Question

# A solid cylinder of mass $$20Kg$$ has length $$1 m$$ and radius $$0.2 m$$. Then its moment of inertia $$(in kg-{ m }^{ 2 })$$ about its geometrical axis is-

Solution

## $$\because \quad I=\dfrac { M{ R }^{ 2 } }{ 2 } \quad for\quad solid\quad cylinder\\ I=\dfrac { 20\times (0.2)\times (0.2) }{ 2 } \\ I=0.4kg{ m }^{ 2 }$$ Physics

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