A small block starts slipping down from a point $B$ on an inclined plane $A B,$ which is making an angle $\theta$ with the horizontal section $BC$ is smooth and the remaining section $CA$ is rough with a coefficient of friction $\mu .$ It is found that the block comes to rest as it reaches the bottom (point $A)$ of the inclined plane. If $B C=2A C$, the coefficient of friction is given by $\mu=k \tan \theta$ .The value of $k$ is
A$13$
B$8$
C$3$
D$17$
JEE MAIN 2020, Diffcult
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C$3$
c Apply work energy theorem
$\operatorname{mg} \sin \theta( AC +2 AC )-\mu mg \cos \theta AC =0$
$\mu=3 \tan \theta$
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