A) \[{{\tan }^{-1}}\left( \frac{1}{2} \right)\]
B) \[{{\tan }^{-1}}\left( \frac{1}{3} \right)\]
C) \[{{\tan }^{-1}}(1)\]
D) 0°
Correct Answer: B
Solution :
For equilibrium of the system torques on M1 and M2 due to BH must counter balance each other i.e. \[{{M}_{1}}\times {{B}_{H}}={{M}_{2}}\times {{B}_{H}}\]. If q is the angle between M1 and BH will be \[(90-\theta )\]; so \[{{M}_{1}}{{B}_{H}}\sin \theta ={{M}_{2}}{{B}_{H}}\sin (90-\theta )\] \[\Rightarrow \tan \theta =\frac{{{M}_{2}}}{{{M}_{1}}}=\frac{M}{3M}=\frac{1}{3}\Rightarrow \theta ={{\tan }^{-1}}\left( \frac{1}{3} \right)\]
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