A) \[3\,kg\,{{m}^{2}}\]
B) \[0.1\,kg\,{{m}^{2}}\]
C) \[4\,kg\,{{m}^{2}}\]
D) \[0.2\text{ }kg\text{ }{{m}^{2}}\]
Correct Answer: A
Solution :
The moment of inertia of the given system (have 5 particles each of mas 2 kg on the rim of circular disc of radius \[0.1m\] and of negligible mass) is given by, = M.I of particles + M.I. of disc. As the mass of the disc is negligible therefore, M.I. of the system = M.I. of particles, therefore, \[=5\times M.I.\]of a particles kept on the rim, \[=5\times 2\times {{(0.1)}^{2}}=0.1kg\,{{m}^{2}}.\]
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