A) zero
B) \[16\times {{10}^{24}}N\]
C) \[25\times {{10}^{16}}N\]
D) \[36\times {{10}^{21}}N\]
Correct Answer: D
Solution :
\[F=mR{{\omega }^{2}}\] \[=(6\times {{10}^{24}})\times (1.5\times {{10}^{8}}\times {{10}^{3}})\times {{(2\times {{10}^{-7}})}^{2}}\] \[=36\times {{10}^{21}}N\]
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