})\dfrac{{2.4}}{{0.3}} = 4$
This is the ratio between the masses of both the spheres.

So the correct answer is option (D).

Note:These problems require clear understanding of the Archimedes’ principle. One can easily make these calculations however they seem lengthy, for your clear understanding. One thing to keep in mind when solving these kinds of problems or any problem related to mechanics is how the opposing forces or supporting forces work after that you can solve problems with ease.