A) \[144{{\pi }^{2}}m/se{{c}^{2}}\]
B) \[144m/se{{c}^{2}}\]
C) \[\frac{144}{{{\pi }^{2}}}m/se{{c}^{2}}\]
D) \[288{{\pi }^{2}}m/se{{c}^{2}}\]
Correct Answer: A
Solution :
Maximum acceleration \[=a{{\omega }^{2}}=a\times 4{{\pi }^{2}}{{n}^{2}}\] \[=0.01\times 4\times {{(\pi )}^{2}}\times {{(60)}^{2}}=144{{\pi }^{2}}m/\sec \]
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