A) \[1\,\,kg\,\,{{m}^{2}}\]
B) \[0.1\,\,kg\,\,{{m}^{2}}\]
C) \[2\,\,kg\,\,{{m}^{2}}\]
D) \[0.2\,\,kg\,\,{{m}^{2}}\]
Correct Answer: B
Solution :
Key Idea: Since disc is of negligible mass moment of inertia is due to particles attached to its rim. The moment of inertia of a rigid body about a given axis is the sum of the products of the masses of its particles by the square of their respective distance from the axis of rotation. \[\therefore \] \[I=\Sigma \,m\,{{r}^{2}}\] \[I=5\times 2\times {{\left( 0.1 \right)}^{2}}=0.1\,kg\,{{m}^{2}}\]
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