A) \[\frac{2}{5}M{{R}^{2}}\]
B) \[\frac{7}{5}M{{R}^{2}}\]
C) \[\frac{5}{3}M{{R}^{2}}\]
D) \[\frac{2}{3}M{{R}^{2}}\]
Correct Answer: B
Solution :
Moment of inertia of solid sphere about an axis passing through its centre of gravity. \[r=\frac{2}{5}M{{R}^{2}}\] where m = mass of sphere R = radius of sphere From theorem of parallel axis, moment of inertia about its tangential axis \[I=IM{{R}^{2}}\] \[=\frac{2}{5}M{{R}^{2}}+M{{R}^{2}}\] \[=\frac{7}{5}M{{R}^{2}}\]
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