• # question_answer A uniform stick of mass M is placed in a frictionless well as shown. The stick makes angle $\theta$ was the horizontal. The force which the vertical wall exerts on right end of stick is - A) $\frac{Mg}{2\cot \theta }$ B) $\frac{Mg}{2\tan \theta }$ C) $\frac{Mg}{2\cos \theta }$ D)  $\frac{Mg}{2\sin \theta }$

 The free body diagram of rod is Where ${{N}_{x}}$ and ${{N}_{y}}$ are horizontal and vertical components of reaction exerted by wall on rod. Net torque on rod about left end A is zero $\therefore Mg\frac{\ell }{2}\cos \theta ={{N}_{x}}\ell \sin \theta$$\Rightarrow {{N}_{x}}=\frac{Mg}{2\tan \theta }$