(c) 2 cm.
Let the internal radius of the shell be r cm.
∴ Internal volume of the shell = \(\frac43πr^3\) cu. cm
External radius of the shell = \(\frac{21}{2}\) cm.
External volume of the shell = \(\frac43\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}\times\frac{21}{2}\) cu. cm
= 4851 cu. cm.
Weight of 1 cu. cm of metal = 10g.
∴ Volume of the metal in the shell = \(22775\frac5{21}\times \frac{1}{10}\) cu. cm
= \(\frac{478280}{21}\times\frac{1}{10}\) cu. cm = \(\frac{47828}{21}\) cu. cm
∴ Internal vol. of the shell = \(4851-\frac{47828}{21}=\frac{54043}{21}\) cu. cm
⇒ \(\frac43πr^3=\frac{54043}{21}\) ⇒ r3 = \(\frac{54043\times3\times7}{21\times4\times22}\) = 614.125
⇒ \(r = 8.5 \) cm
∴ Thickness of shell = 10.5 cm – 8.5 cm = 2 cm