A) 5016 gm.
B) 1416.8 gm.
C) 19948.27 gm.
D) 5667.1 gm.
Correct Answer: C
Solution :
(c): Volume of metal used \[=\frac{4\pi {{R}^{3}}}{3}-\frac{4\pi {{r}^{3}}}{3}\]\[=\frac{4\pi }{3}\left( {{12}^{3}}-{{10}^{3}} \right)=3050.66\,\,c{{m}^{3}}\] Weight = Volume \[\times \] density \[\Rightarrow \] \[4.9\times 3050.66\] \[\Rightarrow \] \[14948.27\,\,gm\]
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