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So, external r= d/2= 5 cm (Let external radius= ro)
Let internal radius= ri
Weight= 1789 \frac{1}{3} g= 5368g
Density (D)= 7g/cm³
Volume of the sphere= Outer vol- Inner vol 
( \frac{4}{3}  \pi r_o^3) - ( \frac{4}{3} \pi r_i^3 )
 \frac{4}{3}  \pi  (r_o^3 - r_i^3)
 \frac{4}{3} *  \frac{22}{7}(5^3-r_i^3)
 \frac{88}{21} (125-r_i^3) Let this be V
Therefore, for 1cm³ of Vol., weight= 7g
So for V cm³, wt =7*V g
=7* \frac{88}{21} (125-r_i^3)
\frac{88}{3} (125-r_i^3) g
But, given wt = 5368 g
So, 5368=\frac{88}{3} (125-r_i^3)
or,  \frac{5368*3}{88} =125-r_i^3
r_i= \sqrt[3]{58} = 3.87 cm
Now, thickness= r_o-r_i= 5-3.87= 1.13cm

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