Question

Sol. [Key Concept: Time down an incline $t \propto 1/\sqrt{a}$; set $t_{rough} = 1.5\ t_{smooth}$ with $a = g(sin\theta - \mu cos\theta)$ vs $gsin\theta$.]

For Rough incline plane

$${a = g(sin\theta - \mu cos\theta) }{t_{1} = \sqrt{\frac{2\mathcal{l}}{\text{ }g(sin\theta - \mu cos\theta)}} }$$For smooth incline plane

$${t_{2} = \sqrt{\frac{2\mathcal{l}}{\text{ }gsin\theta}}\& t_{1} = 1.5t_{2} }{\sqrt{\frac{2\mathcal{l}}{g(sin\theta - \mu cos\theta)}} = \frac{3}{2} \times \sqrt{\frac{2\mathcal{l}}{gsin\theta}} }{(sin\theta - \mu cos\theta) = \frac{4}{9}sin\theta }{\frac{5}{9}sin\theta = \mu cos\theta \Rightarrow \frac{5}{9}tan\theta }{\mu = \frac{5}{9}tan45^{\circ} = \frac{5}{9} }$$