question_answer

A uniform rod AB of length 3r remains in equilibrium on a hemispherical bowl of radius r as shown in figure. Ignoring friction, the inclination of the rod \[\theta \]with the horizontal is

A) \[{{\cos }^{-1}}\left( 1/3 \right)\]

B) \[{{\sin }^{-1}}\left( 1/3 \right)\]

C) \[{{\cos }^{-1}}\left( 0.9 \right)\]

D) \[{{\sin }^{-1}}\left( 0.9 \right)\]

Correct Answer: C

Solution :

[c] From the geometry \[AD=2r;\text{ }AC=2r\cos \theta \] \[CD=2r\sin \theta \] \[AG=1.5r;\text{ GC=2r cos}\]\[\text{tan}\theta \text{=}\frac{GC}{CD}=\frac{2r\cos \theta -1.5r}{2r\sin \theta }\]