The upper half of an inclined plane of inclination $\theta$ is perfectly smooth while lower half is rough. A…
- $\mu=\frac{1}{\tan \theta}$
- $\mu=\frac{2}{\tan \theta}$
- $\mu=2 \tan \theta$
- $\mu=\tan \theta$
Solution

The block may be stationary, when
$m g \sin \theta \cdot L=\mu m g \cos \theta \frac{L}{2}$
$\begin{aligned}
\mu & =\frac{m g \sin \theta \cdot L}{m g \cos \theta \frac{L}{2}} \\
& =2 \frac{\sin \theta}{\cos \theta}=2 \tan \theta \\
\text {or } \mu & =2 \tan \theta
\end{aligned}$
Asked in: NEET 2013 (All India)