Question

A Cubic block rests on a plane of $\mu =\sqrt{3}$. The angle through which the plane be inclined to the horizontal so that the block just slides down will be

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

Given that,

$\mu =\sqrt{3}$

We know that,

$\tan \theta =\mu$

$\tan \theta =\sqrt{3}$

$\tan \theta =\sqrt{3}$

$\theta ={60}^0$

Hence, the angle is ${60}^0$