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 0.1 m and of negligible mass is given by = moment of inertia of disc\[+\]moment of inertia of particle. Since, the mass of the disc is negligible therefore, MI of the system = MI of particle. \[=5\times 2\times {{(0.1)}^{2}}=0.1\text{ }kg-{{m}^{2}}\]
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