When a block moves down a smooth inclined plane of inclination $\theta$, its velocity on reaching the bottom is $v$. When it slides down a rough inclined plane of same inclination, its velocity on reaching the bottom is $v/n$, where $n$ is a number greater than $1$. The coefficient of friction between the block and the rough surface is

• Correct answer
  A

  $\mu =\left(1-\frac{1}{n^2}\right)\tan \theta$
• B

  $\mu =\left(1-\frac{1}{n^2}\right)\cot \theta$
• C

  $\mu ={\left(1-\frac{1}{n^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\tan \mathit{\theta }$
• D

  $\mu ={\left(1-\frac{1}{n^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\cot \mathit{\theta }$

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