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Question: A concrete sphere of radius \[R\] has a cavity of radius \(r\) which is packed with saw-dust. The sp...

A concrete sphere of radius RR has a cavity of radius rr which is packed with saw-dust. The specific gravities of concrete and saw-dust are 2.42.4 and 0.30.3 respectively. The sphere floats with its entire volume submerged under water. Calculate the ratio of mass of concrete and saw-dust.
A) 11
B) 22
C) 33
D) 44

Explanation

Solution

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)(R.D)
Mathematically, R.D=ρ1ρ0R.D = \dfrac{{{\rho _1}}}{{{\rho _0}}}
Where ρ1{\rho _1} and ρ0{\rho _0} are densities of the substance of concern and water respectively. The value of ρ0{\rho _0} is 1gcm31gc{m^{ - 3}} or 1kgm31kg{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 RR and the radius of the carving is rr.
This central carving or cavity is filled with saw-dust.
As the question mentions, the specific gravity of concrete and saw dust are respectively 2.42.4 and 0.30.3.
From the given data and the formula mentioned, we can calculate the densities as follows.
ρ1=2.4kgm3{\rho _1} = 2.4kg{m^{ - 3}} and ρ2=0.3kgm3{\rho _2} = 0.3kg{m^{ - 3}}
Where ρ1{\rho _1} and ρ0{\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 V1=43πR343πr3=43π(R3r3){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 V2=43πr3{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=V1+V2V = {V_1} + {V_2}
V=4π3R3\Rightarrow V = \dfrac{{4\pi }}{3}{R^3} ………. (3)
Let’s calculate the mass of each sphere.
For concrete sphere,
m1=V1ρ1{m_1} = {V_1}{\rho _1}
m1=4π3(R3r3)ρ1\Rightarrow {m_1} = \dfrac{{4\pi }}{3}({R^3} - {r^3}){\rho _1} ……… (4)
Let’s leave density as ρ1{\rho _1} for the ease of calculations later.
For the saw dust sphere,
m2=V2ρ2{m_2} = {V_2}{\rho _2}
m2=4π3r3ρ2\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,
m1m2=(R3r3)ρ1r3ρ2\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=ρgVU = \rho gV
Where UU is the up-thrust
ρ\rho is the density of the fluid
gg is acceleration due to gravity and
VV is the volume of the floating object
So in the given question,
U=4π3R3×1×gU = \dfrac{{4\pi }}{3}{R^3} \times 1 \times g ………. (7)
Weight of the whole sphere is
W=m1g+m2gW = {m_1}g + {m_2}g
Where gg is the acceleration due to gravity.
W=4π3(R3r3)ρ1g+4π3r3ρ2g\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=UW = U
4π3(R3r3)ρ1g+4π3r3ρ2g=4π3R3×1×g\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 4π3g\dfrac{{4\pi }}{3}g both sides, we get
(R3r3)ρ1+r3ρ2=R3\Rightarrow ({R^3} - {r^3}){\rho _1} + {r^3}{\rho _2} = {R^3}
Simplifying
R3ρ1R3=r3ρ1r3ρ2\Rightarrow {R^3}{\rho _1} - {R^3} = {r^3}{\rho _1} - {r^3}{\rho _2}
R3(ρ11)=r3(ρ1ρ2)\Rightarrow {R^3}({\rho _1} - 1) = {r^3}({\rho _1} - {\rho _2})
R3r3=ρ1ρ2ρ11\Rightarrow \dfrac{{{R^3}}}{{{r^3}}} = \dfrac{{{\rho _1} - {\rho _2}}}{{{\rho _1} - 1}}
Applying properties of double ration we get
R3r3r3=ρ1ρ2(ρ11)ρ11\Rightarrow \dfrac{{{R^3} - {r^3}}}{{{r^3}}} = \dfrac{{{\rho _1} - {\rho _2} - ({\rho _1} - 1)}}{{{\rho _1} - 1}}
R3r3r3=1ρ2ρ11\Rightarrow \dfrac{{{R^3} - {r^3}}}{{{r^3}}} = \dfrac{{1 - {\rho _2}}}{{{\rho _1} - 1}}
Multiplying each side of the equation by ρ1ρ2\dfrac{{{\rho _1}}}{{{\rho _2}}} we get,
(R3r3)ρ1r3ρ2=(1ρ2ρ11)ρ1ρ2\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
m1m2=(1ρ2ρ11)ρ1ρ2\Rightarrow \dfrac{{{m_1}}}{{{m_2}}} = (\dfrac{{1 - {\rho _2}}}{{{\rho _1} - 1}})\dfrac{{{\rho _1}}}{{{\rho _2}}}
Substituting the values of ρ1{\rho _1} and ρ2{\rho _2} we get
m1m2=(10.32.41)2.40.3=4\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.