A) (4, 5, 0)
B) (2, 3, -1)
C) (-2, 3, -1)
D) (2, 0, 2)
Correct Answer: B
Solution :
Given, vertices of \[=0.875\times 100=87.5%\] are A (0, 4, 1), B(2, 3, -1) and C(4, 5, 0). Now, \[\frac{{{B}_{centre}}}{{{B}_{axis}}}={{\left( 1+\frac{{{X}^{2}}}{{{R}^{2}}} \right)}^{3/2}}\] \[{{B}_{axis}}=\frac{1}{8}{{B}_{centre}}\] \[\frac{8}{1}={{\left( 1+\frac{{{X}^{2}}}{{{R}^{2}}} \right)}^{3/2}}\] \[4=1+\frac{{{X}^{2}}}{{{R}^{2}}}\] and \[3=\frac{{{X}^{2}}}{{{R}^{2}}}\] \[{{X}^{2}}=3{{R}^{2}}\] \[X=\sqrt{3}R\] \[e=M\frac{di}{dt}\] \[=0.005\times \frac{d}{dt}\left( {{i}_{0}}\sin \omega t \right)\] \[=0.005\times {{i}_{0}}\cos \omega t\] \[{{e}_{\max }}=0.005\times 10\times 100\pi =5\pi \] is a right angled triangle We know that, the orthocentre of a right angled triangle is the vertex containing \[{{v}_{0}}=\frac{1}{2\pi \sqrt{LC}}\] angle. \[=\frac{1}{2\pi \sqrt{1\times {{10}^{-3}}\times 0.1\times {{10}^{-6}}}}\] Orthocentre is point B (2, 3, -1).
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