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 \[=MI\]of disc \[+MI\] of particles Since, the mass of the disc is negligible therefore, \[MI\] of the system \[=MI\] of particles \[=5\times 2\times {{(0.1)}^{2}}=0.1\,\,kg\,\,{{m}^{2}}\]
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