A) 3 MR2
B) 3/2 MR2
C) 5 MR2
D) 7/2 MR2
Correct Answer: D
Solution :
Clearly \[{{I}_{YY'}}={{I}_{1}}({{I}_{2}}+{{I}_{3}})\] where \[{{I}_{1}}=MI\] about diameter \[=\frac{1}{2}M{{R}^{2}}\]and \[{{I}_{2}}={{I}_{3}}=MI\]about tangent \[{{I}_{2}}={{I}_{3}}=\frac{1}{2}M{{R}^{2}}+M{{R}^{2}}\] (By parallel axis theorem) \[\therefore \] \[{{I}_{2}}={{I}_{3}}=\frac{3}{2}M{{R}^{2}}\] So, \[{{I}_{YY'}}=\frac{1}{2}M{{R}^{2}}+\int_{{}}^{{}}{\left( \frac{3}{2}M{{R}^{2}}+\frac{3}{2}M{{R}^{2}} \right)}=\frac{7}{2}M{{R}^{2}}\]
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