A) \[25\,m/{{s}^{2}}\]
B) \[36\,m/{{s}^{2}}\]
C) \[5\,m/{{s}^{2}}\]
D) \[15\,m/{{s}^{2}}\]
Correct Answer: C
Solution :
Given,\[r=5cm=5\times {{10}^{-2}}m\]and \[T=0.2\,\pi s\] We know that acceleration \[a=r{{\omega }^{2}}\] \[=\frac{4{{\pi }^{2}}}{{{T}^{2}}}r\] \[=\frac{4\times {{\pi }^{2}}\times 5\times {{10}^{-2}}}{{{(0.2\pi )}^{2}}}=5\,m{{s}^{-2}}\]
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