A block of mass $m$ is lying on the edge having inclination angle $\alpha=\tan ^{-1}\left(\frac{1}{5}\right)$. Wedge is moving with a constant acceleration, $a=2 \mathrm{~ms}^{-2}$. The minimum value of coefficient of friction $\mu$, so that $m$ remains stationary with respect to wedge is

• A

  $\frac{2}{a}$
• Correct answer
  B

  $\frac{5}{12}$
• C

  $\frac{1}{5}$
• D

  $\frac{2}{5}$

Solution

FBD of $m$ in frame of wedge
$N=m g \cos (\alpha)-m a \sin (\alpha)$ Now, $f=\mu N=m a \cos \alpha+m g \sin \alpha$ $\begin{aligned} & \Rightarrow \quad \mu=\frac{a \cos \alpha+g \sin \alpha}{g \cos \alpha-a \sin \alpha} \\ & \Rightarrow \quad \mu=\frac{a+g \tan \alpha}{g-a \tan \alpha}=\frac{5}{12} \Rightarrow \mu=\frac{5}{12}\end{aligned}$

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