If a block moving up an inclined plane at $30^{\circ}$ with a velocity of $5 \,m / s$, stops after $0.5 s$, then coefficient of friction will be nearly
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(b)
Using $v=u+a t$
retardation will be provided by friction as well as gravitational force
$a=\frac{u}{t}$
$g \sin 30^{\circ}+\mu g \cos 30^{\circ}=\frac{5}{0.5}=10$
$\mu=\frac{1}{\sqrt{3}} \equiv 0.6$
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