A) 0.2N
B) 0.1 N
C) 2 N
D) 1 N
Correct Answer: A
Solution :
Given, \[m=0.1\,kg,\,r=\frac{1.0}{2}=0.5m\] \[T=\frac{31.4}{10}3.14\,s=\pi \sec \] \[\therefore \] \[F=mr{{\omega }^{2}}=mr{{\left( \frac{2\pi }{T} \right)}^{2}}=\frac{4{{\pi }^{2}}mr}{{{T}^{2}}}\] \[=\frac{4\times {{\pi }^{2}}\times 0.1\times 0.5}{{{\pi }^{2}}}=0.2\,N\]
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