A) \[\frac{1}{2}M{{R}^{2}}\]
B) \[\frac{3}{2}M{{R}^{2}}\]
C) \[M{{R}^{2}}\]
D) \[M{{R}^{2}}/9\]
Correct Answer: B
Solution :
The tangent to the ring in the plane of the ring is parallel to one of the diameters of the ring. |
The distance between these two parallel axes (in figure) is R, the radius of the ring. |
Using the parallel axes theorem. |
\[{{i}_{\tan gent}}={{I}_{dia}}+M{{R}^{2}}=\frac{M{{R}^{2}}}{2}+M{{R}^{2}}\] |
\[=\frac{3}{2}M{{R}^{2}}\] |
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