A) \[\frac{3\sqrt{3}}{8}{{I}_{0}}\]
B) \[\frac{{{I}_{0}}}{8}\]
C) \[\frac{3}{8}{{I}_{0}}\]
D) \[\frac{\sqrt{3}}{8}{{I}_{0}}\]
Correct Answer: A
Solution :
From the geometry of the figure \[{{p}_{1}}{{p}_{2}}=2a\sin {{60}^{o}}\] so, \[{{I}_{{{P}_{2}}}}=\frac{L}{{{p}_{1}}p_{2}^{2}}\] \[=\frac{L}{{{(2a\sin {{60}^{o}})}^{2}}}\]\[=\frac{L}{3{{a}^{2}}}\] and \[{{I}_{{{P}_{3}}}}=\frac{L}{({{P}_{1}}P_{2}^{2}+{{a}^{2}})}\cos {{30}^{o}}\] =\[\frac{L}{[{{(2a\sin {{60}^{o}})}^{2}}+{{a}^{2}}]}\frac{\sqrt{3}}{2}\]\[=\frac{\sqrt{3}\,L}{8\,{{a}^{2}}}\] \[\Rightarrow {{I}_{{{P}_{3}}}}=\frac{3\sqrt{3}}{8}{{I}_{{{P}_{2}}}}=\frac{3\sqrt{3}}{8}{{I}_{0}}\] All options are wrong.You need to login to perform this action.
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