question_answer

A triangular block of mass M with angles\[30{}^\circ \],\[~60{}^\circ \], and \[90{}^\circ \] rests with its \[30{}^\circ -90{}^\circ \]side on a horizontal table. A cubical block of mass m rests on the \[60{}^\circ -30{}^\circ \] side. The acceleration which M must have relative to the table to keep m stationary relative to the triangular block assuming frictionless contact is     

A) \[g\]

B) \[\frac{g}{\sqrt{2}}\]

C) (c)\[\frac{g}{\sqrt{3}}\]

D) \[\frac{g}{\sqrt{5}}\]

Correct Answer: C

Solution :

[c] \[ma\cos \theta \text{ }30{}^\circ =mg\sin 30{}^\circ \]             \[\therefore a=\frac{g}{\sqrt{3}}\]