A) \[\frac{M{{L}^{2}}}{18}\]
B) \[\frac{M{{(x+y)}^{2}}}{9}\]
C) \[\frac{M({{L}^{2}}+{{x}^{2}})}{9}\]
D) \[\frac{M({{L}^{2}}+2{{y}^{2}})}{18}\]
Correct Answer: B
Solution :
We have\[\frac{x}{y}=\frac{2}{1}\]or\[\frac{x+y}{y}=\frac{3}{1}\] or \[y=\frac{x+y}{3}=\frac{L}{3}\] \[\Rightarrow \] \[x=2y=\frac{2L}{3}\] Now, \[{{I}_{P}}={{I}_{com}}+M{{(d)}^{2}}\] \[=\frac{M{{(L)}^{2}}}{12}+M{{\left( \frac{L}{2}-\frac{L}{3} \right)}^{2}}\] \[=\frac{M{{L}^{2}}}{12}+\frac{M{{L}^{2}}}{36}\] \[=\frac{4M{{L}^{2}}}{36}=\frac{M{{L}^{2}}}{9}=\frac{M{{(x+y)}^{2}}}{9}\]You need to login to perform this action.
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