A) \[(-\infty ,-3)\]
B) \[x=(2\times {{10}^{-2}})\cos \pi t\]
C) \[a=2\times {{10}^{-2}}m=2cm\]
D) \[t=0,\]
Correct Answer: A
Solution :
In this question distance of centre of mass of new disc from the centre of mass of remaining disc is\[\omega =\frac{2\pi }{T}=\pi \]. Mass of remaining disc \[\Rightarrow \] \[T=2s\] \[=\frac{T}{4}=\frac{2}{4}=0.5s\] \[\pi /2\] \[{{v}_{a}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{({{10}^{-3}})}{OA}\] NOTE The given distance must be\[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{({{10}^{-3}})}{\sqrt{{{(\sqrt{2})}^{2}}+{{(\sqrt{2})}^{2}}}}\]for real approach to the solution.
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