Question

a spherical metallic shell 10cm external diameter weighs 1789 ⅓ g. find the thickness of the shell if the density of the metal is 7 g /cm³.

Answer 1

Solution:-

Mass of the spherical metallic shell = 1789 1/3 g
= 5368/3 g
Density = 7 g/cm³ 
Volume of the spherical metallic shell = Mass/density
⇒ 5368/7 ÷ 7
= 255.61 cm³
Now,
External diameter = 10 cm
So, external radius = 5 cm
Let the internal radius be 'r'
Now,
According to the question.
4/3π(R³ - r³) = 255.61 cm³
⇒ 5³ - r³ = (255.61*3)/(4*3.14)
⇒ 125 - r³ = 766.83/12.56
⇒ 125 - r³ = 61.05
⇒ r³ = 125 - 61.05
⇒ r³ = 63.95
⇒ r = ∛63.95
⇒ r = 3.9989 or 4 (Approx)
So, the thickness = 5 - 4 
= 1 cm
Answer.

Answer 2

Concept we will  be using:
i) Volume of a spherical shell  =$\frac{4}{3} \pi (R^3-r^3)$, where R is the outer radius and r is the inner radius
ii) $Density= \frac{Mass}{Volume}$
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Solution:
Outer diameter = 10cm
Outer radius = R=  $\frac{Diameter}{2}= \frac{10}{2}=5$

Let the inner radius be r.

Then, volume of spherical shell is given by,
Volume
$= \frac{4}{3} \pi (R^3-r^3)$
Plug in R=5:
$= \frac{4}{3} \pi (5^3-r^3)$
$=\frac{4}{3} \pi (125-r^3)$

Mass of the shell = [tex]1789 \frac{1}{3} = \frac{5368}{3} [/tex]gm

Therefore, density of the shell is given by,
$Density\\ = \frac{Mass}{Volume} = \frac{ \frac{5368}{3} }{\frac{4}{3} \pi (125-r^3)}=\frac{1342}{\pi (125-r^3)}$

But, given density is 7g/³
Therefore,
[tex]\frac{1342}{\pi (125-r^3)}=7\\ \text{Reciprocate both sides:}\\ \frac{\pi (125-r^3)}{1342}= \frac{1}{7}\\ \text{Multiply both sides by 1342:}\\ \pi(125-r^3)= \frac{1342}{7}\\ \text{Plug in } \pi = \frac{22}{7}:\\ \frac{22}{7}(125-r^3)=\frac{1342}{7}\\ \text{Multiply both sides by } \frac{22}{7}:\\ (125-r^3)=\frac{1342}{7}\times\frac{7}{22}\\ 125-r^3=61\\ \text{Subtract 125 from both sides:}\\ -r^3= -64\\ r^3=64\\ \text{Cuberoot both sides:} \\r=4[/tex]

Therefore,
Thickness of the shell = outer radius - inner radius = R - r = 5 - 4=1cm

Answer : Required thickness is 1cm