A) \[m<{{m}_{0}}\]
B) \[m={{m}_{0}}\]
C) \[m>{{m}_{0}}\]
D) None of theses
Correct Answer: A
Solution :
\[m=\frac{{{m}_{0}}}{\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}}=\frac{{{m}_{0}}}{\sqrt{1-\frac{{{(0.6c)}^{2}}}{{{c}^{2}}}}}\] \[=\frac{{{m}_{0}}}{\sqrt{0.64}}=\frac{{{m}_{0}}}{0.8}\]
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