A small block of mass $m$ slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration $a_{0}$. The angle between the inclined plane and ground is $ heta$ and its base length is $L$. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is . . . . . . . .

• A
  $\sqrt{\frac{2 L}{g \sin 2 \theta - a_{0} (1 + \cos 2 \theta)}}$
• B
  $\sqrt{\frac{4 L}{g \sin 2 \theta - a_{0} (1 + \cos 2 \theta)}}$
• C
  $\sqrt{\frac{4 L}{g \cos 2 \theta - a_{0} \sin \theta \cos \theta}}$
• D
  $\sqrt{\frac{2 L}{g \sin \theta - a_{0} \cos \theta}}$