A) \[1\,{{m}^{3}}\]
B) \[\frac{1}{2}\,{{m}^{3}}\]
C) \[\frac{1}{3}\,{{m}^{3}}\]
D) \[\frac{1}{6}\,{{m}^{3}}\]
Correct Answer: D
Solution :
\[OA=O{{A}^{2}}-A{{P}^{2}}=\frac{1}{3}\] \[AB=BC=CA=\sqrt{2}\] \[AQ=\sqrt{2}\,\,\sin \,{{60}^{o}}\] \[=\sqrt{\frac{3}{2}}\] \[AP=\frac{2}{3}.\frac{\sqrt{3}}{2}=\frac{\sqrt{2}}{3}\] and \[OP=O{{A}^{2}}-A{{P}^{2}}=\frac{1}{3}\] \[\therefore \] Required volume \[=\frac{1}{3}\times \]Area of \[\Delta \,ABC\times OP\] \[=\frac{1}{6}\,{{m}^{3}}\]You need to login to perform this action.
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