A) \[25\text{ }m/{{s}^{2}}\]
B) \[36\text{ }m/{{s}^{2}}\]
C) \[5\text{ }m/{{s}^{2}}\]
D) \[15\text{ }m/{{s}^{2}}\]
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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