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 :
The moment of inertia of the given system that contains 5 particles each of mass = 2 kg on the rim of circular disc of radius Q.I m and of negligible mass is given by = M.I of disc + M.I of particle Since the mass of the disc is negligible therefore, M.I of the system = M.I of particle \[=5\times 2\times {{(0.1)}^{2}}=0.1\,kg\,{{m}^{2}}\]
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