A block is kept on a frictionless inclined surface with angle of inclination $\alpha$. The incline is given an acceleration a to keep the block stationary. Then a is equal to

• A

  $\mathrm{g} / \tan \alpha$
• B

  $\mathrm{g} \operatorname{cosec} \alpha$
• C

  g
• Correct answer
  D

  $\mathrm{g} \operatorname{tan} \alpha$

Solution

$m g \sin \alpha=m a \cos \alpha$ $\therefore \mathrm{a}=\mathrm{g} \tan \alpha$

Hence, the correct answer is option: 'D'.