A) \[0.2\,\,N\]
B) \[0.1\,\,N\]
C) \[2\,\,N\]
D) \[2\,\,N\]
Correct Answer: A
Solution :
Given,\[m=0.1\,\,kg,\,\,r=\frac{1.0}{2}=0.5\,\,m\] \[T=\frac{31.4}{10}3.14=\pi \sec \] \[\therefore \] \[F=mr{{\omega }^{2}}=mr{{\left( \frac{2\pi }{T} \right)}^{2}}=\frac{4{{\pi }^{2}}mr}{{{T}^{2}}}\] \[=\frac{4\times {{\pi }^{2}}\times 0.1\times 0.5}{{{\pi }^{2}}}\] \[=0.2\,\,N\]
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