A) 0.15
B) 0.17
C) 0.20
D) 0.23
E) 0.25
Correct Answer: E
Solution :
The situation is shown below The moment of inertia of disc about the wire is \[I=\frac{m{{v}^{2}}}{2}=\frac{0.2\times 5\times {{10}^{-2}}}{2}\] \[=2.5\times {{10}^{-4}}kg\text{-}{{m}^{2}}\] The time period, \[T=2\pi \sqrt{\frac{I}{K}}\] \[\Rightarrow \] \[K=\frac{4{{\pi }^{2}}I}{{{T}^{2}}}\] \[=\frac{4{{\pi }^{2}}(2.5\times {{10}^{-4}})}{0.2}\] \[=0.25\,kg\text{-}{{m}^{2}}\text{/}{{s}^{2}}\]You need to login to perform this action.
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